# Displacement operator for coherent states?

1. Jun 28, 2010

### Orbb

Hi everyone,

I have the following question about coherent states: It is known that the creation operator has no eigenket. However, the action of a creation operator $$a^{\dagger}$$ on a coherent ket $$|\alpha\rangle$$ can be written as

$$a^{\dagger}|\alpha\rangle = \left( \frac{\partial}{\partial \alpha} + \frac{\alpha^*}{2}\right)|\alpha\rangle.$$

My question now concerns

$$e^{\lambda a^{\dagger}} |\alpha\rangle} = e^{\lambda a^*/2}e^{\lambda\partial_{\alpha}}|\alpha\rangle,$$

which follows from the equation above. I wish to find an explicit expression for that one. It came to my mind that there may be an analogy with the displacement operator acting on position eigenstates

$$e^{\lambda \partial_x}|x\rangle = |x+\lambda\rangle.$$

But does it hold? Or is there another way? Writing down the explicit form of a coherent state doesn't help me much because this way I can't get rid of the creation operator in the exponential.

Thank you very much for any thoughts!

2. Jun 29, 2010

### strangerep

I'm not sure what you're getting at here. A "coherent ket" is already the result of
acting on the vacuum by a generalized displacement operator:
$$|\alpha\rangle ~:=~ e^{\alpha\, a^\dagger} \; |0\rangle ~.$$
You're probably more used to seeing something like
$$e^{\alpha\, a^\dagger - \bar\alpha\, a} \; |0\rangle ~$$
instead, but in this case they turn out to be the same except for a
normalization factor involving $\exp(-|\alpha|^2)$ .

Try something like this:
$$e^{\lambda a^{\dagger}} |\alpha\rangle} ~=~ e^{\lambda a^{\dagger}} \, e^{\alpha\, a^\dagger} \; |0\rangle ~=~ e^{(\lambda + \alpha) a^{\dagger}} \; |0\rangle ~=~ |\lambda + \alpha\rangle} ~,$$

(Again, if you want normalized states you'll need to include something similar
to the gaussian factor I mentioned earlier.)

Indeed, there is a very deep analogy. In both cases you're acting on states
with a group element: in the position-displacement case it's the boring old
translation group, whereas for the coherent states it's the Heisenberg group.

These coherent states are perhaps the oldest and simplest such states, sometimes
named after Glauber who studied them extensively in the context of coherent
states of the EM field.

But the concept is much more general. Given a Hamiltonian, one can usually
find a dynamical group which is stable under the action of the Hamiltonian.
One can then construct generalized coherent states by acting on the vacuum
with (representations of) this group. The states so-constructed form a subspace
which is stable under the Hamiltonian. (I.e., the Hamiltonian transforms these
states amongst themselves). There's a textbook by Perelomov that gives the
general construction. Also a fairly recent book by Gazeau. A classic review
paper is this one:

W-M Zhang, D. H. Feng, R.Gilmore,
"Coherent states: Theory and some applications",
Rev. Mod. Phys., Vol. 62, No. 4, Oct 1990, pp 867-927.

The quantum optics "bible" of Mandel & Wolf also has extensive material
on coherent states.

3. Jun 29, 2010

### Orbb

This is actually what I was looking for. I feel a little dumb because it's actually easy to see . Thank you, strangerep, also for giving some more background on this!

4. Nov 21, 2010

Sorry for digging up this topic, but can anyone please explain how can I get
$$|\alpha\rangle ~:=~ e^{\alpha\, a^\dagger} \; |0\rangle ~.$$

5. Nov 21, 2010

Hi,

I am interested in showing the normalization constant of the coherent state $$|\alpha>$$ without using the knowledge about Fock states (although I don't know if it's possible).

So $$|\alpha> = ce^{\alpha a^\dagger}$$

and $$<\alpha|\alpha> = c^2 e^{(\alpha a^\dagger)^\dagger} e^{\alpha a^\dagger} = c^2 e^{a \alpha^\star} e^{\alpha a^\dagger} = 1$$

But then I don't know where to go.

