# Expectation of Momentum in a Classical (Infinite) Potential Well

• ja07019
In summary, the conversation discusses the calculation of <p> for an infinite potential well for a classical particle. The expectation value is found using the wavefunction u(z,t) = Aexp(iu) and the concept of normalization. The conversation also touches on the necessity of boundary conditions and the use of sine and cosine waves to represent the wavefunction in the well. The concept of uniform velocity is also discussed in relation to the classical particle and its wave nature.
ja07019
Homework Statement
Calculate <p> for an Infinite Potential Well for a classical particle
Relevant Equations
u(z,t) = Aexp{iu} where u=wt-kz
Okay so I begin first by mentioning the length of the well to be L, with upper bound, L/2 and lower bound, -L/2 and the conjugate u* = Aexp{-iz}

First I begin by writing out the expectation formula:
## \langle p \rangle = \int_{\frac{L}{2}}^{ \frac{L}{2} } Aexp(-iu) -i \hbar \frac{ \partial }{ \partial z} Aexp(iu) dz = \int_{\frac{L}{2}}^{ \frac{L}{2} } Aexp(-iu)Aexp(iu)-i \hbar i \frac{ \partial }{ \partial z} (wt-kz)dz = -A^2(i)^2\hbar \int_{\frac{L}{2}}^{ \frac{L}{2} } exp(-iu+iu) \frac{ \partial }{ \partial z} (wt-kz)dz = -A^2(-1) \hbar \int_{\frac{L}{2}}^{ \frac{L}{2} } exp(0) -k dz = -A^2(-1) \hbar (-k) \int_{\frac{L}{2}}^{ \frac{L}{2} } dz = -A^2 \hbar k (\frac{L}{2} - \frac{L}{2}) = -A^2 \hbar k (2L/2) = -A^2 \hbar k L ##
After normalizing the wavefunction, I ended up with ##A^2 = \frac{1}{L}##
Thus the entire thing reduces to ## -\frac{1}{L} \hbar k L = -\hbar k = -mv = p ##
The solution given by the book is <p> = 0 and the correct answer is also in agreement with the QM derived solution.

Does your wavefunction vanish at the boundaries (as it should) for an infinite potential well? After you have the correct wavefunction, then you can use it to calculate the expectation value ##\langle p \rangle##. Your normalization constant is OK.

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This is a classical Potential Well however. Would that still be necessary? I cannot see how exactly that would be done however, as the velocity is given to be constant and the wave has certain required components(frequency and wave number). Thus the wavefront must be complex exponential resulting in cosx+isinx. Thus when taking the absolute value both of the components are squared and the result is simply 1(so no damping can occur which makes sense since the particle will aways be at uniform velocity.

It doesn't matter if the well (or the particle) is classical or not; if the well is infinite, that means that the particle cannot be found outside the well boundaries.

kuruman said:
It doesn't matter if the well (or the particle) is classical or not; if the well is infinite, that means that the particle cannot be found outside the well boundaries.
Okay I can see why this is necessary for QM(solving Schrodinger's Eq). How would I go about setting up the boundary conditions here? Do I just split the solution into a sine and cosine wave and use the asymmetrical vs symmetrical? The problem I have with that is that the wavefunction is no longer the complex exponential. If it is no longer a complex exponential, then suppose the velocity is given by the derivative of the either sine or cosine(whichever is used) which means that the velocity will no longer be uniform(note, it was required that velocity is uniform) and instead will be either sine or cosine. Sine or cosine means the velocity is changing at different points along a well, whereas the complex exponential maintains a constant velocity.

