# Harmonic oscillator

#### aaaa202

A particle has its wave function as the ground state of the harmonic oscillator. Suddenly the spring constant doubles (so the angular frequence dobules). Find the propability that the particle is afterwards in the new ground state. I did solve this, it was quite easy. But doing so I encountered a problem I could really explain - I think it's pretty basic problem but I can't explain it to myself.
You can do two things: You can express the wave function (the old ground state in terms of the new angular frequence) in terms of the new angular frequency and then integrate that with the new ground state to get the probability amplitude.
Or you can express the new ground state in terms of the old angular frequence and integrate up to get the probability amplitude. Either way you get the correct answer.
What my problem was, is that I can't explain that the two procedures are equivalent. It should be pretty basic, after all it is just a matter of which angular frequence you choose to work with. But I still find it a bit miracolous that in the end you get the same... Please explain :)

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#### tom.stoer

What I am understanding is the following: you have two ground state wave functions

$$\psi^{0}_{\omega}(x) = \langle x|0\rangle$$
$$\psi^{0}_{\omega^\prime}(x) = \langle x|0^\prime\rangle$$

for the two frequencies ω and ω'.

What you want to calculate is the probability to find a particle that has been prepared in the old ground state again in the new ground state

$$p = \langle 0^\prime|0\rangle= \int_{-\infty}^{+\infty} dx\,\langle 0^\prime|x\rangle\langle x|0\rangle = \int_{-\infty}^{+\infty} dx\,\psi^{0}_{\omega^\prime}(x)\,\psi^{0}_{\omega}(x)$$

I do not see what you mean by
You can express the wave function (the old ground state in terms of the new angular frequence) in terms of the new angular frequency and then integrate that with the new ground state to get the probability amplitude.
Or you can express the new ground state in terms of the old angular frequence and integrate up to get the probability amplitude. Either way you get the correct answer.

#### aaaa202

sorry yeh, I should have put in some formulas.
What I meant was:

You can either take ψ(x,0) to be a function of ω and integrate that with the new ground state $\psi$0 also expressed through the old ω and use that to determine:
cn = ∫ψ(x,0)$\psi$0dx (I omitted the boundaries at infinity because they were too hard to write properly).
Or you could take ψ(x,0) and $\psi$0 and express them through ω', do the integral and get the same propability amplitude.
Why is it that you must necessarily get the same? I feel it must be trivial, but I can't come up with a trivial argument.

#### tom.stoer

I don't really understand your notation. Look at mine: what's unclear? where's the problem?

#### aaaa202

what's not to understand. To find the coefficient in the expansion of the initial wave function you do the integral as stated above. You can either express those two functions in the integrals as functions of ω or ω'. Either way you will get the same since the dependence in the end cancels out to provide a number. My question is just, whether it should be intuitive whether I should express the functions through ω or ω'..

#### tom.stoer

I dont express any function in terms of the other function; I calculate a scalar product, a projection of |0> to |0'>, nothing else.

$$\langle 0^\prime|0\rangle= \int_{-\infty}^{+\infty} dx\,\psi^{0}_{\omega^\prime}(x)\,\psi^{0}_{\omega}(x) = \left(\frac{m}{\pi\hbar}\right)^{1/2} (\omega\,\omega^\prime)^{1/4}\int_{-\infty}^{+\infty} dx\,\exp\left[ -\frac{m}{2\hbar}(\omega + \omega^\prime)x^2\right]$$

And there are no
functions in the integrals as functions of ω or ω'
they are functions of x depending on two different parameters ω or ω'.

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#### aaaa202

maybe this confusion would disappear if I actually showed my computations:

$\psi$0(x,0) = (mω'/(2h$\pi$))1/4exp(-mω'/(4h)x2)

while the new ground state is:

ψ0 = (mω'/(h$\pi$))1/4exp(-mω'/(2h)x2)

So you calculate the expansion coefficient using:

cn = ∫$\psi$(x,0)ψ0dx

But you could also express our functions through the old angular frequency:

$\psi$0(x,0) = (mω/(h$\pi$))1/4exp(-mω/(2h) x2)

while the new ground state is:

ψ0 = (2mω/(h$\pi$))1/4exp(-mω/x2)

And calculate the same integral for cn and get the same. Basically, I don't understand why the choice of constant ω or ω' is bound to yield the same in the end. I guess it makes sense because it cancels out but it just seems rather weird since the constant alters the exponential and well that's just weird. Hope it made a little sense.

Bear in mind that h means hbar - couldnt find that symbol.

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#### tom.stoer

sorry, but you seem to calculate exactly the same integral

$$\langle 0^\prime|0\rangle= \int_{-\infty}^{+\infty} dx\,\psi^{0}_{\omega^\prime}(x)\,\psi^{0}_{\omega}(x) = \left(\frac{m}{\pi\hbar}\right)^{1/2} (\omega\,\omega^\prime)^{1/4}\int_{-\infty}^{+\infty} dx\,\exp\left[ -\frac{m}{2\hbar}(\omega + \omega^\prime)x^2\right]$$

What is your problem?

#### aaaa202

ugh nevermind. I guess my problem was just that in the first integral the functions contain ω' and the other ω. But I guess working with either of them just represents writing the constant as something else, so yes that certainly shouldn't alter anything. Im sorry I took so much of your time.

no problem ;-)

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