# The Wave Function 05

— 1.6. The Uncertainty Principle —

Imagine that is holding a rope in one’s hand at that the rope is tied at the end to brick wall. If one suddenly jerks the rope it would cause a pulse formation that would travel along the rope until hitting the brick wall. At every instant of time you could fairly reasonably ascribe a position to this wave pulse but on the other hand if you would be asked to calculate its wavelength you wouldn’t know how to do it since this phenomenon isn’t periodic.

Imagine now that, instead of just producing one jerk, you continuously wave the rope so that you end up producing a standing wave. In this case the wavelength is perfectly defined, since this a phenomenon that is periodic, but the wave position loses its meaning.

Quantum mechanics, as we’ll see in later posts, asks for a particle description that is given in terms of wave packets. Roughly speaking, a wave packet is the result of summing an infinite number of waves (with different wave numbers and phases) that exhibit constructive interference in just a small region of space. An infinite number of waves with different momenta is needed to ensure constructive and destructive interference in the appropriate regions of space.

Wave Packet

Hence we see that by summing more and more waves we are able to make the position of the particle more and more defined while simultaneously making its momentum less and less defined (remember that the waves that we are summing all have different momenta).

In a more formal language one would say that one is working in two different spaces. The position space and the momentum space. What we’re seeing is that in the wave packet formalism it is impossible to have a phenomenon that is perfectly localized in both spaces at the same time.

More physically speaking this means that for a particle its position and momentum have an inherent spread. One can theoretically make the spread of one of the quantities as small as one wants but that would cause the spread in the other quantity to get larger and larger. That is to say the more localized a particle is the more its momentum is spread and the more precise a particle’s momentum is the more fuzzy is its position.

This result is known as Heisenberg’s uncertainty principle and one can make it a mathematically rigorous, but for now this handwaving argument is enough. With it we can already see that Quantum Mechanics needs a radical new way of confronting reality.

For now we’ll just put this result in a quantitative footing and leave its proof for a later post

$\displaystyle \sigma_x \sigma_p \geq \frac{\hbar}{2} \ \ \ \ \ (31)$

One can interpret the uncertainty principle in the language of measurements being made on an ensemble of identically prepared systems. Imagine that you prepare an ensemble whose position measurements are very defined. That is to say that every time you measure the position of a particle the results are very much alike. Well, in this case if you were to also measure the momentum of each particle you would see that the values of momentum you’d end up measuring would be wildly different.

On the other hand you could possibly want to have an ensemble of particles whose momentum measurements would end up with values that have small differences between them. In this case the price to pay would be that the positions of the particles would be scattered all over the place.

Evidently that between those two extremes there is a plethora of possible results. The only limitation that the uncertainty principle stipulates is that the product of the spreads of the two quantities has to be bigger than ${\dfrac{\hbar}{2}}$.

 Exercise 5 A particle of mass ${m}$ is in the state $\displaystyle \Psi(x,t)=Ae^{-a\left[\dfrac{mx^2}{\hbar}+it\right]} \ \ \ \ \ (32)$ where ${A}$ and ${a}$ are positive constants. Find A To find the value of ${A}$ one has to normalize the wave function {\begin{aligned} 1 &= \int_{-\infty}^{+\infty} |\Psi(x,t)|^2\,dx\\ &= |A|^2\int_{-\infty}^{+\infty} e^{2a\dfrac{mx^2}{\hbar}}\, dx\\ &= |A|^2 \sqrt{\dfrac{\hbar\pi}{2am}} \end{aligned}} Thus $\displaystyle A=\sqrt[4]{\frac{2am}{\hbar\pi}}$ For what potential energy function ${V(x)}$ does ${\Psi}$ satisfy the Schroedinger equation? The Schroedinger equation is $\displaystyle i\hbar\frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi \ \ \ \ \ (33)$ For the first term it follows $\displaystyle \frac{\partial \Psi}{\partial t}=-ia\Psi$ For the first ${x}$ derivative it is $\displaystyle \frac{\partial \Psi}{\partial x}=-\frac{2amx}{\hbar}\Psi$ For the second order ${x}$ derivative it is {\begin{aligned} \frac{\partial ^2 \Psi}{\partial x^2} &= -\frac{2am}{\hbar}\Psi+ \dfrac{4a^2m^2x^2}{\hbar ^2}\Psi\\ &= -\dfrac{2am}{\hbar}\left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi \end{aligned}} Replacing these expressions into the Schroedinger equation yields {\begin{aligned} V\Psi &= i\hbar\dfrac{\partial \Psi}{\partial t}+\dfrac{\hbar ^2}{2m}\dfrac{\partial^2 \Psi}{\partial x^2}\\ &= a\hbar\Psi+\dfrac{\hbar ^2}{2m}\left[ -\dfrac{2am}{\hbar} \left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi \right]\\ &= a\hbar\Psi-a\hbar\Psi+\hbar a\dfrac{2amx^2}{\hbar}\Psi\\ &= 2ma^2x^2\Psi \end{aligned}} Thus $\displaystyle V=2ma^2x^2$ Calculate the expectation values of ${x}$, ${x^2}$, ${p}$ and ${p^2}$. The expectation value of ${x}$ $\displaystyle =|A|^2\int_{-\infty}^{+\infty}xe^{-2ax\frac{x^2}{\hbar}}\, dx=0$ The expectation value of ${p}$ $\displaystyle =m\frac{d}{dt}=0$ The expectation value of ${x^2}$ {\begin{aligned} &= |A|^2\int_{-\infty}^{+\infty}x^2e^{-2ax\frac{x^2}{\hbar}}\, dx\\ &= 2|A|^2\dfrac{1}{4(2m/\hbar)}\sqrt{\dfrac{\pi\hbar}{2am}}\\ &= \dfrac{\hbar}{4am} \end{aligned}} The expectation value of ${p^2}$ {\begin{aligned} &= \int_{-\infty}^{+\infty}\Psi ^* \left( \dfrac{\hbar}{i}\dfrac{\partial }{\partial x} \right)^2\Psi\, dx\\ &= -\hbar ^2\int_{-\infty}^{+\infty}\Psi ^* \dfrac{\partial ^2 \Psi}{\partial x^2}\, dx\\ &= -\hbar ^2\int_{-\infty}^{+\infty}\Psi ^* \left[ -\dfrac{2am}{\hbar} \left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi \right]\, dx\\ &= 2am\hbar\int_{-\infty}^{+\infty}\Psi ^* \left( 1-\dfrac{2amx^2}{\hbar} \right)\Psi\, dx\\ &= 2am\hbar\left[ \int_{-\infty}^{+\infty}\Psi ^*\Psi\, dx -\dfrac{2am}{\hbar}\int_{-\infty}^{+\infty}\Psi ^* x^2 \Psi\, dx\right]\\ &= 2am\hbar\left[ 1-\dfrac{2am}{\hbar} \right]\\ &= 2am\hbar\left[ 1-\dfrac{2am}{\hbar}\dfrac{\hbar}{4am}\right]\\ &=2am\hbar\left( 1-1/2 \right)\\ &=am\hbar \end{aligned}} Find ${\sigma_x}$ and ${\sigma_p}$. Is their product consistent with the uncertainty principle? $\displaystyle \sigma_x=\sqrt{-^2}=\sqrt{\dfrac{\hbar}{4am}}$ $\displaystyle \sigma_p=\sqrt{-^2}=\sqrt{am\hbar}$ And the product of the two previous quantities is $\displaystyle \sigma_x \sigma_p=\sqrt{\dfrac{\hbar}{4am}}\sqrt{am\hbar}=\frac{\hbar}{2}$ The product is consistent with the uncertainty principle.