ja07019 said:
Okay I can see why this is necessary for QM(solving Schrodinger's Eq). How would I go about setting up the boundary conditions here?
Do it the usual way, demand that the wavefunction vanish at the boundaries.
ja07019 said:
Do I just split the solution into a sine and cosine wave and use the asymmetrical vs symmetrical?
Yes.
ja07019 said:
The problem I have with that is that the wavefunction is no longer the complex exponential.
So what? It is always true that with the correct choice of coefficients ##A## and ##B##, ##C_1e^{iu}+C_2e^{-iu}=A \sin u+B\cos u##. Note that you need both ##e^{iu}## and ##e^{-iu}## for a particle in a box bouncing back and forth. If you just use ##e^{iu}##, that means that the particle moves "forth", i.e. propagates in the +x direction but never "back" in the -x direction.
ja07019 said:
If it is no longer a complex exponential, then suppose the velocity is given by the derivative of the either sine or cosine(whichever is used) which means that the velocity will no longer be uniform(note, it was required that velocity is uniform) and instead will be either sine or cosine. Sine or cosine means the velocity is changing at different points along a well, whereas the complex exponential maintains a constant velocity.
There is no such thing as the instantaneous value of the particle in QM, so uniformity is moot. The best you can do is find the expectation of the velocity from the expectation value of the position,$$\langle v \rangle =\frac{d\langle x \rangle}{dt}.$$

ja07019 said:
Problem Statement: Calculate <p> for an Infinite Potential Well for a classical particle
Relevant Equations: u(z,t) = Aexp{iu} where u=wt-kz
What does "classical particle" mean for the purposes of this problem? How does it have a wave associated with it?

The problem statement creates an image in my mind, of a ball bouncing elastically back and forth between two rigid walls, moving at constant speed and reversing its direction when it hits a wall. Are you thinking of something else?

vela
jtbell said:
What does "classical particle" mean for the purposes of this problem? How does it have a wave associated with it?

The problem statement creates an image in my mind, of a ball bouncing elastically back and forth between two rigid walls, moving at constant speed and reversing its direction when it hits a wall. Are you thinking of something else?
Classical in the sense that there is no probability involved, more precisely the problem simply says to attempt to treat the wave function of a classical particle(deterministic states) as a stochastic state. Naturally, it makes no sense to treat the propagation of a deterministic wave as a probability density function but the point the book tries to make is that when a quantum electron has a "sufficiently high" energy, the quantum electron behaves like a classical electron(wave).

kuruman said:
Do it the usual way, demand that the wavefunction vanish at the boundaries.

Yes.

So what? It is always true that with the correct choice of coefficients ##A## and ##B##, ##C_1e^{iu}+C_2e^{-iu}=A \sin u+B\cos u##. Note that you need both ##e^{iu}## and ##e^{-iu}## for a particle in a box bouncing back and forth. If you just use ##e^{iu}##, that means that the particle moves "forth", i.e. propagates in the +x direction but never "back" in the -x direction.

There is no such thing as the instantaneous value of the particle in QM, so uniformity is moot. The best you can do is find the expectation of the velocity from the expectation value of the position,$$\langle v \rangle =\frac{d\langle x \rangle}{dt}.$$
All of what you said is definitely correct. At this point I am not so sure if I am using the correct Probability Density Function. However, what I would like to point out is that the problem outright says that the particle bounces around with a uniform velocity.

The only way to have uniform velocity is to have a complex exponential, because if you have only a sine or cosine wave, then the derivative will be once again a sine or cosine wave. This is problematic of course because cos(pi) != cos(pi/3). However if the wave function is a complex exponential, then the derivative will also be a complex exponential. Now, when you want to find the velocity of such wave, you will end up doing
##|e^{ix}| = \sqrt{cos^2s+sin^2s}## which allows for uniform velocity which is true for an electron not experiencing any forces.

So the next issue at hand is that 0 = exp{} will never be true no matter what so I can't find any boundary(ies)

ja07019 said:
At this point I am not so sure if I am using the correct Probability Density Function.
The correct wavefuntions for your symmetric box are $$\psi_n(x) = \begin{cases} \sqrt{\frac{2}{L}}\cos\left(\frac{n\pi x}{L} \right)& n=1,3,5, \dots \\ \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L} \right)& n=2,4,6, \dots \end{cases}$$and the corresponding energies are$$E_n=\frac{n^2 \pi^2}{2mL^2}$$Note that the wavefunctions vanish at ##x=\pm \frac{L}{2}.##
ja07019 said:
However, what I would like to point out is that the problem outright says that the particle bounces around with a uniform velocity.
Sure.
ja07019 said:
However if the wave function is a complex exponential, then the derivative will also be a complex exponential.
Why are you so concerned with the derivative of the wavefunction? What does it have to do with uniform velocity? Remember that ##\psi^*\psi## is a probability density. If you take the derivative of that multiplied by some constants, you get a probability current.

