# 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:

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

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

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

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

# Quantum Mechanics Introduction

— 1. Motivation —

Unlike we did so far in the previous posts first we’ll begin to study Quantum Mechanics by mentioning some of the experiments that motivated the reformulation of Classical Physics into the body of knowledge that is Quantum Mechanics.

Also we’ll spend more time explaining our initial formulations.

— 1.1. New results, new conceptions —

Anyone that has ever realized an experiment before knows that if one wants to know anything about a system that is being studied one needs to interact with it. In a more formal language one should say: “The act of measuring a system introduces a disturbance in the system”.

Thus far we’ve used the concept of mechanical state on our analysis of physical systems. A more critical eye to the concept of reveals the following:

1. In principle the disturbance can, in some cases, be made as small as one wants. The fact that exists a limit in our measuring apparatus lies in the nature of the apparatus and not in the nature of the theory being used.
2. There exists some disturbances whose effect can’t be neglected. Nevertheless one can always calculate exactly the value of the effect of the disturbance and compensate it in the value of the quantity that is being measured afterwards.

Thus one can say that our very successful Classical Theory of Physics is casual and deterministic.

Despite its many successes our Classical Theory of Physics had a few dark clouds hovering over it:

The persistence of these experimental results and the failure of accommodating them in the classical world view indicated that a revision of concepts was needed:

• Corpuscular entities were demonstrating wavelike behavior.
• Wave entities were demonstrating pointlike behavior.
• There exists a statistical nature in atomic and subatomic phenomena that seems to be an essential characteristic of Nature.
• Nature’s atomic nature (no pun intended) implies that the process of measure has to be thought over: some disturbances can’t be made arbitrarily small since there exists a natural limit to how small a disturbance can be made.

— 1.2. Double-slit experiment —

In order to make our previous discussion more concrete we’ll look into a particular experiment that makes it clear how disparate are the two world views that we’ve been discussing.

— 1.2.1. Double-slit and particles —

Suppose an experimental apparatus which consists of a wall with two opening and a second wall that will serve as target. A beam of particles is fired upon the first wall. Some of those particles will hit the wall and of some will pass through the slits and hit the second wall.

Double slit experiment with particles

The particles that pass through slit ${1}$ are responsible for the probability curve ${P_1}$ while the

particles that pass through slit ${2}$ are responsible for the probability curve ${P_2}$. The resulting probability curve, ${P_{12}}$, results from the algebraic sum of the curves ${P_1}$ and ${P_2}$.

— 1.2.2. Double-slit and waves —

Now if we direct a beam of waves on the same experimental apparatus the pattern one observes in the second wall is:

Double slit experiment with waves

Now what we’ll study is the intensity curve. The intensity curve ${I_1}$ is the result one would observe if only slit ${1}$ was open, while the intensity curve ${I_2}$ is what one would observe if only slit ${2}$ was open.

The resulting intensity is given by the mathematical expression ${I_{12}=|h_1+h_2|^2= I_1+2I_1I_2 \cos \theta}$. The last term is responsible for the interaction between the waves that come from slit ${1}$ with the waves that come from slit ${2}$ which ultimately is what causes the interference pattern one observes in the second wall.

— 1.2.3. Double slit and electrons —

Now that we familiarized ourselves with the behavior of particles and waves under the double slit we’ll study what one observes when we subject an electron beam through the same apparatus.

Experimental results show conclusively that electrons are particles so we’re expecting to see the same pattern that we saw for macroscopic particles.

Nevertheless this is what Nature has for us!:

Double slit experiment with electrons

In the case of electrons Nature forces to think again in terms of curves related to wave phenomena. But unlike macroscopic wave phenomena this aren’t curves of intensity but they are curves of probability. This introduces an apparent oxymoron since probability curves are a concept that is intrinsic to particles while an interference pattern is intrinsic to waves…

In order to explain what we’re observing one has to assume that each probability curve ${P_i}$ is associated to a probability amplitude ${\phi_i}$. In order to calculate the probability one has to calculate the modulus squared of the probability amplitude, ${P_i=\phi_i^2}$. Thus one has to first calculate the sum of probability amplitudes of an electron passing through slit ${1}$ or passing through slit ${2}$ and only then should we calculate the modulus squared of the probability amplitude of an electron passing through slit ${1}$ or slit ${2}$:

$\displaystyle P_{12}=|\phi_1+\phi_2|^2$

— 2. Basic concepts and preliminary definitions —

After our initial discussion I suppose that you understand why physicists had the need to introduce a new paradigm that made sense of what’s happening in the atomic and sub- atomic level. Now it’s time to introduce our usual initial definitions.

 Definition 1 An operator is a mathematical operations that carries a given function into another function.

As an example of an operator we have ${\dfrac{d}{dx}}$ which a transforms a functions into its ${x}$ derivative. As another example of an operator we have ${2\times}$ that transforms a function into its double.

