Quantum Physics Quantum Physics Help Forum  2Likes  1 Post By topsquark
 1 Post By Isabela
Jun 11th 2019, 09:38 PM

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 I can not solve this quantum harmonic oscillator question in Cohen's book
If anyone can help me, I'm very grateful. 
 
Jun 12th 2019, 06:12 PM

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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 13th 2019, 06:23 PM

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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 deadend.
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 "CohenTannoudji"? They are very good texts but I'm surprised someone is still using them. I was using them about 20 years ago!
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Jun 13th 2019, 10:26 PM

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The passage in letter a) where $\displaystyle U^{\dagger}$ (t, 0) in $\displaystyle \psi_{n1}>$, 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: $\displaystyle <xU^{\dagger}\psi_n>$?
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 "CohenTannoudji".
Last edited by Isabela; Jun 13th 2019 at 10:31 PM.

 
Jun 16th 2019, 01:41 PM

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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.
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Jun 16th 2019, 10:02 PM

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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
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