Free particle at time t

[najima]

Homework Statement

a free particle of mass m moving in one dimension is known to be in the initial state
ψ(x,0)=sin(k_0 x)
1. what value of p (momentum) will measurement yield at the time t,and with what probabilities will these values occur?
2. suppose that p is measurement at t=3 s and the value (h/2pi)(k_0) is found. what is ψ(x,0)
at t>3 s?

Homework Equations

quantum mechanics by Liboff chapter 6

The Attempt at a Solution

I don't know, must I normalize it? for part 1 when I integrate the expectation value of the momentum that is infinite because we have this
∫_(-∞)^∞▒〖sink_0 x) cos〖k_0 x〗 dx〗
and another question what probabilities occur?
for parti 2. I don't know what can I do. can I write( exp i(k_0 x)-exp i(k_0 x))/2 and we know for free particle we have A exp i(k_0 x)-Bexp i(k_0 x) and I use it?

 

[buffordboy23]

Hi najima,

Free particles don't exist in stationary states, so there is no definite energy of the particle. Have you learned about wave packets and Fourier transforms yet?

 

[najima]

yes,so you mean I use delta function .I try but I can't integrate.

 

[buffordboy23]

Yes, you will integrate, but "not" using the delta function. You are given the initial wave function. From here you can compute its Fourier transform to construct the wave packet function, which is a function of k. Does this help?

 

[najima]

could you help me more?

 

[najima]

so you mean b(k)=∫_(-∞)^∞ sin (k_0 x)exp i(kx)/(2pi)^1/2 is it infinite?

 

[buffordboy23]

I think I made mistake. Let's look at this from the beginning. The initial wave function is

$$\Psi\left(x,0\right) = sin\left(k_{0}x\right)$$

Are you given any boundary conditions for this wave function, or is it over the whole domain of space?

EDIT: Actually, there is no mistake. It just leads to same result.

 

[najima]

No,I am not given any boundary condition. I wrote everything that are given by the question.

 

[buffordboy23]

Okay, good. So, what does this tell you about the localization of the particle? Think about the initial wave function when making this determination.

EDIT: Once you determined this, think about what this means in regards to the momentum of the particle.

 

[najima]

I can't find out.you mean the initial state determined boundary conditions?

 

[buffordboy23]

Yes. The initial state is the given by wavefunction. You said it's not normalizable. Why is this? It's wavefunction does not vanish at +/- infinity. What does this say about the localization of the particle? The particle has an ill-defined position. If this still confuses you graph $$\Psi\left(x,0\right)^{2}$$, the probability density.

Hint: Think about what this means according to the uncertainty principle. And then, how is k related to the momentum? =)

 

[turin]

a free particle of mass m moving in one dimension is known to be in the initial state
ψ(x,0)=sin(k_0 x)

This is bad (imprecise) notation, and you are right to be confused. Basically, the problem is that this is not the wavefunction; it is proportional to the wavefunction (so this doesn't exactly describe the state). And, in particular, the state of a free particle is not a member of the Hilbert space, but we are sloppy in physics and we ignore this "subtle" problem. Basically, buffordboy is hinting as the resolution to the problem, but it can be handled without going there if you are willing to tolerate some physicist's slop.

... must I normalize it?

Well, strictly speaking, if they give you a wave function, then THEY must normalize it, or at least put a normalization factor out front, or they should not call it a state. (See my previous comments.).

for part 1 when I integrate the expectation value of the momentum that is infinite ...

Yes, because you are not using the true state in your calculation. You can use the more applicable definition of the expectation value:
$$\langle{}\Omega\rangle{}_\psi=\frac{\langle{}\psi|\Omega|\psi\rangle}{\langle{}\psi|\psi\rangle}$$
This allows you to generalize the concept of "state" to any normalization.

can I write( exp i(k_0 x)-exp i(k_0 x))/2 and we know for free particle we have A exp i(k_0 x)-Bexp i(k_0 x) and I use it?

Something like that, but don't forget all of the i's.

 

[najima]

for part 1 you mean I must write ∫_(-∞)^∞〖sink_0 x) cos〖k_0 x〗 dx〗/∫_(-∞)^∞〖sink_0 x) sin〖k_0 x〗 dx〗?am I right?

 

[najima]

bufford means that particle can exist everywhere because of sin function? so it is infinite.yes?

 

[buffordboy23]

Yes, because you are not using the true state in your calculation. You can use the more applicable definition of the expectation value:
$$\langle{}\Omega\rangle{}_\psi=\frac{\langle{}\psi|\Omega|\psi\rangle}{\langle{}\psi|\psi\rangle}$$
This allows you to generalize the concept of "state" to any normalization.

Thanks Turin. I had some confusion on this as well about the initial wave function.

Najima have you learned about the bra/ket notation to understand Turin's remark.

 

[buffordboy23]

bufford means that particle can exist everywhere because of sin function? so it is infinite.yes?

