How Does Expanding a Quantum Box Affect Particle Ground State Probability?

[amjad-sh]

Homework Statement

A particle is in the ground state of a box of length L.Suddenly the box expands (symmetrically) to twice its size, leaving the wave function undisturbed.Show that the probability of finding the particle in the ground state of the new box is (8π/3).

Homework Equations

If n is odd : \psi_{n}(x)=√(2/L)*cos(nπx/L)
If n is even: \psi_{n}(x)=√(2/L)*sin(nπx/L)
\int_{-L}^{L}\psi_{n}(x)dx=1

The Attempt at a Solution

If someone can give me a hint because I don't know where to begin.

 

[DuckAmuck]

You want to compute the transition probability of the particle going from one state to the other.
State 1: ground state of box size L
State 2: ground state of box size 2L

So you basically want to find the overlap between the two ground states by taking an integral of their product.

 

[amjad-sh]

Sorry for being late, but I want to raise 2 questions related to the problem above.

1. Suppose that the box didn't expand and the particle is in the ground state, we can still find the probability of the particle being in the ground state of the new box,by making an integral of their product,and the probability will be (8π/3). But how this comes? shouldn't the probability be zero in this case since the ground state of the new box may be not an allowed state for the system?or does this mean that the ground state of the new box is an acceptable state in the previous system?2.when we expand the box did the particle transits "instantaneously" to a new state? Is this behavior called the "quantum jump"?

 

[DuckAmuck]

1. Think about it practically. If you change the size of a box, the particle inside can't just vanish. It can't continue occupying *only* the old volume of the box. It's still going to be in the box, but it will adjust to the new boundaries.

Also, in general, it will be able to hop into a number of states, so going from ground state to ground state isn't a guaranteed 100% chance thing. It could wind up in an excited state.

2. The wavefunction changing is not instantaneous. There is a characteristic decay time. Look up "time evolution".

 

[amjad-sh]

1. Think about it practically. If you change the size of a box, the particle inside can't just vanish. It can't continue occupying *only* the old volume of the box. It's still going to be in the box, but it will adjust to the new boundaries. In general, it will be able to hop into a number of states, so going from ground state to ground state isn't a guaranteed 100% chance thing. It could wind up in an excited state.

Suppose that the ground state of the new box does not match the allowed states of the previous box, and we didn't change anything concerning the old box. If we want to find the probability that the particle is in the ground state of the new box(still being in the old box and the boundaries are the same), shouldn't it be zero?

2. The wavefunction changing is not instantaneous. There is a characteristic decay time. Look up "time evolution".

If the particle is initially in the ground state and then we change the box. Suppose that the ground state of the previous box does not match the allowed states of the new box. Shouldn't the particle in this case jump instantaneously to a new energy state, since it is in a forbidden energy state.

 

[DuckAmuck]

No it will not be zero. Transition probability of zero would imply that the particle just stays in its original state, even after the box is changed. To transition, the states don't have to *match*, they just have to overlap in Hilbert space. They don't even have to have the same energy eigen values.

 

[amjad-sh]

No it will not be zero. Transition probability of zero would imply that the particle just stays in its original state, even after the box is changed. To transition, the states don't have to *match*, they just have to overlap in Hilbert space. They don't even have to have the same energy eigen values.

So do you mean that the transition probability from a state1 to another state2 could be not equal to zero, even if the particle can't possibly move to state2 in reality(or it can move to state 2 but in other cases like changing the box)?
I can't understand how the energy eigenstates are overlapping here.

 

[DuckAmuck]

can't possibly move to state2 in reality

A transition is only "impossible" if the transition probability is 0. All you have are state1 and state2, and those determine your transition probability. You don't impose a probability by hand.

 

[amjad-sh]

To transition, the states don't have to *match*, they just have to overlap in Hilbert space

do you mean here that the energystates of the new box will overlap with the energy states of the old box?
but here what is confusing: if we are in the old box, can the transition probability to a new state be not equal to zero even if this state is forbidden in the old box?

 

[DuckAmuck]

I think I see where the confusion is. The old state isn't *forbidden* in the new box. The old state is actually a linear combination of new box states, for a little while, and then it eventually evolves into state2.

Maybe seeing it in a picture will help. See attached.

