Coherent states of a Quantum Harmonic Oscillator

[Piano man]

Homework Statement

Given that
$$a^+|n\rangle=\sqrt{n+1}|n+1\rangle$$
$$a|n\rangle=\sqrt{n}|n-1\rangle$$
and that the other eigenstates |n> are given by
$$|n\rangle=\frac{(a^+)^n}{\sqrt{n!}}|0\rangle$$
where |0> is the lowest eigenstate.
Define for each complex number z the coherent state
$$|z\rangle=e^{-\frac{|z|^2}{2}}\sum^\infty_{n=0}\frac{z^n}{\sqrt{n!}}|n\rangle$$

Q. Show that |z> is an eigenstate of a:
$$a|z\rangle=z|z\rangle (\text{Hint: use } |n=-1\rangle=0)$$


2. The attempt at a solution

I've tried subbing in a few things but have got nothing substantive.
I would really appreciate if someone could point me in the right direction.

 

[G01]

So, you are told how the a operator acts on the state |n>. Using this info, if you explicitly work out a|z>, what do you get? I need to see some work to see where you're stuck.

 

[Piano man]

okay well I'm not sure if I should go back to the definition of a, which was

a=\frac{ip+mwx}{\sqrt{2m\hbar w}}

or apply

a|n>=\sqrt{n}|n-1> giving

\sqrt{n}e^{\frac{-|z-1|^2}{2}}\sum^\infty_{n=0}\frac{(z-1)^n}{\sqrt{n!}}|n>

but neither approach is yielding anything useful.

(I'm sorry about the formatting of this, but the tex commands aren't coding properly for whatever reason.)

 

[G01]

Maybe I can try posting the latex? :

$$a=\frac{ip+mwx}{\sqrt{2m\hbar w}}$$

$$a|n>=\sqrt{n}|n-1>$$

$$\sqrt{n}e^{\frac{-|z-1|^2}{2}}\sum^\infty_{n=0}\frac{(z-1)^n}{\sqrt{n!}}|n>$$

 

[G01]

OK. Your approach in writing equation 3 is correct, but you have one important error:

Look at equation 2: $$a|n>=\sqrt{n}|n-1>$$

Now if you apply the above to equation 3: $|n>\rightarrow|n-1>$ not $z^n\rightarrow(z-1)^n$

The operator a, does not affect the complex number $z^n$. After you change this, see how far you can get.

HINT: After you simplify you'll have to change the limits on the sum to get a|z> to look like |z>. It is in this step that you make use of the hint they gave you.

 

[Piano man]

right, so
a|z>=\sqrt{n}e^{\frac{-|z|^2}{2}}\sum^\infty_{n=0}\frac{z^n}{\sqrt{n!}}|n-1>

(again, sorry about the display, for some reason every time I use the $$wraps it just gives the first equation I wrote in the OP)$$

 

[Piano man]

$$a|z>=\sqrt{n}e^{\frac{-|z|^2}{2}}\sum^\infty_{n=0}\frac{z^n}{\sqrt{n!}}|n-1>$$

edit: okay it's working now. I think it's just the preview post that isn't working.

so where can I go from here?

 

[Piano man]

Ah! I got it, thanks a million for your help :D

 

[G01]

Ah! I got it, thanks a million for your help :D

Sorry, I haven't responded! Been at a conference all day today. Glad you figured it out!