What is infinite?

 

[najima]

yes .I've learned. :)
is it this∫_(-∞)^∞〖sink_0 x) cos〖k_0 x〗 dx〗/∫_(-∞)^∞〖sink_0 x) sin〖k_0 x〗 dx〗?but I can integrate it yet.

 

[najima]

probability of particle existence .

 

[buffordboy23]

najima, here was my thinking. The position of the particle is ill-defined, just like you said. The uncertainty in its position approaches infinity. But it has a well-defined wavelength, which is related to $$k_{0}$$. And how is this related to momentum?

I am going to grab a cup of coffee and will be right back on trying to work out the same result using Turin's suggestion.

EDIT: Using the exponential terms for sine, Maple says that the integral is equal to zero when computing the expectation value for p. The expectation value makes sense physically, and could be deduced without any calculations. But, I don't think this is what the problem is asking for, now is it? The momentum of the particle is well-defined but is equally likely to have positive or negative direction, so this is why the expectation value is zero. You shouldn't need to any integrals to determine what the momentum of the particle is.

Going back to what I said about using Fourier transforms, you should find that $$\Psi\left(x,t\right)$$ is constant, which may useful for the second part of the problem.

 

[turin]

editting; please wait.

for part 1 you mean I must write ∫_(-∞)^∞〖sink_0 x) cos〖k_0 x〗 dx〗/∫_(-∞)^∞〖sink_0 x) sin〖k_0 x〗 dx〗?am I right?

Not exaclty. This is hard for me to read. This is what I see:

$$\frac{\int_{-\infty}^{+\infty}\sin\left(k_0x\right)\cos\left(k_0x\right)dx}{\int_{-\infty}^{+\infty}\sin\left(k_0x\right)\sin\left(k_0x\right)dx}$$

Be careful about what the derivative of sine is, and what is the exact definition of P as an operator on the coordinate basis. Also, it may be easier for you to see the answer (i.e. do the integral) if you use the complex exponential decomposition that you suggested in another post. However, I suggest that you first identify why I say that this integral expression for the expectation value of P is wrong.

 

[buffordboy23]

2. suppose that p is measurement at t=3 s and the value (h/2pi)(k_0) is found. what is ψ(x,0)
at t>3 s?

I think this question is confusing as well. The very act of making a measurement changes the wavefunction--the wavefunction before a measurement will be different from the wavefunction after the measurement. I think that this question wants you to think that this is a hypothetical measurement, so the initial wavefunction is not modified. Basically, it would be better to ask, How does the initial wavefunction evolve over time?"

 

[buffordboy23]

Free particles don't exist in stationary states, so there is no definite energy of the particle. Have you learned about wave packets and Fourier transforms yet?

In reviewing this problem, this is an exceptional case. The particle appears to have a definite momentum, since the initial wavefunction has a precisely defined wavelength, related to k0. There is only one value of k. Usually, for problems about the free particle, the variable k is a continuous variable, so there is a range of k's, which comprise a wave-packet. Therefore, you have to use Fourier transforms to determine the evolution of the wavefunction. This threw me off earlier.

 

[najima]

thanks a lot turin and buffordboy for your helping,it helps me,

 

[najima]

now I try it again ,by using your suggestion.this is my homework for tomorrow.:(

 

[turin]

Sorry, people. I got distracted (Fri night and all) and forgot to remove the "editting" warning at the top of my last post. Apparently you can't edit a post from last week? Anyway ...

The particle appears to have a definite momentum, since the initial wavefunction has a precisely defined wavelength, related to k0. There is only one value of k.

However, momentum also has a direction ...

 

[buffordboy23]

However, momentum also has a direction ...

It appears to me that you think zero, the expectation value for momentum, is the answer.

Consider the original question:

1. what value of p (momentum) will measurement yield at the time t,and with what probabilities will these values occur?
2. suppose that p is measurement at t=3 s and the value (h/2pi)(k_0) is found. what is ψ(x,0)
at t>3 s?

The question does not seem to require the expectation value (the average value returned from a series of measurements on the same system) for the answer.

Instead, it seeks the values that the momentum can have. There are two possibilities, each with equal magnitude, but different in direction.

I agree with your point earlier about this problem; there are many ambiguities to confuse the student.

 

[turin]

It appears to me that you think zero, the expectation value for momentum, is the answer.

The question does not seem to require the expectation value ...

I had already forgotten (and don't care so much) about the original question (because, as you say, it is not so good, anyway). But, no. I was merely pointing out that this is NOT a momentum eigenstate. But it is good that you brought it up, for the benefit of whomever; this is one of those common misconceptions.

 

[buffordboy23]

No, your right that it's not a momentum eigenstate, because the wavefunction is actually a superposition of two momentum eigenstates, exp(ik0x) and exp(-ik0x).

I just wanted to make things clearer for whomever reads this post in the future.