Infinite Square Well -- Instantaneous expansion of the Well

[JD_PM]

Homework Statement

My doubts are on c)

Homework Equations

$$< H > = \int \Psi^* \hat H \Psi dx = \frac{2}{a} \int_{0}^{a} sin (x\frac{\pi}{a}) \hat H sin (x\frac{\pi}{a}) dx$$

The Attempt at a Solution

I understand that mathematically the following equation yields (which is the right answer):

$$< H > = \frac{2}{a} \int_{0}^{a} sin (x\frac{\pi}{a}) \hat H sin (x\frac{\pi}{a}) dx = \frac{\pi^2 \hbar^2}{2ma^2} $$

But before seeing this method I did the following:

We know that the expectation value of the energy is:

$$< H > = \sum_{n=1}^{\infty} |c_n|^2 E_n$$

So after the well expands to twice its original size I would have said that:

$$< H > = \sum_{n=1}^{\infty} |c_n|^2 E_n = \frac{n^2 \pi^2 \hbar^2}{8ma^2}$$

Actually, Griffiths recommends not using the series method but I do not see why, as $|c_n|^2=1$

Why am I wrong?

 

[DrClaude]

Actually, Griffiths recommends not using the series method but I do not see why, as $|c_n|^2=1$

That's not correct, it is $\sum_n |c_n|^2=1$.

 

[JD_PM]

That's not correct, it is $\sum_n |c_n|^2=1$.

My bad, should I edit it?

 

[DrClaude]

My bad, should I edit it?

You don't need to. The problem is not just the typo. How did you find the equality

$$< H > = \sum_{n=1}^{\infty} |c_n|^2 E_n = \frac{n^2 \pi^2 \hbar^2}{8ma^2}$$

without knowing the $c_n$'s, apart from the fact that $\sum_n |c_n|^2=1$?

 

[JD_PM]

You don't need to. The problem is not just the typo. How did you find the equality

without knowing the $c_n$'s, apart from the fact that $\sum_n |c_n|^2=1$?

The limited values for the energy in the infinite square well with width $a$ are:

$$E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$$

Now the width is $2a$ so:

$$E_n = \frac{n^2 \pi^2 \hbar^2}{8ma^2}$$

My intuition tells me that I should get $\frac{n^2 \pi^2 \hbar^2}{8ma^2}$ because we make the measurement (I know this word is extremely delicate in QM) once the well has been expanded. However the provided answer states $\frac{n^2 \pi^2 \hbar^2}{2ma^2}$, which the energy before the well expansion!

What is going on here?

I want to stress that I am interested in the physical meaning not mathematical (I understand the integral method).

 

[DrClaude]

First, you shouldn't expect to get a result with $n$ in it. The quantum number simply labels the energy eigenstates, so it should disappear when calculating the actual energy of the system.

The key word in the problem is "suddenly." If the change is sudden, then the state hasn't time to adapt to the new potential. The wave function will remain unchanged from what it was previously. Therefore, it does not correspond to an eigenstate of the new Hamiltonian, so the answer will be more complicated.

It is always valid to use
$$\langle \hat{H} \rangle = \sum_{n=1}^{\infty} |c_n|^2 E_n$$
but for that you need to calculate the coefficients $c_n$ for the eigenstates of the new Hamiltonian. This can be done, but it is not the easiest way to solve the problem. This is why Griffiths warns about not taking that route.

 

[JD_PM]

First, you shouldn't expect to get a result with $n$ in it. The quantum number simply labels the energy eigenstates, so it should disappear when calculating the actual energy of the system.

The key word in the problem is "suddenly." If the change is sudden, then the state hasn't time to adapt to the new potential. The wave function will remain unchanged from what it was previously. Therefore, it does not correspond to an eigenstate of the new Hamiltonian, so the answer will be more complicated.

It is always valid to use
$$\langle \hat{H} \rangle = \sum_{n=1}^{\infty} |c_n|^2 E_n$$
but for that you need to calculate the coefficients $c_n$ for the eigenstates of the new Hamiltonian. This can be done, but it is not the easiest way to solve the problem. This is why Griffiths warns about not taking that route.

Mmm I see what you mean thanks.

But, the sudden change disturbs me... I mean does not that imply instantaneous change? (which goes against Relativity)?

 

[DrClaude]

But, the sudden change disturbs me... I mean does not that imply instantaneous change? (which goes against Relativity)?

I don't why instantaneous would go against relativity. For a real system, the change only need to be much faster than the normal dynamics of the system.

 

[JD_PM]

I don't why instantaneous would go against relativity.

Because instantaneous information travels faster than light

 

[JD_PM]

For a real system, the change only need to be much faster than the normal dynamics of the system.

Even though the wells just expand much faster than the wavefunction, in the end it is different. See my discussion about this:

https://physics.stackexchange.com/q...ell/453285?noredirect=1#comment1017637_453285

https://chat.stackexchange.com/rooms/88089/discussion-between-aaron-stevens-and-jd-pm

 

[PeroK]

Because instantaneous information travels faster than light

QM is non-relativistic. It treats time as a parameter separate from space.

Classical mechanics is non-relativistic. A rigid body is non-relativistic.

In any case, "instantaneous" simply means "fast enough to ignore the transition phase". It doesn't necessarily imply an instantaneous change in reality.

 

[DrClaude]

Because instantaneous information travels faster than light

The information is not instantaneous, the change in potential is. One could imagine a situation where the information travels at the speed of light, but at all points where the quantum system is located (and remember that the problem is 1D) the change on potential happens at the same time.

Again, while in practice shutting down a potential in no time is not possible, it doesn't violate any law of physics.