Klein-Gordon Momentum Question

  • #1

Main Question or Discussion Point

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

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

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

[tex] \displaystyle P = \sum_{n} k_n a_n^{\dagger} a_n [/tex] "

The KG field is given by: [tex] \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] [/tex]

The following relations are true:
[tex] \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 [/tex]

and [tex] E_n^2 = k_n^2 + m^2 [/tex]

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

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

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

[tex] \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][/tex]

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

Any help would be greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
2,111
16
How did you deal with the time derivative [tex]\frac{\partial\phi}{\partial t}[/tex]? You should first solve the classical momentum to be a function of [tex]\phi[/tex], the canonical momenta field [tex]\Pi[/tex], and possibly their spatial derivatives but not time derivatives.

You already have the operator [tex]\phi[/tex] written in terms of [tex]a_n[/tex] and [tex]a_n^{\dagger}[/tex]. (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 [tex]\Pi[/tex]. You get the momentum operator in terms of [tex]a_n[/tex] and [tex]a_n^{\dagger}[/tex] when you substitute the formulas for operators [tex]\phi[/tex] and [tex]\Pi[/tex].
 
  • #3
I worked the time derivative out to be:

[tex] \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] [/tex]

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.
 
  • #4
2,111
16
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

[tex]
P=-\int dx\; \Pi(x)\nabla\phi(x)
[/tex]

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

You don't struggle with quantization of quantity [tex]\dot{x}[/tex] in single particle QM either. Instead the quantities are written as functions of position [tex]x[/tex] and canonical momenta [tex]p[/tex], and the position and the canonical momenta are promoted to become operators then.
 
  • #5
410
0
Whether you use [tex]\Pi[/tex] or [tex]\dot\phi[/tex] 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 [tex]\phi[/tex] as a single sum (after getting rid of time dependence), by replacing [tex]a_{n}[/tex] with [tex]a_{-n}[/tex] and factoring out the exponential.

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

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