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

Since the derivative vanishes one can conclude that the integral is constant.

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.

  1. 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(x<a)&=\int_0^a|\Psi|^2\,dx\\ &=\dfrac{|A|^2}{a^2}\int_0^a x^2\,dx\\ &=\dfrac{|A|^2}{a^2}\left[ \dfrac{x^3}{3} \right]_0^a\\ &=\dfrac{|A|^2}{3}a\\ &=\dfrac{3}{3b}a\\ &=\dfrac{a}{b} \end{aligned}}At 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<a)_{b=a}=1} 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<a)_{b=2a}=1/2} which is indeed the correct result.

  • What is the expectation value of {x}?{\begin{aligned} <x>&= \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.

  1. 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 {<x>} and {<x^2>}{\begin{aligned} <x>&=\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} <x^2>&=\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 {[<x>-\sigma,<x>+\sigma]}?

    \displaystyle \sigma ^2=<x^2>-<x>^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 {[<x>-\sigma, <x>+\sigma ]} is

    {\begin{aligned} P(<x>-\sigma, <x>+\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}}

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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