# The Wave Function 04

— 1.5. Momentum and other Dynamical quantities —

Let us suppose that we have a particle that is described by the wave function ${\Psi}$ then the expectation value of its position is (as we saw in The Wave Function 02):

$\displaystyle =\int_{-\infty}^{+\infty}x|\Psi(x,t)|^2\, dx$

Neophytes interpret the previous equations as if it was saying that the expectation value coincides with the average of various measurements of the position of a particle that is described by ${\Psi}$. This interpretation is wrong since the first measurement will make the wave function collapse to the value that is actually obtained and if the following measurements of the position are done right away they’ll just be of the same value of the first measurement.

Actually ${}$ is the average of position measurements of particles that are all described by the state ${\Psi}$. That is to say that we have two ways of actually accomplishing what is implied by the previous interpretation of ${}$:

1. We have a single particle. Then after a position measurement is made we have to able to make the particle to return to its ${\Psi}$ state before we make a new measurement.
2. We have a collection – a statistical ensemble is a more respectable name – of a great number of particles (in order for it to be statistically significant) and we arrange them all to be in state ${\Psi}$. If we perform the measurement of the position of all this particles, then average of the measurements should be ${}$.

To put it more succinctly:

The expectation value is the average of repeated measurements on an ensemble of identically prepared systems.

Since ${\Psi}$ is a time dependent mathematical object it is obvious that ${}$ also is a time dependent quantity:

{\begin{aligned} \dfrac{d}{dt}&= \int_{-\infty}^{+\infty}x\dfrac{\partial}{\partial t}|\Psi|^2\, dx \\ &= \dfrac{i\hbar}{2m}\int_{-\infty}^{+\infty}x\dfrac{\partial}{\partial x}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}-\dfrac{\partial \Psi^*}{\partial x}\Psi \right)\, dx \\ &= -\dfrac{i\hbar}{2m}\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}-\dfrac{\partial \Psi^*}{\partial x}\Psi \right)\,dx \\ &= -\dfrac{i\hbar}{m}\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}\right)\,dx \end{aligned}}

where we have used integration by parts and the fact that the wave function has to be square integrable which is to say that the function is vanishingly small as ${x}$ approaches infinity.

(Allow me to go on a tangent here but I just want to say that rigorously speaking the Hilbert space isn’t the best mathematical space to construct the mathematical formalism of quantum mechanics. The problem with the Hilbert space approach to quantum mechanics is two fold:

1. the functions that are in Hilbert space are necessarily square integrable. The problem is that many times we need to calculate quantities that depend not on a given function but on its derivative (for example), but just because a function is square integrable it doesn’t mean that its derivative also is. Hence we don’t have any mathematical guarantee that most of the integrals that we are computing actually converge.
2. The second problem is that when we are dealing with continuous spectra (later on we’ll see what this means) the eigenfunctions (we’ll see what this means) are divergent

The proper way of doing quantum mechanics is by using rigged Hilbert spaces. A good first introduction to rigged Hilbert spaces and their use in Quantum Mechanics is given by Rafael de la Madrid in the article The role of the rigged Hilbert space in Quantum Mechanics )

The previous equation doesn’t express the average velocity of a quantum particle. In our construction of quantum mechanic nothing allows us to talk about the velocity of particle. In fact we don’t even know what the meaning of

velocity of a particle

is in quantum mechanics!

Since a particle doesn’t have a definitive position prior to is measurement it also can’t have a well defined velocity. Later on we’ll see how how to construct the probability density for velocity in the state ${\Psi}$.

For the purposes of the present section we’ll just postulate that the expectation value of the velocity is equal to the time derivative of the expectation value of position.

$\displaystyle =\dfrac{d}{dt} \ \ \ \ \ (24)$

As we saw in the lagrangian formalism and in the hamiltonian formalism posts of our blog it is more customary (since it is more powerful) to work with momentum instead of velocity. Since ${p=mv}$ the relevant equation for momentum is;

$\displaystyle

=m\dfrac{d}{dt}=-i\hbar\int_{-\infty}^{+\infty}\left( \Psi^*\dfrac{\partial \Psi}{\partial x}\right)\,dx \ \ \ \ \ (25)$

Since ${x}$ represents the position operator operator we can say in an analogous way that

$\displaystyle \frac{\hbar}{i}\frac{\partial}{\partial x}$

represents the momentum operator. A way to see why this definition makes sense is to rewrite the definition of the expectation value of the position

$\displaystyle =\int \Psi^* x \Psi \, dx$

and to rewrite equation 25 in a more compelling way

$\displaystyle

= \int \Psi^*\left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \Psi \, dx$

After knowing how to calculate the expectation value of these two dynamical quantities the question now is how can one calculate the expectation value of other dynamical quantities of interest?

The thing is that all dynamical quantities can be expressed as functions of of ${x}$ and ${p}$. Taking this into account one just has to write the appropriate function of the quantity of interest in terms of ${p}$ and ${x}$ and then calculate the expectation value.

In a more formal (hence more respectable) way the equation for the expectation value of the dynamical quantity ${Q=Q(x,p)}$ is

$\displaystyle =\int\Psi^*Q\left( x,\frac{\hbar}{i}\frac{\partial}{\partial x} \right)\Psi\, dx \ \ \ \ \ (26)$

As an example let us look into what would be the relevant expression for the kinetic energy the relevant definition can be found at Newtonian Mechanics 01. Henceforth we’ll use ${T}$ to denote the kinetic energy instead of ${K}$ in order to use the same notation that is used in Introduction to Quantum Mechanics (2nd Edition).

$\displaystyle T=\frac{1}{2}mv^2=\frac{p^2}{2m}$

Hence the expectation value is

$\displaystyle =-\frac{\hbar ^2}{2m}\int\Psi^*\frac{\partial ^2\Psi}{\partial x^2}\, dx \ \ \ \ \ (27)$

 Exercise 3 Why can’t you do integration by parts directly on $\displaystyle \frac{d}{dt}=\int x\frac{\partial}{\partial t}|\Psi|^2 \, dx$ pull the time derivative over onto ${x}$, note that ${\partial x/\partial t=0}$ and conclude that ${d/dt=0}$? Because integration by parts can only be used when the differentiation and integration are done with the same variable.

