Klein-Gordon Momentum Question

[div curl F= 0]

Dear all, I'd be very grateful for some help on this question:

"The momentum operator is defined by: \displaystyle P = - \int_{0}^{L} dz \left(\frac{\partial \phi}{\partial t}\right) \left( \frac{\partial \phi}{\partial z} \right)

Show that P can be written in terms of the operators a_n and a^{\dagger}_n as:

\displaystyle P = \sum_{n} k_n a_n^{\dagger} a_n "

The KG field is given by: \displaystyle \phi(t,z) = \sum_{n} \frac{1}{\sqrt{2E_n L}} \left[a_n e^{-i(E_n t - k_n z)} + a^{\dagger}_n e^{+i(E_n t - k_n z)} \right]

The following relations are true:
\displaystyle k_n = \frac{2 \pi n}{L} \;;\; \left[a_n, a_m^{\dagger} \right] = \delta_{nm} \;;\; \left[a_n, a_m \right] = \left[a_n^{\dagger}, a_m^{\dagger} \right] = 0

and E_n^2 = k_n^2 + m^2

\displaystyle \int_{0}^{L} dz e^{iz(k_n-k_m)} = L \delta_{nm}

----------------------

I've fed all this information into the definition of the momentum operator and have the result:

\displaystyle P = \sum_{n} \frac{k_n}{2} \left[1 + 2 a_n^{\dagger} a_n - a_n a_{-n} - a^{\dagger}_n a^{\dagger}_{-n} \right]

but I am unsure of how to reduce this down even further.

Any help would be greatly appreciated.

 

[jostpuur]

How did you deal with the time derivative \frac{\partial\phi}{\partial t}? You should first solve the classical momentum to be a function of \phi, the canonical momenta field \Pi, and possibly their spatial derivatives but not time derivatives.

You already have the operator \phi written in terms of a_n and a_n^{\dagger}. (I don't think the Heisenberg picture time evolution is relevant now. Operators with fixed time should be enough.) There exists similar formula for the operator \Pi. You get the momentum operator in terms of a_n and a_n^{\dagger} when you substitute the formulas for operators \phi and \Pi.

 

[div curl F= 0]

I worked the time derivative out to be:

\displaystyle \frac{\partial \phi}{\partial t} = - i \sum_n \sqrt{\frac{E_n}{2L}} \left[ a_n e^{-i(E_n t - k_n z)} - a_n^{\dagger} e^{+i(E_n t - k_n z)} \right]

Whilst integrating the whole expression I set t = 0 to remove the time dependence (as seems to be the trick played in a few books I've read) and used the orthogonality between the complex exponentials to arrive at were I'm stuck.

 

[jostpuur]

I'm not sure what happens when you compute time derivatives of the operators in Heisenberg picture like that. It could be that you can get correct answer, but I wouldn't count on it. If it gives the correct result, IMO it should be proven as a result of its own. Anyway, I would suggest using the classical formula

<br /> P=-\int dx\; \Pi(x)\nabla\phi(x)<br />

for momentum in start. You get the momentum operator when you promote \phi and \Pi to operators.

You don't struggle with quantization of quantity \dot{x} in single particle QM either. Instead the quantities are written as functions of position x and canonical momenta p, and the position and the canonical momenta are promoted to become operators then.

 

[lbrits]

Whether you use \Pi or \dot\phi I don't think it'll affect the Hamiltonian that you calculate, so I think your method is correct. I don't know if you've made any mistakes getting to that point, however, another useful trick is to rewrite the sums in \phi as a single sum (after getting rid of time dependence), by replacing a_{n} with a_{-n} and factoring out the exponential.

Also, for \phi to be real, I recall that a_{-n} = a^\dagger_n, but you'll have to check this, as I'm prone to talking out of my ass =)