Exploring the Probability of a Harmonic Oscillator in a Changed Spring Constant

[L-x]

Homework Statement

In the time interval (t + δt, t) the Hamiltonian H of some system varies in such a way that |H|ψi>| remains finite. Show that under these circumstances |ψi> is a continuous function of time.
A harmonic oscillator with frequency ω is in its ground state when the stiffness of the spring is instantaneously reduced by a factor f^4 < 1, so its natural frequency becomes (f^2)ω. What is the probability that the oscillator is subsequently found to have energy (3/2)h(f^2)ω? Discuss the classical analogue
of this problem.

Homework Equations

ih dψ/dt = Hψ (part 1)
Er = ( r+ 1/2)hω for a 1D harmonic oscillator

The Attempt at a Solution

The first part of the question is trivial (using the TDSE) and I had no problems answering it, I include it only because, to me, it doesn't seem to help answer the second part of the question, which is unusual (but not unheard of) for this textbook, so i wish to make sure I'm not missing something.

As for answering the second part of the question: I think the oscillator must now have energy hω/2 > E > (f^2)hω/2 as hω/2 is the ground state energy of the oscillator before the change, and (f^2)hω/2. My thought process went along the lines of "If the oscillator has close to no potential energy before the change, the change in the spring constant (and therefore the potential) will have close to no effect, which gives us the maximum energy possible, and vice versa to obtain the minimum"

This means the oscillator is now going to be in a superposition of energy eigenstates |r> ((f^-2) -1)/2 > r > 0. However unless these eigenstates each have equal probability I can't find an expression for their amplitudes, and i can't convince myself that they do have equal probability.

Is my working so far correct, and more importantly is there an equal probability to find the system in any of the possible eigenstates. If not- how should i proceed?

Thanks for the help
-Alex

 

[vela]

Apply the first part of the question to the second. What does it say about the state of the system immediately after the spring constant changes?

 

[L-x]

As |ψ> is continuous wrt to time, an instantaneous change in H will be |ψ> is the same as before. I'm really confused about what this means though. |ψ> before is |0>, a state where you are certain to measure E = hω/2 if you act with the original hamiltonian. I realize are still in this state after the change in hamiltonian, so my earlier thinking was flawed and the energy of the state must be the same, but I'm now really confused about what that means for how we represent the state in the eigenbasis of our new hamiltonian.

If f^2 (r+1/2) hω = hω/2 has no solutions for integer r (which, in general, will be the case) there must be a non-zero amplitude to measure more than one value of energy: we must have a superposition of eigenstates of the new hamiltonian, and I am faced with the same problem as before- I really don't understand how we begin assigning amplitudes to them.

 

[ideasrule]

|ψ> before is |0>, a state where you are certain to measure E = hω/2 if you act with the original hamiltonian. I realize are still in this state after the change in hamiltonian, so my earlier thinking was flawed and the energy of the state must be the same, but I'm now really confused about what that means for how we represent the state in the eigenbasis of our new hamiltonian.

No, the energy of the state isn't the same. In fact, the particle is no longer in any state; suddenly changing the Hamiltonian separates the wavefunction into a linear combination of stationary states, which are different from the stationary states of the previous Hamiltonian.

When H changes, it's the wavefunction that (momentarily) stays the same, not the energy and not the stationary states.

If f^2 (r+1/2) hω = hω/2 has no solutions for integer r (which, in general, will be the case) there must be a non-zero amplitude to measure more than one value of energy: we must have a superposition of eigenstates of the new hamiltonian, and I am faced with the same problem as before- I really don't understand how we begin assigning amplitudes to them.

Do you know about Fourier's trick? If not, I'm very sure that it's described in your QM textbook. Remember that the the bases of both the original and the new Hamiltonian and orthonormal.

 

[vela]

If a system is in a state $|\psi\rangle$, the amplitude of finding it in the state $|\phi\rangle$ is given by $\langle \phi | \psi \rangle$.

 

[L-x]

Thanks for your help so far. I have an answer but wanted to check my justification of it.

if we call the energy eigenstates after the collision |0'>, |1'>, |2'> etc. we wish to find:
<1'|0>.

Inserting the identity in the form
$\int|x><x| dx$
we obtain:
$\int<1'|x><x|0> dx$

which will be 0, as the integral is of the form $xe^{ax^2}$ between +/- infinity

Can we justify this by saying the state before the change in the hamiltonian had even parity, and this will still be the case afterwards (as the state is unchanged) hence only the even parity states of the new oscillator have a chance of being excited?