 Exercise 4 Calculate $\displaystyle \frac{d}{dt}$ First lets us remember the the Schroedinger equation: $\displaystyle \frac{\partial \Psi}{\partial t}=\frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}-\frac{i}{\hbar}V\Psi \ \ \ \ \ (28)$ And its complex conjugate $\displaystyle \frac{\partial \Psi^*}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2}+\frac{i}{\hbar}V\Psi^* \ \ \ \ \ (29)$ for the time evolution of the expectation value of momentum is {\begin{aligned} \dfrac{d}{dt} &= \dfrac{d}{dt}\int\Psi ^* \dfrac{\hbar}{i}\dfrac{\partial \Psi}{\partial x}\, dx\\ &= \dfrac{\hbar}{i}\int \dfrac{\partial}{\partial t}\left( \Psi ^* \dfrac{\partial \Psi}{\partial x}\right)\, dx\\ &= \dfrac{\hbar}{i}\int\left( \dfrac{\partial \Psi^*}{\partial t}\dfrac{\partial \Psi}{\partial x}+\Psi^* \dfrac{\partial}{\partial x}\dfrac{\partial \Psi}{\partial t} \right) \, dx\\ &= \dfrac{\hbar}{i}\int \left[ \left( -\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi^*}{\partial x^2}+\dfrac{i}{\hbar}V\Psi^* \right)\dfrac{\partial \Psi}{\partial x} + \Psi^*\dfrac{\partial}{\partial x}\left( \dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi \right)\right]\, dx\\\ &= \dfrac{\hbar}{i}\int \left[ -\dfrac{i\hbar}{2m}\left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3} \right)+\dfrac{i}{\hbar}\left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x}\right)\right]\, dx \end{aligned}} First we’ll calculate the first term of the integral (ignoring the constant factors) doing integration by parts (remember that the boundary terms are vanishing) two times {\begin{aligned} \int \left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\right)\, dx &= \left[ \dfrac{\partial \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x}\right]-\int\dfrac{\partial \Psi^*}{\partial x}\dfrac{\partial ^2 \Psi}{\partial x^2}\, dx- \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &=-\left[ \Psi ^*\dfrac{\partial ^2 \Psi}{\partial x^2} \right]+\int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx - \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &= 0 \end{aligned}} Then we’ll calculate the second term of the integral {\begin{aligned} \int \left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x} \right)\, dx &= \int \left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^* \dfrac{\partial V}{\partial x}\Psi-\Psi ^*V\dfrac{\partial \Psi}{\partial x} \right)\, dx\\ &= -\int\Psi ^* \dfrac{\partial V}{\partial x}\Psi\, dx\\ &=<-\dfrac{\partial V}{\partial x}> \end{aligned}} In conclusion it is $\displaystyle \frac{d}{dt}=<-\dfrac{\partial V}{\partial x}> \ \ \ \ \ (30)$ Hence the expectation value of the momentum operator obeys Newton’s Second Axiom. The previous result can be generalized and its generalization is known in the Quantum Mechanics literature as Ehrenfest’s theorem

# The Wave Function 03

— 1.4. Normalization —

The Scroedinger equation is a linear partial differential equation. As such, if ${\Psi(x,t)}$ is a solution to it, then ${A\Psi(x,t)}$ (where ${A}$ is a complex constant) also is a solution.

Does this mean that a physical problem has an infinite number of solutions in Quantum Mechanics? It doesn’t! The thing is that besides the The Scroedinger equation one also has condition 11 to take into account. Stating 11 for the wave function:

$\displaystyle \int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=1 \ \ \ \ \ (15)$

The previous equations states the quite obvious fact that the particle under study has to be in some place at a given instant.

Since ${A}$ was a complex constant the normalization condition fixes ${A}$ in absolute value but can’t tell us nothing regarding its phase. Apparently once again one is haunted with the perspective of having an infinite number of solutions to any given physical problem. The things is that this time the phase doesn’t carry any physical significance (a fact that will be demonstrated later) and thus we actually have just one physical solution.

In the previous discussion one is obviously assuming that the wave function is normalizable. That is to say that the function doesn’t blow up and vanishes quickly enough at infinity so that the integral being computed makes sense.

At this level it is customary to say that these wave functions don’t represent physical states but that isn’t exactly true. A wave function that isn’t normalizable because integral is infinite might represent a beam of particles in a scattering experiment. The fact that the integral diverges to infinity can then be said to represent the fact that beam is composed by an infinite amount of particles.

While the identically null wave function represents the absence of particles.

A question that now arises has to do with the consistency of our normalization and this is a very sensible question. The point is that we normalize the Schroedinger equation for a given time instant, so how does one know that the normalization holds for other times?

Let us look into the time evolution of our normalization condition 15.

$\displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=\int_{-\infty}^{+\infty}\frac{\partial}{\partial t}|\Psi (x,t)|^2dx \ \ \ \ \ (16)$

Calculating the derivative under the integral for the right hand side of the previous equation

{\begin{aligned} \frac{\partial}{\partial t}|\Psi (x,t)|^2&=\frac{\partial}{\partial t}(\Psi^* (x,t)\Psi (x,t))\\ &=\Psi^* (x,t)\frac{\partial\Psi (x,t)}{\partial t}+\frac{\partial \Psi^* (x,t)}{\partial t}\Psi (x,t) \end{aligned}}

The complex conjugate of the Schroedinger equation is

$\displaystyle \frac{\partial \Psi^*(x,t)}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial^2\Psi^*(x,t)}{\partial x^2}+\frac{i}{\hbar}V\Psi^*(x,t) \ \ \ \ \ (17)$

Hence for the derivative under the integral

{\begin{aligned} \frac{\partial}{\partial t}|\Psi (x,t)|^2&=\frac{\partial}{\partial t}(\Psi^* (x,t)\Psi (x,t))\\ &=\Psi^* (x,t)\frac{\partial\Psi (x,t)}{\partial t}+\frac{\partial \Psi^* (x,t)}{\partial t}\Psi (x,t)\\ &=\frac{i\hbar}{2m}\left( \Psi^*(x,t)\frac{\partial^2\Psi(x,t)}{\partial x^2}-\frac{\partial^2\Psi^*(x,t)}{\partial x^2}\Psi (x,t)\right)\\ &=\frac{\partial}{\partial x}\left[ \frac{i\hbar}{2m}\left( \Psi^*(x,t)\frac{\partial\Psi(x,t)}{\partial x}-\frac{\partial\Psi^*(x,t)}{\partial x}\Psi(x,t) \right) \right] \end{aligned}}

Getting back to 16

$\displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=\frac{i\hbar}{2m}\left[ \Psi^*(x,t)\frac{\partial\Psi(x,t)}{\partial x}-\frac{\partial\Psi^*(x,t)}{\partial x}\Psi(x,t) \right]_{-\infty}^{+\infty} \ \ \ \ \ (18)$

Since we’re assuming that our wave function is normalizable the wave function (and its complex conjugate) must vanish for ${+\infty}$ and ${-\infty}$.

In conclusion

$\displaystyle \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi (x,t)|^2dx=0$

In conclusion one can say that if one normalizes the wave equation for a given time interval it stays normalized for all time intervals.

