Position and Momentum probability for +x direction

[tanaygupta2000]

Homework Statement

A quantum particle is confined in a one-dimentional box defined by the potential,
V{x) = 0 for 0 < x < L, and infinity otherwise. The particle is in the ground state. What is the
probability (P+) that an experimenter will find it moving along the positive-x direction?
What would be the magnitude of momentum (p) of that motion?
(a) P+ = 0 and p = 0
(b) P+ = 1/2 and p=n/L
(c) P+=1/2Landp=n/L
(d) P+ = 1/2 and p = 0
(e) P+ = 0 and p = 1t / L

Relevant Equations

wave function, psi = sqrt(2/L) sin(pi x/L)
Energy,E = pi^2 hbar^2/2mL^2

For the region where V = 0, solving the schrodinger equation leads to the above value of wave function, psi = sqrt(2/L) sin(pi x/L)
Since in the qus. it is not stated about the 'direction of movement' only restricted to +x direction, I think that the probability will be 1/2.
And finding the expectation value of momentum for this wave function, i am getting something like -i hbar/2L, which is nearly equal to 0.
So, I think that the correct option will be option-(d).
Am I right? Help will be appreciated!

 

[TSny]

For the region where V = 0, solving the schrodinger equation leads to the above value of wave function, psi = sqrt(2/L) sin(pi x/L)

Good

Since in the qus. it is not stated about the 'direction of movement' only restricted to +x direction, I think that the probability will be 1/2.

Ok. To see this concretely, express ##\sin (\pi x/L)## in terms of ##e^{i \pi x/L}## and ##e^{-i \pi x/L}##.

And finding the expectation value of momentum for this wave function, i am getting something like -i hbar/2L, which is nearly equal to 0.

You should get exactly zero for the expectation value of ##p##.

The expectation value of ##p## does not represent the result of a measurement of ##p##. The only possible results of a measurement of ##p## are the eigenvalues of the momentum operator ##\hat p##.

They want you to find the value of ##p## that you would get if you measure ##p## and find that the measured value corresponds to movement in the +x direction. Expressing the wavefunction in terms of the complex exponentials will be helpful.

 

[tanaygupta2000]

By this approach, I am getting probability from 0 to L is exactly '1' (obvious since V = infinity otherwise). However, the expectation value of momentum is correctly coming '0'. There is no option in the question for Probability = 1.
Thanks !

 

[kuruman]

The choices that you are given leave room for interpretation. First, quantity "n" in options (b) and (c) is not specified in the problem. Of course we are told that the particle is in the ground state, so we may justifiably assume that n = 1. Option (e) has the confusing p = 1t / L. Presumably the 1 is from n = 1. What is t? Is it the time independent variable?

I think this problem can be answered without math. Note that ##\langle p \rangle =0## and invoke a symmetry argument.

 

[tanaygupta2000]

quantity "n" in options (b) and (c) is not specified in the problem

I have attached the image for clarity

 

[TSny]

By this approach, I am getting probability from 0 to L is exactly '1' (obvious since V = infinity otherwise).

The probability that a measurement of position, ##x##, will yield a value in the range ##0 \leq x \leq 1## is exactly 1. But the question isn't asking for this.

I think the question is asking about a measurement that determines which direction the particle is traveling. For this one-dimensional system, the direction of motion is determined by the sign of the momentum. So, they want the probability that a measurement of momentum will have a positive value for the momentum.

 

[tanaygupta2000]

By looking at the options, probability can either be 0 or 1/2. And since the value of momentum is coming to be precisely zero, can we assume that equal probability of measurement goes above 0 as the same as below 0 (i.e., 1/2 on both sides)?
This argument leads the probability to be 1/2 towards +x direction.

 

[haruspex]

Option (e) has the confusing p = 1t / L.

Judging from the other options, I suggest 1t is a misreading of (a handwritten?) n.

 

[tanaygupta2000]

I apologise for it. Have attached an image of the question for clarity. Thanks !

 

[kuruman]

Suppose you and I look at this particle bouncing back and forth. We are facing each other and the box is between us. What's "to the left" (negative) for you is "the right" (positive) for me. Do you see the symmetry?

Judging from the other options, I suggest 1t is a misreading of (a handwritten?) n.

I think that's a good suggestion.

 

[kuruman]

I apologise for it. Have attached an image of the question for clarity. Thanks !

I see. n is actually ##\pi##. Thanks for the clarification.

 

[tanaygupta2000]

Suppose you and I look at this particle bouncing back and forth. We are facing each other and the box is between us. What's "to the left" (negative) for you is "the right" (positive) for me. Do you see the symmetry?

I think that's a good suggestion.

Yes I can clearly see it. But I am still confused how can the particle enter the region -x, since V = infinite there.

 

[tanaygupta2000]

Yes I can clearly see it. But I am still confused how can the particle enter the region -x, since V = infinite there.

I think probability = 1/2 only comes as an assumption by observing the value of momentum (0).

 

[haruspex]

I think probability = 1/2 only comes as an assumption by observing the value of momentum (0).

The probability is 1/2 from a simple symmetry argument. The expected value is also zero by the same argument.
But then you want the (expected?) value of p given that it is positive. Or equivalently, the value of |p|. This will not be zero.

 

[tanaygupta2000]

Thank You so much for the help !