I can not solve this quantum harmonic oscillator question in Cohen's book

topsquark

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Apr 2008
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A few quick question for you and about the notation.

1) \(\displaystyle | \psi _n >\) is the oscillator state with n oscillators, correct?

2) That being the case, do you know that \(\displaystyle a | \psi _n > = \sqrt{n} ~ | \psi _{n - 1} >\) and \(\displaystyle a^{\dagger} | \psi _{n + 1} > = \sqrt{n + 1} ~ | \psi _{n + 1} >\) ?

3) Do you know that the number operator \(\displaystyle N | \psi _n > = a^{\dagger} a | \psi _n > = n | \psi _n > \) ?

Okay. We have to calculate
\(\displaystyle \overline{a} (t) | \psi _n > = U^{\dagger} (t, 0) ~ a ~ U(t, 0) | \psi _n >\).
(The tilde doesn't come out so well, so I'm using the overline.)

I'm going to do this the "long way." I think it gives a better idea of how to do this correctly. So step by step:
\(\displaystyle U^{\dagger} (t, 0) ~ a ~ e^{-iHt/ \hbar} | \psi _n > = U^{\dagger} (t, 0) ~ a ~ e^{-i \hbar \omega (n + 1/2) t/ \hbar } | \psi _n > = e^{-i \hbar \omega (n + 1/2) t/ \hbar } U^{\dagger} (t, 0) ~ a | \psi _n > \)

\(\displaystyle = e^{-i \hbar \omega (n + 1/2) t/ \hbar } U^{\dagger} (t, 0) \sqrt{n} | \psi _{n - 1} > = \sqrt{n} ~ e^{-i \hbar \omega (n + 1/2) t/ \hbar } U^{\dagger} (t, 0) | \psi _{n - 1} >\)

\(\displaystyle = \sqrt{n} ~ e^{-i \hbar \omega (n + 1/2) t/ \hbar } e^{i \hbar \omega ((n - 1) + 1/2) t/ \hbar} | \psi _{n -1} >\)

\(\displaystyle = \sqrt{n} ~ e^{-i \hbar \omega t/ \hbar } | \psi _{n - 1} >\)

Now to pretty it up a bit:
\(\displaystyle \overline{a} (t) | \psi _n > = U^{\dagger} (t, 0) ~ a ~ U(t, 0) | \psi _n > = = \sqrt{n} ~ e^{-i \hbar \omega t/ \hbar } | \psi _{n - 1} >\)

\(\displaystyle \overline{a} (t) | \psi _n > = e^{-i \hbar \omega t/ \hbar } \sqrt{n} | \psi _{n - 1} > = e^{-i \hbar \omega t/ \hbar } a ~ | \psi _n >\)

You can similarly show that
\(\displaystyle \overline{a ^{\dagger}} (t) | \psi _n > = e^{i \hbar \omega t/ \hbar } a ^{\dagger} ~ | \psi _n >\)

b) and c) can be done in a similar fashion. See if you can finish these. If you have problems, just let us know.

For d) this is just an application of U(t) on the wavefunction. So calculate \(\displaystyle U(t) | \psi _n >\). (Hint: What is \(\displaystyle | \psi _n (t = 0) >\)?

Let's let e) and f) wait until you have a better idea about these.

-Dan
 
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Jun 2019
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1
A few quick question for you and about the notation.

1) \(\displaystyle | \psi _n >\) is the oscillator state with n oscillators, correct?

2) That being the case, do you know that \(\displaystyle a | \psi _n > = \sqrt{n} ~ | \psi _{n - 1} >\) and \(\displaystyle a^{\dagger} | \psi _{n + 1} > = \sqrt{n + 1} ~ | \psi _{n + 1} >\) ?

3) Do you know that the number operator \(\displaystyle N | \psi _n > = a^{\dagger} a | \psi _n > = n | \psi _n > \) ?

Okay. We have to calculate
\(\displaystyle \overline{a} (t) | \psi _n > = U^{\dagger} (t, 0) ~ a ~ U(t, 0) | \psi _n >\).
(The tilde doesn't come out so well, so I'm using the overline.)

I'm going to do this the "long way." I think it gives a better idea of how to do this correctly. So step by step:
\(\displaystyle U^{\dagger} (t, 0) ~ a ~ e^{-iHt/ \hbar} | \psi _n > = U^{\dagger} (t, 0) ~ a ~ e^{-i \hbar \omega (n + 1/2) t/ \hbar } | \psi _n > = e^{-i \hbar \omega (n + 1/2) t/ \hbar } U^{\dagger} (t, 0) ~ a | \psi _n > \)

\(\displaystyle = e^{-i \hbar \omega (n + 1/2) t/ \hbar } U^{\dagger} (t, 0) \sqrt{n} | \psi _{n - 1} > = \sqrt{n} ~ e^{-i \hbar \omega (n + 1/2) t/ \hbar } U^{\dagger} (t, 0) | \psi _{n - 1} >\)

\(\displaystyle = \sqrt{n} ~ e^{-i \hbar \omega (n + 1/2) t/ \hbar } e^{i \hbar \omega ((n - 1) + 1/2) t/ \hbar} | \psi _{n -1} >\)

\(\displaystyle = \sqrt{n} ~ e^{-i \hbar \omega t/ \hbar } | \psi _{n - 1} >\)