 Exercise 1 At time ${t=0}$ a particle is represented by the wave function $\displaystyle \Psi(x,0)=\begin{cases} Ax/a & \text{if } 0\leq x\leq a\\ A(b-x)/(b-a) & \text{if } a\leq x\leq b \\ 0 & \text{otherwise}\end{cases} \ \ \ \ \ (19)$   where ${A}$, ${a}$ and ${b}$ are constants. Normalize ${\Psi}$. {\begin{aligned} 1&=\int_{-\infty}^{+\infty} |\Psi|^2\,dx\\ &=\int_0^a|\Psi|^2\,dx+\int_a^b|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a|x^2\,dx+\dfrac{|A|^2}{(b-a)^2}\int_a^b(b-x)^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^3}{3} \right]_0^a+\dfrac{|A|^2}{(b-a)^2}\left[ \dfrac{(b-x)^3}{3} \right]_a^b\\ &=\dfrac{|A|^2a}{3}+\dfrac{|A|^2}{(b-a)^2}\dfrac{(b-a)^3}{3}\\ &=\dfrac{|A|^2a}{3}+|A|^2\dfrac{b-a}{3}\\ &=\dfrac{b|A|^2}{3} \end{aligned}} Hence for ${A}$ it is $\displaystyle A=\sqrt{\dfrac{3}{b}}$ Sketch ${\Psi(x,0)}$In ${0\leq x \leq a}$ ${\Psi(x,0)}$ is a strictly increasing function that goes from ${0}$ to ${A}$.In ${a \leq x \leq b}$ ${\Psi(x,0)}$ is strictly decreasing function that goes from ${A}$ to ${0}$.Hence the plot of ${\Psi(x,0)}$ is (choosing the following values ${a=1}$, ${b=2}$ and ${A=\sqrt{b}=\sqrt{2}}$):   Where is the particle most likely to be found at ${t=0}$? Since ${x=a}$ is maximum of the ${\Psi}$ function the most likely value for the particle to be found is at ${x=a}$. What is the probability of finding the particle to the left of ${a}$? Check the answers for ${b=a}$ and ${b=2a}$.{\begin{aligned} P(xAt first let us look into the ${b=a}$ limiting case. We can imagine that this is the end result of ${b}$ getting nearer and nearer to ${a}$. That is to say that the domain of the strictly decreasing part of ${\Psi(x,0)}$ is getting shorter and shorter and when finally ${b=a}$ ${\Psi(x,0)}$ doesn’t have a domain where its is strictly decreasing and ${\Psi(x,0)}$ is defined by its strictly increasing and vanishing features (in the appropriate domains). That is to say that to the right of ${a}$ the function is ${0}$. Hence the probability of the particle being found to the left of ${a}$ is ${1}$.From the previous calculation ${P(x which is indeed the correct result. The ${b=2a}$ case can be analyzed in a different way. In this case: ${x=a}$ is the half point of the domain of ${\Psi(x,0)}$ where ${\Psi(x,0)}$ is non vanishing (end points of the domain are excluded). ${\Psi(x,0)}$ is strictly increasing in the first half of the domain (${0\leq x\leq a}$). ${\Psi(x,0)}$ is strictly decreasing in the second half of the domain (${a\leq x\leq b}$). ${\Psi(x,0)}$ is continuous. Thus one can conclude that ${\Psi(x,0)}$ is symmetric around ${a}$ and consequently the probability of the particle being found to the left of ${a}$ has to be ${1/2}$. From the previous calculation ${P(x which is indeed the correct result. What is the expectation value of ${x}$?{\begin{aligned} &= \int_a^b x|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a x^3\,dx+\dfrac{|A|^2}{(b-a)^2}\int_a^b x(b-x)^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^4}{4} \right]_0^a+\dfrac{|A|^2}{(b-a)^2}\left[ 1/2x^2b^2-2/3x^3b+x^4/4 \right]_a^b\\ &=\dfrac{2a+b}{4} \end{aligned}}
 Exercise 2 Consider the wave function $\displaystyle \Psi(x,t)=Ae^{-\lambda |x|}e^{-i\omega t} \ \ \ \ \ (20)$   where ${A}$, ${\lambda}$ and ${\omega}$ are positive real constants. Normalize ${\Psi}$ {\begin{aligned} 1&=\int_{-\infty}^{+\infty} |\Psi|^2\,dx\\ &=\int_{-\infty}^{+\infty} |A|^2e^{-2\lambda |x|}\,dx\\ &=2|A|^2\int_0^{+\infty}e^{-2\lambda |x|}\,dx \\ &=2|A|^2\int_0^{+\infty}e^{-2\lambda x}\,dx \\ &=-\dfrac{|A|^2}{\lambda}\left[ e^{-2\lambda x} \right]_0^{+\infty}\\ &=\dfrac{|A|^2}{\lambda} \end{aligned}} Hence it is $\displaystyle A=\sqrt{\lambda}$ Determine ${}$ and ${}${\begin{aligned} &=\int_{-\infty}^{+\infty} x|\Psi|^2\,dx\\ &=|A|^2\int_{-\infty}^{+\infty} xe^{-2\lambda |x|}\,dx\\ &=0 \end{aligned}}The integral is vanishing because we’re calculating the integral of an odd function between symmetrical limits.{\begin{aligned} &=\int_{-\infty}^{+\infty} x^2|\Psi|^2\,dx\\ &=2\lambda\int_0^{+\infty} x^2e^{-2\lambda x}\,dx\\ &=2\lambda\int_0^{+\infty} \dfrac{1}{4}\dfrac{\partial^2}{\partial \lambda ^2}\left( e^{-2\lambda x} \right)\,dx\\ &=\dfrac{\lambda}{2} \dfrac{\partial^2}{\partial \lambda ^2}\int_0^{+\infty}e^{-2\lambda\,dx} x \,dx\\ &= \dfrac{\lambda}{2} \dfrac{\partial^2}{\partial \lambda ^2} \left[ -\dfrac{e^{-2\lambda\,dx}}{2\lambda} \right]_0^{+\infty}\\ &= \dfrac{\lambda}{2}\dfrac{\partial^2}{\partial \lambda ^2}\left(\dfrac{1}{2\lambda} \right)\\ &=\dfrac{\lambda}{2}\dfrac{\partial}{\partial \lambda}\left(-\dfrac{1}{\lambda ^2} \right)\\ &=\dfrac{\lambda}{2}\dfrac{1}{\lambda^3}\\ &= \dfrac{1}{2\lambda^2} \end{aligned}} Find the standard deviation of ${x}$. Sketch the graph of ${\Psi ^2}$. What is the probability that the particle will be found outside the range ${[-\sigma,+\sigma]}$? $\displaystyle \sigma ^2=-^2=\frac{1}{2\lambda ^2}-0=\frac{1}{2\lambda ^2}$ Hence the standard deviation is $\displaystyle \sigma=\dfrac{\sqrt{2}}{2\lambda}$ The square of the wave function is proportional to ${e^{-2\lambda |x|}}$. Dealing for piecewise definitions of the square of the wave function, its first derivative in order to ${x}$ and its second derivative in order to ${x}$ $\displaystyle |\Psi|^2=\begin{cases} e^{2\lambda x} & \text{if } x < 0\\ e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (21)$   $\displaystyle \dfrac{\partial}{\partial x}|\Psi|^2=\begin{cases} 2\lambda e^{2\lambda x} & \text{if } x < 0\\ -2\lambda e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (22)$   $\displaystyle \dfrac{\partial ^2}{\partial x ^2}|\Psi|^2=\begin{cases} 4\lambda ^2 e^{2\lambda x} & \text{if } x < 0\\ 4\lambda ^2 e^{-2\lambda x} & \text{if } x \geq 0 \end{cases} \ \ \ \ \ (23)$   As we can see the first derivative of ${|\Psi|^2}$ changes its sign on ${0}$ from positive to negative. Hence it was strictly increasing before ${0}$ and it is strictly decreasing after ${0}$. Hence ${0}$ is a maximum of ${|\Psi|^2}$. The second derivative is always positive so ${|\Psi|^2}$ is always concave up (convex). Hence its graphical representation is: The probability that the particle is to be found outside the range ${[-\sigma, +\sigma ]}$ is {\begin{aligned} P(-\sigma, +\sigma)&= 2\int_\sigma^{+\infty}|\Psi|^2\,dx\\ &= 2\lambda\int_\sigma^{+\infty}e^{2\lambda x}\\ &= 2\lambda\left[ -\dfrac{e^{2\lambda x}}{2\lambda} \right]_\sigma^{+\infty}\\ &=\lambda \dfrac{e^{2\lambda x}}{2\lambda}\\ &=e^{-2\lambda\dfrac{\sqrt{2}}{2\lambda}}\\ &=e^{-\sqrt{2}} \end{aligned}}

