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 $$E_n^2 = k_n^2 + m^2$$

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

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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

$$P=-\int dx\; \Pi(x)\nabla\phi(x)$$

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 =)