Here is what I suggest following @jtbell's post. Consider a classical particle in a box and give it the quantum mechanical treatment. If you have trouble imagining this, think of a superball bouncing elastically in zero gravity between two walls in one-dimensional motion. Assume reasonable values for its speed, mass and the wall separation. Find a value for the quantum number ##n## assuming that the superball is in a pure stationary quantum state.

Find the probability density ##\psi^* \psi##. Clearly it will be zero at some values of ##x## and non-zero at other values. This means that the superball will not be found where its probability is zero. What is the separation ##\delta x## between two successive zeroes as a fraction of the separation between walls ##L##? What does this say about any "classical" measurement you can make of the particle's position?

Find the energy separation between adjacent values of ##n## in Joules. What does this say about any "classical" measurement you can make of the particle's energy?

All this might give you a better understanding of what's going on, but is not necessary for answering the original question, "Calculate <p> for an Infinite Potential Well for a classical particle." The calculation is the same for a classical particle as for a quantum particle and gives the same result.

PeroK
kuruman said:
Why are you so concerned with the derivative of the wavefunction?

I might be doing this wrong. But the point I would like to get across is once again that the magnitude of the wavefunction is given by |u(x,t)|, where u is the wavefunction. Now if the wavefunction is a real sinusoid, then the wavefunction's derivative(the velocity) will be sinusoidal. This means that ##|u_x(x,t)| = |acos(f(x,t))|## However, this means that the velocity will be zero at some points and a at other points, whereas the complex exponential will not: ##|aexp(iv)| = \sqrt{a^2cos^2v +a^2 sin^2v} = a##. Here, the wavefunction maintains a constant velocity no matter where.
kuruman said:
Consider a classical particle in a box and give it the quantum mechanical treatment
Okay take the variance of the position, ##\langle (z- \langle z \rangle )^2 \rangle##. For a classical electron, this is simply ## \frac{a^2}{3}##. However, this is the case when n is really large(still discrete) but large enough that you can ignore it. Thus a classical electron is really a high energy electron. In a low energy electron, or quantum electron, n can no longer be ignored and the variance in the position is ## \frac{a^2}{3} [1-(-1)^n\frac{6}{n^2 \pi ^2}] ##. Now if you notice carefully, you will notice the low energy electron actually has a ##\pm## value, which is a result of the sinusoidal. And again I want to stress that the ##\pm## is a result of normalizing the wave function by applying the absolute value.

jtbell said:
What does "classical particle" mean for the purposes of this problem? How does it have a wave associated with it?

The problem statement creates an image in my mind, of a ball bouncing elastically back and forth between two rigid walls, moving at constant speed and reversing its direction when it hits a wall. Are you thinking of something else?
Sorry. It's not a ball it's a particle(which also behave as a wave). It's not a quantum particle in that there is no probability involved

ja07019 said:
Sorry. It's not a ball it's a particle(which also behave as a wave). It's not a quantum particle in that there is no probability involved
Why is there no probability involved? I believe this problem is an illustration of the Correspondence Principle. Read about it, do what I outlined in post #10 and you will see how it works.

kuruman said:
Why is there no probability involved?
The problem explicitly says that this is a classical particle(ie no Probability involved) but everything else should work the same.

ja07019 said:
The problem explicitly says that this is a classical particle(ie no Probability involved) but everything else should work the same.
The problem as you posted it says it's a classical particle. The explication "no Probability involved" is yours. In case you didn't go to the Wikipedia link I suggested

"In physics, the correspondence principle states that the behavior of systems described by the theory of quantum mechanics (or by the old quantum theory) reproduces classical physics in the limit of large quantum numbers. In other words, it says that for large orbits and for large energies, quantum calculations must agree with classical calculations."

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kuruman said:
The problem as you posted it says it's a classical particle. The explication "no Probability involved" is yours. In case you didn't go to the Wikipedia link I suggested
I think I should be a little bit more clear, I have found the quantum mechanical solution per Probability theory, but now I am trying to arrive to the correspondence principle via another route, the classical route. I can see how I can reach the same conclusion by applying such limits to the quantum mechanical result, but I want to reach the same endpoint using classical mechanics.