# The Wave Function Exercises 01

Exercise 1

• ${^2=21^2=441}$

${=1/N\sum j^2N(J)=\dfrac{6434}{14}=459.6}$

• Calculating for each ${\Delta j}$
 ${j}$ ${\Delta j=j-}$ 14 ${14-21=-7}$ 15 ${15-21=-6}$ 16 ${16-21=-5}$ 22 ${22-21=1}$ 24 ${24-21=3}$ 25 ${25-21=4}$

Hence for the variance it follows

${\sigma ^2=1/N\sum (\Delta j)^2N(j)=\dfrac{260}{14}=18.6}$

Hence the standard deviation is

$\displaystyle \sigma =\sqrt{18.6}=4.3$

• ${\sigma^2=-^2=459.6-441=18.6}$

And for the standard deviation it is

$\displaystyle \sigma =\sqrt{18.6}=4.3$

Which confirms the second equation for the standard deviation.

Exercise 2 Consider the first ${25}$ digits in the decimal expansion of ${\pi}$.

• What is the probability of getting each of the 10 digits assuming that one selects a digit at random.

The first 25 digits of the decimal expansion of ${\pi}$ are

$\displaystyle \{3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3\}$

Hence for the digits it is

 ${N(0)=0}$ ${P(0)=0}$ ${N(1)=2}$ ${P(1)=2/25}$ ${N(2)=3}$ ${P(2)=3/25}$ ${N(3)=5}$ ${P(3)=1/5}$ ${N(4)=3}$ ${P(4)=3/25}$ ${N(5)=3}$ ${P(5)=3/25}$ ${N(6)=3}$ ${P(6)=3/25}$ ${N(7)=1}$ ${P(7)=1/25}$ ${N(8)=2}$ ${P(8)=2/25}$ ${N(9)=3}$ ${P(9)=3/25}$
• The most probable digit is ${5}$. The median digit is ${4}$. The average is ${\sum P(i)N(i)=4.72}$.

• ${\sigma=2.47}$

 Exercise 3 The needle on a broken car is free to swing, and bounces perfectly off the pins on either end, so that if you give it a flick it is equally likely to come to rest at any angle between ${0}$ and ${\pi}$. Along the ${\left[0,\pi\right]}$ interval the probability of the needle flicking an angle ${d\theta}$ is ${d\theta/\pi}$. Given the definition of probability density it is ${\rho(\theta)=1/\pi}$. Additionally the probability density also needs to be normalized. $\displaystyle \int_0^\pi \rho(\theta)d\theta=1\Leftrightarrow\int_0^\pi 1/\pi d\theta=1$ which is trivially true. The plot for the probability density is Compute ${\left\langle\theta \right\rangle}$, ${\left\langle\theta^2 \right\rangle}$ and ${\sigma}$. {\begin{aligned} \left\langle\theta \right\rangle &= \int_0^\pi\frac{\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\theta d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^2}{2} \right]_0^\pi\\ &= \frac{\pi}{2} \end{aligned}} For ${\left\langle\theta^2 \right\rangle}$ it is {\begin{aligned} \left\langle\theta^2 \right\rangle &= \int_0^\pi\frac{\theta^2}{\pi}d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^3}{3} \right]_0^\pi\\ &= \frac{\pi^2}{3} \end{aligned}} The variance is ${\sigma^2=\left\langle\theta^2 \right\rangle-\left\langle\theta\right\rangle^2 =\dfrac{\pi^2}{3}-\dfrac{\pi^2}{4}=\dfrac{\pi^22}{12}}$. And the standard deviation is ${\sigma=\dfrac{\pi}{2\sqrt{3}}}$. Compute ${\left\langle\sin\theta\right\rangle}$, ${\left\langle\cos\theta\right\rangle}$ and ${\left\langle\cos^2\theta\right\rangle}$. {\begin{aligned} \left\langle\sin\theta \right\rangle &= \int_0^\pi\frac{\sin\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\sin\theta d\theta\\ &= \frac{1}{\pi} \left[ -\cos\theta \right]_0^\pi\\ &= \frac{2}{\pi} \end{aligned}} and {\begin{aligned} \left\langle\cos\theta \right\rangle &= \int_0^\pi\frac{\cos\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\cos\theta d\theta\\ &= \frac{1}{\pi} \left[ \sin\theta \right]_0^\pi\\ &= 0 \end{aligned}} We’ll leave ${\left\langle\cos\theta^2 \right\rangle}$ as an exercise for the reader. As a hint remember that ${\cos^2\theta=\dfrac{1+\cos(2\theta)}{2}}$.

 Exercise 4 In exercise ${1.1}$ it was shown that the the probability density is $\displaystyle \rho(x)=\frac{1}{2\sqrt{hx}}$ Hence the mean value of ${x}$ is {\begin{aligned} \left\langle x \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{h}{3} \end{aligned}} For ${\left\langle x^2 \right\rangle}$ it is {\begin{aligned} \left\langle x^2 \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\int_0^h x^{3/2}dx\\ &= \frac{1}{2\sqrt{h}}\left[\frac{2}{5}x^{5/2} \right]_0^h\\ &= \frac{h^2}{5} \end{aligned}} Hence the variance is $\displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{h^2}{5}-\frac{h^2}{9}=\frac{4}{45}h^2$ and the standard deviation is $\displaystyle \sigma=\frac{2h}{3\sqrt{5}}$ For the distance to the mean to be more than one standard deviation away from the average we have two alternatives. The first is the interval ${\left[0,\left\langle x \right\rangle+\sigma\right]}$ and the second is ${\left[\left\langle x \right\rangle+\sigma,h\right]}$. Hence the total probability is the sum of these two probabilities. Let ${P_1}$ denote the probability of the first interval and ${P_2}$ denote the probability of the second interval. {\begin{aligned} P_1 &= \int_0^{\left\langle x \right\rangle-\sigma}\frac{1}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\left[2x^{1/2} \right]_0^{\left\langle x \right\rangle-\sigma}\\ &= \frac{1}{\sqrt{h}}\sqrt{\frac{h}{3}-\frac{2h}{3\sqrt{5}}}\\ &=\sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}} \end{aligned}} Now for the second interval it is {\begin{aligned} P_2 &= \int_{\left\langle x \right\rangle+\sigma}^h\frac{1}{2\sqrt{hx}}dx\\ &= \ldots\\ &=1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}} \end{aligned}} Hence the total probability ${P}$ is ${P=P_1+P_2}$ {\begin{aligned} P&=P_1+P_2\\ &= \sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}}+1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}}\\ &\approx 0.3929 \end{aligned}}