Thanks very much for all your help.

 

[vela]

Yes, your reasoning is correct.

 

[L-x]

Sorry to bump this again, but there's something I'm still slightly unclear on something. The particle is now in a linear superposition of states of odd parity, let's call them |n'>, with associated amplitudes n. What I'm wondering is: can an amplitude k be nonzero if the associated state |k'> is a state in which the particle has more energy than it did originally? My reasoning above suggests that that this is indeed the case- however if that is true I don't understand where the energy has come from if the particle ends up in this state. However if it isn't true I don't see how we can say the earlier reasoning is invalid.

 

[vela]

I don't think you can really expect energy to be conserved in a situation like this. For example, classically, suppose you changed the spring constant from k to k' at the moment that the particle is at rest. If A is the amplitude of oscillation, the energy of the system instantaneously changes from 1/2 kA² to 1/2 k'A². It's clearly not conserved.

 

[L-x]

Thanks, unfortunately this has meant that I'm tearing my hair out with a follow on question:P is the probability that at the end of the experiment described in Problem 3.12, the
oscillator is in its second excited state. Show that when f=1/2, P=0.1444 as follows:
First show that the annihilation operator of the original oscillator
$$A=\frac{1}{2} [(f^{-1} + f)A' + (f^{-1} - f)A'^{dagger}]$$ done, no problems
the writing the ground state oscillator as a sum over the energy eigenkets of the final oscillator, impose the condition A|0>=0
I have done this, and obtained a recurrence relation for the amplitudes (lets call them cn) of each eigenket in terms of c0. Unfortunately this reccurence relation gives a really nasty cn, and worse still when i take c0=1 and try to find the sum over all |cn|^2 the summation does not converge, so I can't normalise the amplitudes.

The reccurence relation i obtained is

$$c_{n+2} = \frac{-3}{5}\sqrt{\frac{n-1}{n}}c_{n}$$

EDIT this should be $$\sqrt{\frac{n+1}{n+2}}$$, see below

which gives $$c_{n} = {(\frac{-3}{5}})^{n/2}\sqrt{\frac{2^{-n}n!}{{\frac{n}{2}!}^2}}$$ when we set c0=1

 

[L-x]

I have checked the formula for cn, and got a mathematician friend of mine to check it as well, and have confirmed that it is correct and that the sum over all |cn|^2 doesn't converge.

Am I going about the normalisation in the wrong way? I thought that since we wanted the sum over all the probabilities to equal 1, we should define S= (sum from n=1 to infinity) |cn|^2 and then divide by root S to get normalised amplitudes

 

[vela]

I don't think your recurrence relation is correct. When n=0, it's undefined. It should be

$$c_{n+2} = -\frac{3}{5}\sqrt{\frac{n+1}{n+2}}c_n$$

I also don't see how you concluded the series doesn't converge since, setting c₀=1, you have

$$\begin{align*} |c_2|^2 &= \left(\frac{3}{5}\right)^2\frac{1}{2} < \left(\frac{3}{5}\right)^2 \\ |c_4|^2 &= \left(\frac{3}{5}\right)^4\frac{1\cdot 3}{2\cdot 4} < \left(\frac{3}{5}\right)^4 \\ |c_6|^2 &= \left(\frac{3}{5}\right)^6\frac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6} < \left(\frac{3}{5}\right)^6 \\ &\vdots\end{align*}$$

so the series is bounded above by a geometric series, which does converge. (The ratio test also says the series will converge.)

I don't know if this is the approach you want to take, though. Do you recognize the series or know of some other way to figure out what it will converge to?

 

[L-x]

That is the recurrence relation i have, unfortunately i'd been trying the question for so long that I had it memorised by the time I posted asking for help, and i slightly misremembered it.

I used mathematica to sum the series, and I've just seen where I went wrong, I (stupidly) asked mathematica to sum from n=0 to infinity all |cn|^2, using the formula I gave above for cn. This seemed perfectly sensible at the time, but unfortunately the formula is only valid for even n, cn for odd n are all 0, as i showed before (with your help).

fixing this tells us that the series converges to 5/4, which is a perfectly nice, sensible sort of number.

I'm interested to hear if there is a more sensible method than this that comes to mind though- anything that requires a computer program to solve I'm never happy with, although as I study more statistical mechanics I'm becoming increasingly resigned to the fact that it is sometimes necessary.

 

[vela]

Try comparing the series you obtained to the Maclaurin series for 1/√(1-x).