Did I make a mistake, or how can I show that
$$e^{a \alpha^\star} e^{\alpha a^\dagger} = e^{2\alpha}$$

Thanks,

6. Nov 21, 2010

### strangerep

I'm not sure what you mean by "get". What I wrote was a definition.
(That's what the symbol ":=" means. :-)

7. Nov 21, 2010

### strangerep

That's mathematical nonsense. The LHS is a state, but the RHS is an operator.

Yes, you made a mistake. If you just want to combine a product of exponentials
of noncommuting quantities, you'll need to use the BCH (Baker-Campbell-Hausdorff)
formula. Try Wiki for more detail. (In the case of the Heisenberg algebra that
you're working with here, the formula is much easier because it terminates at the
first commutator.)

(If you can't get enough info on Wiki, search this forum for other references
to the BCH formula. If you still can't get it, let me know and I'll try to find time
to post something better.)

8. Nov 22, 2010

Thank you for helping. I was careless when typing my last post. I read through wiki about BCH but only the last part of the page mentioned about annihilation operators, but then the explanation is not quite clear. So in short, is there a formula to find $$<\alpha|$$ if knowing that $$|\alpha> = e^{\alpha a^\dagger}|0>$$?

9. Nov 22, 2010

### strangerep

Sure. Just take the Hermitian conjugate of the exponential (i.e., expand as
a formal Taylor series and take the Hermitian conjugate term by term).
Then apply to the vacuum bra. Result:

$$\langle \alpha| ~=~ \langle 0| \, e^{\bar{\alpha} a}$$

10. Nov 22, 2010

Yes I think I get that from the previous post.

So the coherent state $$|\alpha> = ce^{\alpha a^\dagger} |0>$$.

Now I want to show that the normalized constant c is $$e^{-\frac{\alpha^2}{2}}$$

So I do the dot product

$$<\alpha|\alpha> = c^2 e^{a \bar{\alpha}}e^{\alpha a^\dagger}$$

We consider
$$e^{a \bar{\alpha}}e^{\alpha a^\dagger}$$

1) Should there be a negative sign? the wiki page of BCH mentioned $$e^{\alpha a^\dagger}e^{-a \bar{\alpha}}$$

2) I understand that since $$a$$ is not hermitian,
$$e^{\alpha a^\dagger}e^{-a \bar{\alpha}}=e^{\alpha a^\dagger-a \bar{\alpha}} \times C$$

but again finding C may not be trivial for me.

11. Nov 22, 2010

### strangerep

You just made the same mistake I mentioned earlier. Your LHS is a scalar but the
RHS is an operator. Where did your vacuum bra and ket go??

OK. I'll put you out of your misery. It's really quite easy.
(I'll change the notation though, because I'm lifting this from
one of my private workfiles and I can't be bothered changing it too much... :-)

Defining a coherent state as $$|z\rangle := e^{\overline z a^*} |0\rangle$$,
we want to calculate $$\langle z' | z\rangle$$ .
The commutation relation is $$[a,a^*] = \hbar$$ .

We use the well-known Baker-Campbell-Hausdorff (BCH) formula:

$$e^{A+B} ~=~ e^A \, e^B \, e^{-[A,B]/2} ~,$$
(if [A,B] commutes with A and B), which implies
$$e^A \, e^B ~=~ e^B\,e^A \, e^{[A,B]} ~.$$

Noting also that $$e^{za}|0\rangle = |0\rangle$$, we find

$$\langle z' | z\rangle ~=~ \langle 0 | e^{z'a} e^{\overline z a^*}|0\rangle ~=~ \langle 0| e^{\overline z a^*}e^{z'a} e^{[z'a,\, \overline z a^*]} |0\rangle ~=~ e^{\hbar z' \overline z} \, \langle 0|e^{\overline z a^*}e^{z'a}|0\rangle ~=~ e^{\hbar z' \overline z} ~.$$

Last edited: Nov 22, 2010
12. Nov 22, 2010

Ya I am out of my misery. A tiny bit of problem was you wrote $$[a, a^\dagger]=\hbar$$