 Exercise 5 The probability density is ${\rho(x)=Ae^{-\lambda(x-a)^2}}$ Determine ${A}$. Making the change of variable ${u=x-a}$ (${dx=du}$) the normalization condition is {\begin{aligned} 1 &= A\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &= A\sqrt{\frac{\pi}{\lambda}} \end{aligned}} Hence for ${A}$ it is $\displaystyle A=\sqrt{\frac{\lambda}{\pi}}$ Find ${\left\langle x \right\rangle}$, ${\left\langle x^2 \right\rangle}$ and ${\sigma}$. {\begin{aligned} \left\langle x \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty ue^{-\lambda u^2}du+a\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &=\sqrt{\frac{\lambda}{\pi}}\left( 0+a\sqrt{\frac{\pi}{\lambda}} \right)\\ &= a \end{aligned}} If you don’t see why ${\displaystyle\int_{-\infty}^\infty ue^{-\lambda u^2}du=0}$ check this post on my other blog. For ${\left\langle x^2 \right\rangle}$ it is {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right) \end{aligned}} Now ${\displaystyle 2a\int_{-\infty}^\infty u e^{-\lambda u^2}du=0}$ as in the previous calculation. For the third term it is ${\displaystyle a^2\int_{-\infty}^\infty e^{-\lambda u^2}du=a^2\sqrt{\frac{\pi}{\lambda}}}$. The first integral is the hard one and a special technique can be employed to evaluate it. {\begin{aligned} \int_{-\infty}^\infty u^2e^{-\lambda u^2}du &= \int_{-\infty}^\infty-\frac{d}{d\lambda}\left( e^{-\lambda u^2} \right)du\\ &= -\frac{d}{d\lambda}\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &=-\frac{d}{d\lambda}\sqrt{\frac{\pi}{\lambda}}\\ &=\frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}} \end{aligned}} Hence it is {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &= \sqrt{\frac{\lambda}{\pi}}\left( \frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}}+0+a^2\sqrt{\frac{\pi}{\lambda}} \right)\\ &=a^2+\frac{1}{2\lambda} \end{aligned}} The variance is $\displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{1}{2\lambda}$ Hence the standard deviation is $\displaystyle \sigma=\frac{1}{\sqrt{2\lambda}}$

— Mathematica file —

The resolution of exercise 2 was done using some basic Mathematica code which I’ll post here hoping that it can be helpful to the readers of this blog.

// N[Pi, 25]

piexpansion = IntegerDigits[3141592653589793238462643]

digitcount = {}

For[i = 0, i <= 9, i++, AppendTo[digitcount, Count[A, i]]]

digitcount

digitprobability = {}

For[i = 0, i <= 9, i++, AppendTo[digitprobability, Count[A, i]/25]]

digitprobability

digits = {}

For[i = 0, i <= 9, i++, AppendTo[digits, i]]

digits

j = N[digits.digitprobability]

digitssquared = {}

For[i = 0, i <= 9, i++, AppendTo[digitssquared, i^2]]

digitssquared

jsquared = N[digitssquared.digitprobability]

sigmasquared = jsquared - j^2

std = Sqrt[sigmasquared]

deviations = {}

deviations = piexpansion - j

deviationssquared = (piexpansion - j)^2

variance = Mean[deviationssquared]

standarddeviation = Sqrt[variance] 

# The Wave Function 02

— 1.3. Probability —

In the previous post we were introduced to the Schrodinger equation (equation 1), stated Born’s interpretation of what is the physical meaning of the wave function and took a little glimpse into some philosophical positions one might have regarding Quantum Mechanics.

Since probability plays such an essential role in Quantum Mechanics it seems that a brief revision of some of its concepts is in order so that we are sure that we have the tools that allows one to do Quantum Mechanics.

— 1.3.1. Discrete variables —

The example used in the book in order to expound on the terminology and concepts of probability is of a set that consists of 14 people in a class room:

• one person has 14 years
• one person has 15 years
• three people have 16 years
• two people 22 years
• five people have 25 years

Let ${N(j)}$ represent the number of people with age ${j}$. Hence

• ${N(14)=1}$
• ${N(15)=1}$
• ${N(16)=3}$
• ${N(22)=2}$
• ${N(25)=5}$

One can represent the previous data points by use of a histogram:

The the total number of people in the room is given by

$\displaystyle N=\sum_{j=0}^{\infty}N(j) \ \ \ \ \ (2)$

Adopting a frequentist definition of probability Griffiths then makes a number of definitions of probability concepts under the assumption that the phenomena at study are discrete ones.

 Definition 1 The probability of an event ${j}$, ${P(j)}$ is proportional to the number elements that have the property ${j}$ and inversely proportional to the total elements (${N}$) under study. $\displaystyle P(j)=\frac{N(j)}{N} \ \ \ \ \ (3)$

It is easy to see that from equation 3 together with equation 2 it follows

$\displaystyle \sum_{j=0}^{\infty}P(j)=1 \ \ \ \ \ (4)$

After defining ${P(j)}$ we can also define what is the most probable value for ${j}$.

 Definition 2 The most value for ${j}$ is the one for which ${P(j)}$ is a maximum.
 Definition 3 The average value of ${j}$ is given by $\displaystyle =\sum_{j=0}^{\infty}jP(j) \ \ \ \ \ (5)$

But what if we are interested in computing the average value of ${j^2}$? Then the appropriate expression must be

$\displaystyle =\sum_{j=0}^{\infty}j^2P(j)$

Hence one can write with full generality that average value for some function of ${j}$, denoted by ${f(j)}$ is given by

$\displaystyle =\sum_{j=0}^{\infty}f(j)P(j) \ \ \ \ \ (6)$

After introducing the definition of maximum of a probability distribution it is time to introduce a couple of definitions that relate t the symmetry and spread of a distribution.

 Definition 4 The median is the value of ${j}$ for which the probability of having a larger value than ${j}$ is the same as the probability of having a value with a smaller value than ${j}$.

After seeing a definition that relates to the the symmetry of a distribution we’ll introduce a definition that is an indication of its spread.

But first we’ll look at two examples that will serve as a motivation for that:

and

Both histograms have the same median, the same average, the same most probable value and the same number of elements. Nevertheless it is visually obvious that the two histograms represent two different kinds of phenomena.

The first histogram represents a phenomenon whose values are sharply peaked about the average (central) value.

The second histogram on the other hand represents a phenomenon represents a more broad and flat distribution.

The existence of such a difference in two otherwise equal distributions introduces the necessity of introducing a measure of spread.

A first thought could be to define the difference about the average for each individual value

$\displaystyle \Delta j=j-$

This approach doesn’t work since that for random distributions one would expect to find equally positive and negative values for ${\Delta j}$.

One way to circumvent this issue would be to use ${|\Delta j|}$, and even though this approach does work theoretically it has the problem of not using a differentiable function.

These two issues are avoided if one uses the squares of the deviations about the average.

The quantity of interest in called the variance of the distribution.

