# Taking up QFT for the first time

1. Sep 4, 2006

### aav

I'm taking up QFT for the first time. Basic question really. Given the free complex Klein-Gordon field $$\phi$$ under the global gauge transformation $$exp(\imath\alpha)$$ where $$\alpha$$ is some real parameter, the Noether current is given by
$$j^{\mu}=\imath \left( \phi \partial^{\mu} \phi\dagger - \phi\dagger \partial^{\mu} \phi \right)$$
Now, the conserved charge is Q, given by
$$Q = \int d^3x j^0 = -\imath \int d^3x \left( \pi \phi - \pi\dagger\phi\dagger \right)$$
where $$\pi$$ is the conjugate field momentum.

Question is: if I want to express Q in terms of the ladder operators
$$\hat{a}$$, $$\hat{a}\dagger$$, $$\hat{b}$$ and $$\hat{b}\dagger$$, do I need to symmetrize the field operators $$\hat{\phi}$$ and $$\hat{\pi}$$, since they do not commute?

I searched the usual books but there seems to be no comment on the subject. I performed the calculation for $$\hat{Q}$$ using symmetrized field operators and obtained the usual result, but perhaps it was unnecessary?

Last edited: Sep 4, 2006
2. Sep 6, 2006

### dextercioby

It's a lot easier if you compute the classical expression for Q using the complex amplitudes a,a*,b,b* and then, when doing the quantization, do the necessary symmetrizations required by the quantization procedure.

So i guess it was unnecessary to do it on phi & pi. It's a lot clearer if it's done on the classical expression of Q.

Daniel.

3. Sep 6, 2006

### simic4

your expression for Q is not correct, since its not hermitian. you need to change the ordering of the pi and phi in one of your terms.

then just plug in the quatnum expression of the phi and pi (in terms of creation operators) into that expression.

4. Sep 6, 2006

### simic4

you can then verify that this IS a charge operator by considering a any definite particle state and noticing that it is an eigenstate of Q with eigenvalue of the form N-M (particles - antiparticles).

The second thing you should check is the Q and H commute.

You also might want to check that Q generates the symmetry transformation that gave you the conserved quantity Q in the first place.

then you will have proved beyond doubt that Q is the conservd charge operator, and that the symmetry of the classical field theory implies a symmetry of the quantum system.

A general thing to keep in mind is that: Noether's theorem is a CLASSICAL result. it does not ALWAYS hold up in the quantum version as quantum effects can break the symmetry. when you study path integrals you will understand this better (the path integral measure is sometimes not invariant under the symmetry in a way which can have physical consequences)

Last edited: Sep 6, 2006
5. Sep 7, 2006

### aav

Thanks for the assist

Yes, that was my mistake. Either version
$$j^0=-\imath\left(\pi\phi-\phi^{\dagger}\pi^{\dagger}\right)$$
$$j^0=-\imath\left(\phi\pi-\pi^{\dagger}\phi^{\dagger}\right)$$
gives the desired result.

I also overlooked Greiner's expression eqn. (4.67) giving the charge in terms of normally ordered operators
$$\hat{Q} = \imath \int d^3x : \phi^{\dagger}\pi^{\dagger}-\pi\phi :$$
(I guess I should stop eyeing the girl from the condensed matter lab who sits next to me and focus more on the book)

Hm. The field Hamiltonian is also expressible in terms of $$N_a$$ and $$N_b$$, so H commutes with Q, so they have a common set of eigenstates, namely the particle number states.

Okay, if $$\mid q \rangle$$ is an eigenstate of Q then $$e^{\imath\hat{Q}}\mid q \rangle=e^{\imath q}\mid q \rangle$$. Assuming a state can be expanded in terms of eigenstates of Q, we can generalize that statement accordingly, in which case $$e^{\imath q'}$$ is the global phase shift, and q' the total charge from all the contributing eigenstates (times some proportionality constant, I suppose). Is this correct?

Last edited: Sep 7, 2006
6. Sep 7, 2006

Staff Emeritus
Why don't ask HER to check your results? Could be the start of something beautiful :)

7. Sep 7, 2006