Quantum Optics

I've discovered a slight hole in my understanding of quantum physics. In all the basic courses on QM it was stressed that all observables must have a Hermitian operator in order to be physical.

However it seems that the coherent states of light are eigenstates of the non-Hermitian annihilation operator. But yet the coherent states is apparently observable (and corresponds the most to "classical" light).

I'm not entirely sure how to reconcile this. Any help would be appreciated.
Thanks

(edited 13 years ago)

Reply 1

TableChair

1

Anything you observe about the coherent state would be an observable that has a Hermitian operator. You couldn't simply measure whether it's in a coherent state or not as you would say measure the position or momentum of a particle. At least I think think that makes sense, I'm pretty tired.

Reply 2

suneilr

OP

3

Original post by TableChair

Anything you observe about the coherent state would be an observable that has a Hermitian operator. You couldn't simply measure whether it's in a coherent state or not as you would say measure the position or momentum of a particle. At least I think think that makes sense, I'm pretty tired.

Why do we define the notion of a coherent state then? Wouldn't it make more sense to define some other states that are eigenstates of a hermitian operator? I don't really understand what the coherent state represents.

Reply 3

Supermerp

2

Original post by suneilr

Why do we define the notion of a coherent state then? Wouldn't it make more sense to define some other states that are eigenstates of a hermitian operator? I don't really understand what the coherent state represents.

As you said, it represents the most classical quantum thing you can get. The reason it's classical is because if you have coherent states then all your operators in the Hamiltonian become just numbers, leaving you with the Hamiltonian you'd have written down for a purely classical system.

The really confusing thing about coherent states (over just any other superposition of states) is that they aren't number states. In fact, in some sense they're the opposite of number states. Number and phase are conjugate like position and momentum are. A coherent state is one with a definite phase. So they're used to describe quantum systems where everything is in phase like a laser or a superfluid (this is then also useful because it reduces the problem to an essentially classical one).

For a laser, the non-number-conservation is fine because (normally) the chemical potential for photons is zero and you can create/destroy them at will. For a superfluid of atoms, it's a bit weird because you can't really have a source/sink of atoms. The normal way people get round it is to say something like "If there are loads of atoms in the same state, changing the number by one doesn't really make a difference so it's all fine". Other people are more careful and do everything slightly differently (Leggett wrote a whole book that was quite careful. And is quite confusing).

Reply 4

suneilr

OP

3

Original post by Supermerp

As you said, it represents the most classical quantum thing you can get. The reason it's classical is because if you have coherent states then all your operators in the Hamiltonian become just numbers, leaving you with the Hamiltonian you'd have written down for a purely classical system.

Sorry can you explain this bit please? Even if you use the number states states |n> your Hamiltonian has the form

\Sigma \hbar\omega (n + \frac{1}{2})

which is just numbers and not operators.

If you try looking at with coherent states you get

\hat{H}|\alpha> = \Sigma_{\lambda} \hbar\omega_{\lambda} (\hat{a}^+\hat{a} + \frac{1}{2})|\alpha>

= \Sigma_{\lambda} \hbar\omega_{\lambda} (\alpha\hat{a}^+ + \frac{1}{2})|\alpha>

but then

\hat{a}^+|\alpha>

isn't defined as far am I'm aware, so I don't see how the coherent states turn the Hamiltonian into just numbers. Unless you look at the expectation value of the the Hamiltonian?

The really confusing thing about coherent states (over just any other superposition of states) is that they aren't number states. In fact, in some sense they're the opposite of number states. Number and phase are conjugate like position and momentum are. A coherent state is one with a definite phase. So they're used to describe quantum systems where everything is in phase like a laser or a superfluid (this is then also useful because it reduces the problem to an essentially classical one).

For a laser, the non-number-conservation is fine because (normally) the chemical potential for photons is zero and you can create/destroy them at will. For a superfluid of atoms, it's a bit weird because you can't really have a source/sink of atoms. The normal way people get round it is to say something like "If there are loads of atoms in the same state, changing the number by one doesn't really make a difference so it's all fine". Other people are more careful and do everything slightly differently (Leggett wrote a whole book that was quite careful. And is quite confusing).

I can kind of understand that the coherent states don't have a definite number, because it's supposed to represent "noise" in a classical EM field. I've completely forgotten what chemical potential is and a quick glance at wiki hasn't helped. As long as the energy/momentum of the system is conserved, does it matter that the particle number isn't definite?

Reply 5

Supermerp

2

Original post by suneilr

Sorry can you explain this bit please? Even if you use the number states states |n> your Hamiltonian has the form

\Sigma \hbar\omega (n + \frac{1}{2})

which is just numbers and not operators.

You're right but the coherent state turns an annihilation operator into a number in any situation, including a more complicated Hamiltonian where you have interactions with something else (like a two level system). Also n is quantized and the coherent state isn't.

\hat{a}^+|\alpha>

isn't defined as far am I'm aware

It is but it's complicated and I can't remember it (I think there's a derivative). As you suggested, you look at the expectation value.

As long as the energy/momentum of the system is conserved, does it matter that the particle number isn't definite?

Well in an abstract sense it doesn't matter but if you're describing a system of 10000 atoms, it's a bit weird if the thing you use to describe them has an arbitrary number of atoms.

Reply 6

suneilr

OP

3

Right that makes more sense. I think I'll post my quantum optics questions in this thread rather than creating a new thread for each question.

In my notes for the Fock/ number states it says that:

"<n|E|n> = 0 where E is the electric field. That is the expectation value of the electric field vanishes in a Fock state, although n photons are present. This is a rather unusual behaviour and show that Fock states behave differently than their classical counterparts."

I don't understand why this is strange. For the classical counterpart, wouldn't the expectation value of a sinusoidal electric field still be 0?

Reply 7

suneilr

OP

3

I'm having trouble understanding operator expansions in phase space/ wigner functions.

In my notes I have

Tr[\hat{D}(\alpha_1)\hat{D}(\alpha_2)] = Tr[e^{\alpha_1\hat{a}^+}e^{-\alpha_1^*\hat{a}}e^{-\alpha_2^*\hat{a}}e^{\alpha_2 \hat{a}^+}]e^{\frac{1}{2}(|\alpha_2|^2-|\alpha_1|^2)}

= Tr[e^{(\alpha_1+\alpha_2)\hat{a}^+}e^{-(\alpha_1^*+\alpha_2^*)\hat{a})}]e^{\frac{1}{2}(|\alpha_2|^2-|\alpha_1|^2)}

= \frac{1}{\pi} \int d^2\alpha <\alpha|e^{(\alpha_1+\alpha_2) \hat{a}^+}e^{-(\alpha_1^*+\alpha_2^*) \hat{a})}|\alpha>e^{\frac{1}{2}(|\alpha_2|^2-|\alpha_1|^2)}

I don't really follow this last line. I know that

\mathbb {I} = \frac{1}{\pi} \int d^\alpha |\alpha><\alpha|

but I don't understand how this has been used in that last line, since it looks like the order of the bras, and kets have been changed, and also sandwiched around an operator.
Any help would be much appreciated.
Thanks

Quantum Optics

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