 Definition 5 The variance of a distribution ,${\sigma ^2}$, that has an average value is given by the expression $\displaystyle \sigma ^2=<(\Delta j)^2> \ \ \ \ \ (7)$
 Definition 6 The standard deviation, ${\sigma}$, of a distribution is given by the square root of its variance.

For the variance it is

$\displaystyle \sigma ^2=-^2 \ \ \ \ \ (8)$

Since by definition 5 the variance is manifestly non-negative then it is

$\displaystyle \geq ^2 \ \ \ \ \ (9)$

where equality only happens when the distribution is composed of equal elements and equal elements only.

— 1.3.2. Continuous variables —

Thus far we’ve always assumed that we are dealing with discrete variables. To generalize our definitions and results to continuous distributions.

One has to have the initial care to note that when dealing with phenomena that allow for a description that is continuous probabilities of finding a given value are vanishing, and that one should talk about the probability of a given interval.

With that in mind and assuming that the distributions are sufficiently well behaved one has that the probability of and event being between ${x}$ and ${x+d}$ is given by

$\displaystyle \rho(x)dx \ \ \ \ \ (10)$

The quantity ${\rho (x)}$ is the probability density.

The generalizations for the other results are:

$\displaystyle \int_{-\infty}^{+\infty}\rho(x)dx=1 \ \ \ \ \ (11)$

$\displaystyle =\int_{-\infty}^{+\infty}x\rho(x)dx \ \ \ \ \ (12)$

$\displaystyle =\int_{-\infty}^{+\infty}f(x)\rho(x)dx \ \ \ \ \ (13)$

$\displaystyle \sigma ^2=<(\Delta x)^2>=-^2 \ \ \ \ \ (14)$

# The Wave Function 01

— 1. The Wave Function —

The purpose of this section is to introduce the wave function of Quantum Mechanics and explain its physical relevance and interpretation.

— 1.1. The Schroedinger Equation —

Classical Dynamics’ goal is to derive the equation of motion, ${x(t)}$, of a particle of mass ${m}$. After finding ${x(t)}$ all other dynamical quantities of interest can be computed from ${x(t)}$.

Of course that the problem is how does one finds ${x(t)}$? In classical mechanics this problem is solved by applying Newton’s Second Axiom

$\displaystyle F=\frac{dp}{dt}$

For conservative systems it is ${F=-\dfrac{\partial V}{\partial x}}$ (previously we’ve used ${U}$ to denote the potential energy but will now use ${V}$ to accord to Griffith’s notation).

Hence for classical mechanics one has

$\displaystyle m\frac{d^2 x}{dt^2}=-\dfrac{\partial V}{\partial x}$

as the equation that determines ${x(t)}$ (with the help of the suitable initial conditions).

Even though Griffith’s only states the Newtonian formalism approach of Classical Dynamics we already know by Classical Physics that apart from Newtonian formalism one also has the Lagrangian formalism and Hamiltonian formalism as suitable alternatives (and most of time more appropriate alternatives) to Newtonian formalism as ways to derive the equation of motion.

As for Quantum Mechanics one has to resort the Schroedinger Equation in order to derive the equation of motion the specifies the Physical state of the particle in study.

$\displaystyle i\hbar\frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi \ \ \ \ \ (1)$

— 1.2. The Statistical Interpretation —

Of course now the question is how one should interpret the wave function. Firstly its name itself should sound strange. A particle is something that is localized while a wave is something that occupies an extense region of space.

According to Born the wave function of a particle is related to the probability of it occupying a region of space.

The proper relationship is ${|\Psi(x,t)|^2dx}$ is the density probability of finding the particle between ${x}$ and ${x+dx}$.

This interpretation of the wave function naturally introduces an indeterminacy to Quantum Theory, since one cannot predict with certainty the position of a particle when it is measured and only its probability.

The conundrum that now presents itself to us is: after measuring the position of a particle we know exactly where it is. But what about what happens before the act of measurement? Where was the particle before our instruments interacted with it and revealed is position to us?

These questions have three possible answers:

1. The realist position: A realist is a physicist that believes that the particle was at the position where it was measured. If this position is true it implies that Quantum Mechanics is an incomplete theory since it can’t predict the exact position of a particle but only the probability of finding it in a given position.
2. The orthodox position: An orthodox quantum physicist is someone that believes that the particle had no definite position before being measured and that it is the act of measurement that forces the particle to occupy a position.
3. The agnostic position: An agnostic physicist is a physicist that thinks that he doesn’t know the answer to this question and so refuses to answer it.

Until 1964 advocating one these three positions was acceptable. But on that year John Stewart Bell proved a theorem, On the Einstein Podolsky Rosen paradox, that showed that if the particle has a definite position before the act of measurement then it makes an observable difference on the results of some experiments (in due time we’ll explain what we mean by this).

Hence the agnostic position was no longer a respectable stance to have and it was up to experiment to show if Nature was a realist or if Nature was an orthodox.

Nevertheless the disagreements of what exactly is the position of a particle when it isn’t bening measured, all three groups of physicists agreed to what would be measured immediately after the first measurement of the particle’s position. If at first one has ${x}$ then the second measurement has to be ${x}$ too.

In conclusion the wave function can evolve by two ways:

1. It evolves without any kind of discontinuity (unless the potential happens to be unbounded at a point) under the Schroedinger Equation.
2. It collapses suddenly to a single value due to the act of measurement.

The interested reader can also take a look at the following book from Bell: Speakable and Unspeakable in Quantum Mechanics

# My semester as a quantum mechanic

Hello all,

I have one class left before having finished my undergraduate coursework in quantum mechanics.  I would like to share my experience in that course and some of what I have learned in that course with you all.  I will be describing the topics we learned in some level of detail, so those of you who have no experience with quantum mechanics can know some of what to expect here, but without the mathematics that our friend Ateixeira intends on going into.

The assigned text, Gasiorowicz, was poor as a pedagogical book in my opinion.  I had to look elsewhere.  I made extensive use of the book by Nouredine Zettili which was fantastic from a pedagogical standpoint.  I also used Schaum’s Outline of Quantum Mechanics as a supplement.  I know many in my class used Griffiths to supplement the class text, which is what I believe we will be following here, and I heard many great things about it, though I did not use it personally.  Despite having a poor text, I currently have an A in the class, and if I do reasonably well on my final that should not change.

So lets get down to the actual physics.  Here is a list of things that we covered in my class:

• The hydrogen atom
• Mathematical formalism of quantum mechanics
• The harmonic oscillator in the number representation
• Orbital angular momentum
• Spin angular momentum
• Time independent perturbation theory
• The application of perturbation theory to the hydrogen atom
• The variational method and its application to the helium atom (To come on Monday, November 28)

Now, before describing any of these topics, I’d like to briefly talk about quantum mechanics in general.  My professor sort of assumed we had a working knowledge of the postulates of quantum mechanics, and the 1-dimensional Schrodinger equation and its application.  On the first day of class he presented the three dimensional Schrodinger equation and over the next three weeks proceeded to solve it for a central potential (the hydrogen atom).  I did not have a working knowledge of the Schroedinger equation, so I had a lot of catching up to do.  Thankfully, for this there was Zettili.  Because of my struggles to understand the material and the beginning of the semester, I don’t want to simply jump into what my coursework was before discussing quantum mechanics in general as an aside.