 

[amjad-sh]

Thanks a lot for your effort.
I understand from this that the position wave function of the particle will evolve in time until it becomes a new position wavefunction and that happens when we change the box.
\psi(x) after we change the box will evolve in time,and it will have a probability to reach "only" one of the position wave functions:\psi_{1'}(x),\psi_{2'}(x),\psi_{3'}(x),...,\psi_{n'}(x) where \psi_{n'}(x) is \langle x | n^{'}\rangle and |1^{'}\rangle ,|2^{'}\rangle,...|n^{'}\rangle are the quantized discrete energy states of the new box. please correct me if I'm wrong.

But I'm still confused about two things:
1.the transition probability from state |1> to state |1'> is not zero. Since when we change the box, |1'> is one of the allowed states.

But suppose that we didn't change the box and |1'> is not an allowed state of the initial box, the transition probability from |1> to |1'> in this case is <1'|1>, inserting the completeness relation we will get \int \psi^{*}_{1&#039;}(x)\psi_{1}(x)dxand the answer will not be zero. How it is not zero while it is impossible for the particle to move to this state?
I predicted an answer but I don't know if it is right, my answer: While the particle in the old box is moving from a state to a new allowed state, the wave function will have the possibility to match a not allowed state while it is evolving in time to reach the allowed wave function.

2. why the transition probability from a state |n> to state |n'> is written like this <n'|n> and it will be like this \int \psi^{*}_{n&#039;}(x)\psi_{n}(x)dx if we insert the completeness relation. Isn't this the amplitude not the probability?

 

[amjad-sh]

<1'|1>, inserting the completeness relation we will get ∫ψ∗1′(x)ψ1(x)dx

sorry I mean |<1'|1>|^2 and the integral then will be also absolute squared,and so part 2 of my question has no sense.

 

[DuckAmuck]

Really you aren't *exactly* transitioning from state1 to state1'. Remember state1 was a hump, but you rapidly expanded the box, so now the state is a hump plus a flat line. This hump + flat line state is what transitions to state1'.

 

[jtbell]

This hump + flat line state is what transitions to state1'.

The "hump plus flat line" state is a linear combination (superpostition) of state1', state2', state3', etc. It will remain in that superposition until we do something (an energy measurement) which collapses the state into an energy eigenstate of the new well. Until that happens, the superposition state "sloshes" back and forth inside the well in a complicated way without settling down into a pure state1'. Note that state1', state2', etc. have different time-variation factors $e^{-iE_k^\prime t / \hbar}$ so they don't oscillate in step with each other.

 

[amjad-sh]

@DuckAmuck @jtbell
I agree with you both about all what you have said above.
but this is not what I'm asking about.
my question is that if we didn't change the box the probability will be not equal to zero.
because the integral is the same in both cases.
As the "hump+flat line" is mathematically equal to the "hump", since the flat line is a wavefunction of zero value.
I don't know if you got my point.

 

[DuckAmuck]

Yeah they are equal integrals. The issue is, you'd be going from state1 to half-state1'. It'd be a half-hump, and not an energy eigenstate of the original box. The only way this transition could happen is if this half-hump corresponds to the eigenstate of some other measurable quantity... but I don't think that's the case.

 

[jtbell]

my question is that if we didn't change the box the probability will be not equal to zero.
because the integral is the same in both cases.

I assume you mean e.g. the integral $$\int {\psi_{1,old}^* \psi_{1,new}} dx$$ where the particle starts in the ground state of the "old" box. If you expand the box then with the limits inserted this is $$\int_0^{L_{new}} {\psi_{1,old}^* \psi_{1,new}} dx$$ which gives you the amplitude for the particle to end up in the ground state of the "new" box.

If you don't expand the box, then you have $$\int_0^{L_{old}} {\psi_{1,old}^* \psi_{1,new}} dx$$ (note the upper limit) which I don't think makes sense because if you "chop off" $\psi_{1,new}$ in the middle, it's not a valid wave function for the "old" box. It doesn't go to zero at both sides of the box.

This raises the question, what do you do if you want to "shrink" the box suddenly? I don't know the answer (yet), but it seems to me that this might be a fundamentally different situation from "expanding" the box suddenly. Consider an analogy in thermo / stat mech. Suppose you have a box divided in half by a removable partition, with gas in one half and vacuum in the other. Yank the partition out and the gas expands freely to fill the entire box. However, you can't make the gas go back into one half by putting the partition back in. You can compress the gas into one half of the box by using a piston, but that's a different process.

Similarly with the particle in a box, if you want to "shrink" the box, I think you have to do it "gradually" in some sense.

 

[amjad-sh]

Thanks all for contribution.