It's a one-way route from QM to classical. Quantum theory is the more complete theory but becomes very cumbersome to use for describing the motion of large objects such as planets. Don't forget that classical theory could not explain the blackbody radiation spectrum without the quantum hypothesis, which Max Planck used as a last resort. If Planck couldn't do it, can you? Good luck.

Well no I am just trying to prove a weak version of the corresponds principle. If you notice something, the wavefunction in QM contains n(energy level) but the classical does not. So there is really no way to let n go to infinity as neither sine nor cosine have some definite value here. However you can go the other route and use classical principles to reach the expectation values and then use the QM method and let the energy be very large. That is all that I am seeking to do here.

ja07019 said:
Classical in the sense that there is no probability involved, more precisely the problem simply says to attempt to treat the wave function of a classical particle(deterministic states) as a stochastic state.
My initial interpretation of the problem was the same as @jtbell's while @kuruman thinks it's about the correspondence principle and you have yet another interpretation based on additional information. Perhaps it would help if you provided the original problem statement in its entirety.

jtbell
This
ja07019 said:
... and then use the QM method and let the energy be very large.
sounds very vague. When you say "large" energy, it's large relative to what? What is a relevant energy scaling parameter here and why?

To me it's obvious that if you want to understand how the Correspondence Principle works, you first need to go from quantum to classical. Once you have done that, then you will understand much better what large (but not infinite) ##n## means in that context. There is mathematical infinity and physical infinity. I offered a suggestion in post #10 about how to proceed. If you choose not to try my suggestion, I will have nothing more to post on this thread.

vela said:
Perhaps it would help if you provided the original problem statement in its entirety.
It might also help if you tell us which textbook you're using. Someone who is watching this thread may be familiar with it.

I haven't responded directly yet because frankly, I'm confused by what you've posted so far. I don't know (yet) whether this is because this is beyond what I've taught at the introductory level of QM, or because you're confused and conflating concepts that don't really belong together.

In particular, I don't understand why you're trying to use a "wave" function to describe a classical particle.

vela said:
My initial interpretation of the problem was the same as @jtbell's while @kuruman thinks it's about the correspondence principle and you have yet another interpretation based on additional information. Perhaps it would help if you provided the original problem statement in its entirety.
I apologize to all of you. I too am not sure what the book is using as a marker for classical particles. As I understand, wave particle duality is still classical mechanics while probability density waves are QM. So it is likely I am wrong here(but that should have nothing to do here).
Anyways the book is Electrical Properties in Materials by Solymar. The question goes as follows"The classical equivalent of the potential well is a particle bouncing between perfectly elastic walls with uniform velocity." Calculate the average position, variance in position, momentum, and variance in momentum.

Now I was planning on doing simple kinematics, but this does provide a tangible result. If you think about it, Probability Density Function must include density of some kind. Kinematics does not seem to lend well to this. So my next choice was to use Harmonic Motion. Simple Harmonic motion cannot be used here, because the velocity must be uniform aka constant. So then I switched over to a complex exponential for the harmonic motion, which was how I ended up with the wave function that I did. And that is where I am now. As far as energies go, there is no limit to high n can go (per the wavefunction) and thus some terms may disappear such as in the variance of position in the well.

Some posters on SE have suggested that the issue lies in the boundaries if that helps.

ja07019 said:
Probability Density Function must include density of some kind. Kinematics does not seem to lend well to this.
Indeed, at any particular time, the momentum of a classical particle is definite, in principle. Nevertheless, there is a way to introduce probability into the picture. Suppose that you have a particle bouncing back and forth between the walls at constant speed. "Look at" the particle at many different randomly-chosen times. What are the possible values that the momentum can have (hint: there are only two!), and what are the probabilities of those values? Then apply the formula for the expectation value of a discrete (not continuous) probability distribution.