 Definition 2 The linear momentum is represented by the operator $\displaystyle p=\dfrac{\hbar}{i}\dfrac{d}{dx} \ \ \ \ \ (1)$
 Definition 3 The energy of a particle is represented by the operator: $\displaystyle E=i\hbar\dfrac{d}{dt} \ \ \ \ \ (2)$
 Definition 4 For a free particle the following mathematical relationships are valid: {\begin{aligned} \label{eq:relacoesdebroglie} k &= \frac{\hbar}{p}\\ \omega &= \frac{E}{\hbar} \end{aligned}}

— 3. Quantum Mechanics Axioms —

After this batch of initial definitions that allows us to know the entities that will take part in the construction of our quantum vision of the world it is time for us to define the rules that govern them

 Axiom 1 The quantum state is defined by the specification of the relevant physical quantities and is represented by a complex function ${\Psi(x,t)}$
 Axiom 2 To every observable in Classical Physics (${A}$) there corresponds a linear, Hermitian operator (${\hat{A}}$) in quantum mechanics.
 Axiom 3 Measurements made to an observable associated with the operator ${\hat{A}}$ done on a quantum system specified by ${\Psi(x,t)}$ will always return ${a}$ where ${a}$ is an eigenvalue of ${a}$. $\displaystyle \hat{A}\Psi(x,t)=a\Psi(x,t) \ \ \ \ \ (3)$
 Axiom 4 The probability that a particle is in the volume element ${dx}$ is denoted by${P(x)dx}$ and is calculated by $\displaystyle P(x)dx=|\Psi(x,t)|^2dx \ \ \ \ \ (4)$
 Axiom 5 The average value of a physical quantity ${A}$, represented by ${}$, is $\displaystyle = \int \Psi^*\hat{A}\Psi \ \ \ \ \ (5)$

Where the integral is calculated in the relevant volume region.

 Axiom 6 The state of a quantum system evolves according to the Schrodinger Equation. $\displaystyle \hat{H}\Psi= i\hbar\frac{\partial \Psi}{\partial t} \ \ \ \ \ (6)$

Where${\hat{H}}$ is the hamiltonian operator and corresponds to the total energy of a quantum system.

# New avenues into Quantum Mechanics

Previously published in here I’ll repost it in this blog so that its contributors may see what we will be discussing at the end of Griffith’s book:

In recent times two articles that can do the seemingly impossible in Quantum Mechanics have been published and they generated some buzz on the interweb.

This first article Observing the Average Trajectories of Single Photons in a Two-Slit Interferometer is notable because the authors show how they were able to observe the trajectories of photons in a double slit experiment and still managed to observe a clear interference pattern. A thing that is impossible to do according to the Complementarity principle.

In the second article, Direct measurement of the quantum wavefunction, the authors computed the the transverse spatial wavefunction of a single photon by means that they consider to be direct. For me, that haven’t read the article, so far it doesn’t seem to be a so direct method as claimed, but nevertheless the level of experimental expertise even to get an indirect computation of the wave function is certainly worthy of respect.

These spectacular achievements were possible because these two teams used weak measurement techniques, together with statistical ensembles of photons and non simultaneous measurements of the complementary variables they set out to determine.

The two abstracts are here (the bold isn’t in the original):

• Direct measurement of the quantum wavefunction

The wavefunction is the complex distribution used to completely describe a quantum system, and is central to quantum theory. But despite its fundamental role, it is typically introduced as an abstract element of the theory with no explicit definition. Rather, physicists come to a working understanding of the wavefunction through its use to calculate measurement outcome probabilities by way of the Born rule. At present, the wavefunction is determined through tomographic methods which estimate the wavefunction most consistent with a diverse collection of measurements. The indirectness of these methods compounds the problem of defining the wavefunction. Here we show that the wavefunction can be measured directly by the sequential measurement of two complementary variables of the system. The crux of our method is that the first measurement is performed in a gentle way through weak measurement so as not to invalidate the second. The result is that the real and imaginary components of the wavefunction appear directly on our measurement apparatus. We give an experimental example by directly measuring the transverse spatial wavefunction of a single photon, a task not previously realized by any method. We show that the concept is universal, being applicable to other degrees of freedom of the photon, such as polarization or frequency, and to other quantum systems ? for example, electron spins, SQUIDs (superconducting quantum interference devices) and trapped ions. Consequently, this method gives the wavefunction a straightforward and general definition in terms of a specific set of experimental operations. We expect it to expand the range of quantum systems that can be characterized and to initiate new avenues in fundamental quantum theory.

• Observing the Average Trajectories of Single Photons in a Two-Slit Interferometer

A consequence of the quantum mechanical uncertainty principle is that one may not discuss the path or ?trajectory? that a quantum particle takes, because any measurement of position irrevocably disturbs the momentum, and vice versa. Using weak measurements, however, it is possible to operationally define a set of trajectories for an ensemble of quantum particles. We sent single photons emitted by a quantum dot through a double-slit interferometer and reconstructed these trajectories by performing a weak measurement of the photon momentum, postselected according to the result of a strong measurement of photon position in a series of planes. The results provide an observationally grounded description of the propagation of subensembles of quantum particles in a two-slit interferometer.

And an excellent explanation of why this was possible can be found in this blog post Watching Photons Interfere: “Observing the Average Trajectories of Single Photons in a Two-Slit Interferometer”