Quantum mechanics, as presented to me, is the study of the Schrodinger equation.  It comes in many forms, though the most familiar is probably the time-independent Schrodinger equation in the position representation:

$\frac{\hbar^{2}}{2m} \nabla^{2} \psi \left( r \right) + V(r) \psi (r) = E \psi (r)$.

Quantum theory gives us a stochastic description of nature, and does away with Newtonian determinism.  From a philosophical perspective, this to me is the most stunning implication of quantum mechanics, though I do not wish to take much time in discussing philosophy here.  In quantum mechanics, a particle is completely described by its wavefunction $\psi (r)$.  Observable information about a particle can be extracted from the wavefunction by applying an operator to it.  In the Schrodinger equation above, $\frac{\hbar^{2}}{2m} \nabla^{2}$ is the kinetic energy operator whereas V(r) represents a generic potential energy operator.  E is a numerical value for energy.  Basically, the version of the Schrodinger equation I have presented above allows you to calculate the possible energy values that a particle is allowed to take for a given potential.  Once the differential equation is solved for the wavefunction and the energy values that its allowed to take the probability of finding the particle at any of those energy values is easily computed.  As mentioned, this is not the only version of the Schrodinger equation, and others are more appropriate for different physical situations.  I should mention, however, that there are a very limited number of cases that the Schrodinger method is exactly solvable, and for most physical situations approximation techniques need to be applied.

With this information, I think we’re ready to proceed on to the hydrogen atom.  This is one of the few physical situations for which the Schrodinger equation is exactly solvable.  The most common isotope of hydrogen consists of a proton and an electron.  The proton exerts a central Coulomb force on the electron.  Associated with this force is the Coulomb potential, which becomes V(r) in the Schrodinger equation:

$V(r) = \frac{-e^{2}}{4 \pi \epsilon_0 r}$.

To find solutions to the Schrodinger equation for the electron with the above potential the Schrodinger equation is represented in spherical coordinates since the Coulomb potential is isotropic, and separation of variables is applied.  The calculation is involved, and took the professor about 7 hours to solve, however there were some interesting results that emerged.  The wavefunction solutions to the hydrogen atom are known as orbital.  Three numbers, known as quantum numbers, define each orbital.  There is the principal quantum number which is associated with the energy associated with an orbital, the angular momentum quantum number which determines the magnitude of the angular momentum of an orbital, and the magnetic quantum number which determines the projection of the angular momentum operator.  Once the wavefunctions that solve the Schrodinger equation for the Coulomb potential are known, the energy spectrum of the hydrogen atom is also known.  When an electron transitions from one orbital to another, the energy that it has lost is emitted as a photon.  Since the energy associated with a photon determines the color of the photon, knowing the wavefunctions of the hydrogen atom allow us to calculate the entire spectrum of the hydrogen atom based on the electron’s transitions between orbitals.  Of everything I have learned, I believe this is the most striking demonstration of the power of quantum theory.  With it we can now explain the entire hydrogen spectrum.

I should stress here that though the Schrodinger equation is exactly solvable for the Coulomb potential, that does not mean we have a complete description of the hydrogen atom.  The Schrodinger equation is nonrelativistic.  Though the relativistic corrections to the Schrodinger equation are minor, they are important.  For instance, solutions to the Schrodinger equation with the Coulomb potential do not say anything about spin angular momentum.  To get a full picture of spin, relativity needs to be considered.  There are other corrections as well, some of which I will discuss later.

After finishing our discussion of the hydrogen atom, the professor went ahead to introduce the mathematical formalism of quantum mechanics.  This is where we learn about kets and bras; abstract representations of the wavefunction.  We also learned some about the Hilbert space, the abstract, infinite dimensional vector space upon which quantum mechanics is performed.  We learned a little bit about commutators.  We learned about different types of operators, and how they can change wavefunctions.  We learned about a specific class of operators in particular, known as Hermitian operators, which are associated with every physical observable.  We learned about the Schrodinger perspective and the Heisenberg perspective on how a quantum system evolves through time.  We also proved the Heisenberg uncertainty principle in a much more general fashion.  There may be things I am forgetting.  In this portion of the course we didn’t learn much about physics itself, however we learned a lot about the background mathematics necessary for understanding quantum mechanics.  Since I suppose describing physics is a lot easier than describing abstract mathematical ideas, I will leave this section as is for now, since this section of the course can not really be done justice without introducing the mathematics itself.

Next we looked at the quantum harmonic oscillator.  The harmonic oscillator has a potential of $\frac{1}{2} m \omega^{2} \hat{x}^2$, where $\hat{x}$ is the position operator.  We approached this using what our professor termed the number basis.  We found an energy spectrum of the quantum oscillator to be

$E_n = \hbar \omega \left( n + \frac{1}{2} \right)$.

One of the most interesting results in our analysis of the quantum oscillator is the existence of zero-point energy, the amount of energy the oscillating particle has in its ground state (when n=0).  The ground state energy of the quantum oscillator is nonzero unlike its classical counterpart.  There were two operators introduced in our analysis of the quantum harmonic oscillator: the creation operator and the annihilation operator.  The annihilation operator, when acting on a wavevector lowered its energy by 1 state, and the creation operator raised the energy of the wavevector by 1 state.  I believe we studied this in preparation for a more detailed study of angular momentum, which used ideas very similar to creation and annihilation.

In classical mechanics angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$, where r is the position of the particle with respect to some origin and p is the linear momentum of the particle.  In quantum mechanics, the angular momentum is defined in the same way, except r and p are replaced with the position operator and momentum operator respectively.  Based on this we were able to conclude that

$[L_i, L_j] = L_i L_j - L_j L_i = i \varepsilon_{ijk} L_k$.

It turns out this equation allows integer values and half-integer values for the angular momentum.  The integer values are associated with orbital angular momentum.  As it turns out, there is a physical quantity with half-integer angular momentum: spin angular momentum for particles known as fermions, such as electrons.  This is a purely quantum quantity, without any real classical analogue.  Furthermore, spin cannot be derived from first principles without accounting for relativity.  We investigated angular momentum in some mathematical detail using tools very similar to those developed for the quantum oscillator.  We also investigated spin operators represented as what is known as Pauli matrices.

As I mentioned before, very few problems in quantum mechanics are solvable exactly.  There are two approximation methods learned in my class which allow us to say something about a physical system that is not exactly solvable.  We are currently working on the variational method.  This method allows us to put an upper bound on the lowest energy expectation value for a particle, and make estimations for some excited states.  We will be applying this to the helium atom today in class.

The other method we have learned for approximating quantum systems is known as perturbation theory.  If we have a system that varies slightly from an exactly solvable system, perturbation theory can be used to approach it.  One application of perturbation theory is in the study of the Stark effect.  Here, a hydrogen atom is placed in an external electric field.  In this case, the spectrum of the hydrogen atom shifts and splits.  Perturbation theory allows you to approach this effect from an analytical perspective.  Perturbation theory also allows us to make some relativistic corrections to the hydrogen atom, correct for the interaction between the magnetic field of the proton and the spin of the electron, and other corrections to the hydrogen atom that the Schrodinger equation could not completely account for.