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

#### Isabela

If anyone can help me, I'm very grateful.

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#### topsquark

Forum Staff
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

1 person

#### Isabela

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

Forum Staff
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!

#### Isabela

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

Forum Staff
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

Forum Staff
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