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

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 Jun 11th 2019, 09:38 PM #1 Junior Member   Join Date: Jun 2019 Posts: 2 I can not solve this quantum harmonic oscillator question in Cohen's book If anyone can help me, I'm very grateful. Attached Thumbnails
 Jun 12th 2019, 06:12 PM #2 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,730 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 Isabela likes this. __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Jun 13th 2019, 06:23 PM #3 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,730 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! __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Jun 13th 2019, 10:26 PM #4 Junior Member   Join Date: Jun 2019 Posts: 2 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: $\displaystyle$? 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". topsquark likes this. Last edited by Isabela; Jun 13th 2019 at 10:31 PM.
 Jun 16th 2019, 01:41 PM #5 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,730 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. __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Jun 16th 2019, 10:02 PM #6 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,730 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 __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.

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