For position, you can reason similarly, except that now you have a continuous probability distribution, because at randomly-chosen times, you can find the particle anywhere between the walls.

ja07019 said:
"The classical equivalent of the potential well is a particle bouncing between perfectly elastic walls with uniform velocity." Calculate the average position, variance in position, momentum, and variance in momentum.
This problem description clarifies what is being asked, namely a classical treatment of a classical problem. OP's attempt to use a wavefunction gave me the impression that the question was about a quantum mechanical treatment of a classical problem.

Orodruin
jtbell said:
"Look at" the particle at many different randomly-chosen times.
Thank you for this. Classically the system will have two values, but there magnitudes are equal.
The only thing I am now concerned about is lacking some kind of pdf to bring about the variances for momentum and position in a way that both values are found using the same pdf

But I have to say this was very illuminating.

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ja07019 said:
Thank you for this. Classically the system will have two values, but there magnitudes are equal.
The only thing I am now concerned about is lacking some kind of pdf to bring about the variances for momentum and position in a way that both values are found using the same pdf

But I have to say this was very illuminating.
What's a "pdf" in the context of your last post?

kuruman said:
What's a "pdf" in the context of your last post?
The probability density function so I can find <value>

The expectation value in the classical setting is a time-average over a full period of the classical system.

jtbell
ja07019 said:
some kind of pdf to bring about the variances for momentum and position in a way that both values are found using the same pdf
I don't think you can do that for the classical case, because the pdf's for momentum and position are in terms of different variables, ##p## and ##x##: ##P_p(p)## and ##P_x(x)##. It's not obvious to me that there's a general relationship between them.

The pdf's for position and momentum are different functions in QM, too. For position: $$P_x(x) = \psi^*(x)\psi(x)$$ For momentum, $$P_p(p) = \phi^*(p)\phi(p)$$ where $$\phi(p) = \frac 1 {\sqrt{2\pi\hbar}} \int^{+\infty}_{-\infty} {\psi(x)e^{-ipx/\hbar}\,dx}$$ The two pdf's are related, but definitely different.

Are you perhaps thinking of the fact that in the QM case you can calculate the expectation value of momentum from the probability amplitude function for position, ##\psi(x)##: $$\langle p \rangle = \int^{+\infty}_{-\infty} {\psi^*(x) \left( -i\hbar \frac {\partial}{\partial x} \right) \psi(x)\,dx}$$ as well as from the probability distribution function for momentum: $$\langle p \rangle = \int^{+\infty}_{-\infty} {p P_p(p)\,dp}$$ for a continuous pdf, or $$\langle p \rangle = \sum_i {p_i P_p(p_i)}$$ for a discrete pdf?

To elaborate on @jtbell #24. The expectation value, as you know, is an average.

Suppose you performed ##N## measurements of the momentum at random intervals and you got ##N_1## values at ##+p## and ##N_2## at ##-p##, where ##N=N_1+N_2##. What is the average value of the momentum?

Suppose you flip a fair coin ##N## times and you get ##N_1## tails and ##N_2## heads. You assign +1 point to tails and -1 point to heads. How would you express the average of the flips as a number? What do you think the a priori average could be before you do any measurements?

## 1. What is the concept of expectation of momentum in a classical potential well?

The expectation of momentum in a classical potential well is a measure of the average momentum of a particle confined within the well. It is calculated by taking the integral of the momentum operator over the wave function of the particle.

## 2. How is the expectation of momentum related to the size of the potential well?

The expectation of momentum is inversely proportional to the size of the potential well. This means that as the size of the well increases, the expectation of momentum decreases. This is because a larger well allows for more possible momentum states, leading to a broader distribution of momentum values.

## 3. Can the expectation of momentum be negative?

No, the expectation of momentum cannot be negative. This is because momentum is a vector quantity and its magnitude cannot be negative. However, the expectation of momentum can be zero if the particle is in a stationary state within the well.

## 4. How does the expectation of momentum change with the energy of the particle?

The expectation of momentum increases with the energy of the particle. This is because higher energy states allow for larger momentum values, leading to a broader distribution of possible momentum values and a higher expectation value.

## 5. Is the expectation of momentum a constant value?

No, the expectation of momentum can vary depending on the state of the particle within the potential well. It is a statistical measure of the average momentum and can change as the particle's energy or position within the well changes.

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