Now to pretty it up a bit:
\(\displaystyle \overline{a} (t) | \psi _n > = U^{\dagger} (t, 0) ~ a ~ U(t, 0) | \psi _n > = = \sqrt{n} ~ e^{-i \hbar \omega t/ \hbar } | \psi _{n - 1} >\)

\(\displaystyle \overline{a} (t) | \psi _n > = e^{-i \hbar \omega t/ \hbar } \sqrt{n} | \psi _{n - 1} > = e^{-i \hbar \omega t/ \hbar } a ~ | \psi _n >\)

You can similarly show that
\(\displaystyle \overline{a ^{\dagger}} (t) | \psi _n > = e^{i \hbar \omega t/ \hbar } a ^{\dagger} ~ | \psi _n >\)

b) and c) can be done in a similar fashion. See if you can finish these. If you have problems, just let us know.

For d) this is just an application of U(t) on the wavefunction. So calculate \(\displaystyle U(t) | \psi _n >\). (Hint: What is \(\displaystyle | \psi _n (t = 0) >\)?

Let's let e) and f) wait until you have a better idea about these.

-Dan
The passage \(\displaystyle U ^{\dagger}(t,0)\) to exponential: n - 1 came from comutation \(\displaystyle [a, a{^\dagger}] = 1\) or applying the in \(\displaystyle |{\psi_n}>\) with n became n - 1?

In c) I would have to apply the operator P in this way:
\(\displaystyle | P | U {\dagger} | x>\) to show the relation?
Or something like: \(\displaystyle <x | U {\dagger} | \psi_n>\)? I have sketched some things but in c) I have no idea what to do and I have reviewed my old class notes and I have nothing as far as operator evolution applied in | x> or | p>.
 

topsquark

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Apr 2008
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On the dance floor, baby!
I'm still working on it. I'll get back to you as I can.

Yes, I've gone over to using the momentum space wavefunction myself. By doing this I've been able to work out an expression or two (which weren't nearly as helpful as I thought they would be.)

Let me give you my general direction of attack. You might be able to use it.

I keep running into expressions like \(\displaystyle < p' | H |x' >\). This will evolve into a term proportional to
\(\displaystyle < p' | \dfrac{d^2}{dx'^2} |x' >\)

This appears to be a dead-end.

I can also derive an expression for \(\displaystyle < p'| H | \psi _n >\) that deals directly with the Hermite polynomial form for the wavefunction using a change of basis. It's very messy and I don't know if I can put it into any kind of intelligible form for a solution.

I'll keep working on it from this end. This is one of those problems that looks easy when reading and gets all bollixed up when trying to solve it.

-Dan

Addendum: Is "Cohen" short for "Cohen-Tannoudji"? They are very good texts but I'm surprised someone is still using them. I was using them about 20 years ago!
 
Jun 2019
3
1
The passage in letter a) where \(\displaystyle U^{\dagger}\) (t, 0) in \(\displaystyle |\psi_{n-1}>\), is the n - 1 in the exponential comes from the commutator \(\displaystyle [a, a^{\dagger}] = 1 \) or are defined by the time evolution operator?

In c) I would have to apply the operator P in this way: \(\displaystyle | P | U^{\dagger} | x>\) to show the relation? Or something like: \(\displaystyle <x|U^{\dagger}|\psi_n>\)?
I thought of trying to apply commutator in |x> like \(\displaystyle [\overline{P}(t), U^{\dagger}] | x>\), but I block in this and don't know what use.
I have sketched some things but in c) I have no idea what to do and I have reviewed my classes notes and I have nothing as far as operator evolution applied in | x> or | p>.

Yes, "Cohen" is for "Cohen-Tannoudji".
 
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topsquark

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Apr 2008
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On the dance floor, baby!
I'm not giving up on this, but it is quite possible you will run out of time to turn it in before I can be done with it.

Basically this is an easy problem:
1) We have \(\displaystyle U^{\dagger}(T, 0) |x>\), which boils down to how to find H |x>.

2) Change the basis to energy eigenkets by inserting \(\displaystyle \sum _n | \psi _n >< \psi _n |\)

3) Evaulate \(\displaystyle H | \psi _n >\) and put it back in terms of \(\displaystyle U^{\dagger} (T, 0)\) by putting it back into the exponential.

4) Change the basis back to position eigenkets by inserting \(\displaystyle \int | x' >< x' | dx'\)

In theory this last step should give a delta function letting us do the integral by susbstituting x' with x, giving back the |x> ket.

But I can't make it work!! (I really should be able to do this. I'm missing something elementary so I might not find it quickly.)

If you are given the answer or find it on your own, please let me know. This is driving me crazy!

-Dan

(Addendum: If you like I can provide more details on the steps I listed. I can't do this right now but I can post it later if you like.
 

topsquark

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Apr 2008
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On the dance floor, baby!
Arrrggghh!

I didn't read the whole of part c). I was trying to show that \(\displaystyle U^{\dagger} | x >\) is an eigenstate of |x >, which it clearly cannot be. (I'm playing an idle game and realized that \(\displaystyle U^{\dagger} |x>\) can't be an eigenstate according to the properties of a SHO.)

I'll get back to you tomorrow.

-Dan