Derive expression for Klein-Gordon annihilation operator

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[/tex]1. Homework Statement [/b]

Given: [tex]\phi(x)=\phi^+(x)+\phi^-(x)[/tex]
Where:
[tex]\phi^+(x)=\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]
and
[tex]\phi^-(x)=\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx}[/tex]

Show that:

[tex]a(\mathbf{k})=\frac{1}{\sqrt{2 \hbar c^2 V \omega_{\mathbf{k}}}} \int\mathrm{d}^3 \mathbf{x} \mathrm{e}^{\mathrm{i}kx} (\mathrm{i}\dot{\phi}(x)+\omega_{\mathbf{k}} \phi(x))[/tex]

Homework Equations



See above

The Attempt at a Solution



First differentiate [tex]\phi(x)[/tex] w.r.t. time:

[tex]\dot{\phi}(x) = -\mathrm{i} \omega_{\mathbf{k}} \phi^+(x) + \mathrm{i} \omega_{\mathbf{k}} \phi^-(x) = -\mathrm{i} \omega_{\mathbf{k}} (\phi^+(x)-\phi^-(x))[/tex]

Then add [tex]\mathrm{i}\dot{\phi}(x)[/tex] to [tex]\omega_\mathbf{k}\phi(x)[/tex] to give:

[tex]\omega_{\mathbf{k}} \phi(x)+\mathrm{i} \dot{\phi}(x) = 2 \omega_{\mathbf{k}} \sum_{\mathbf{k}} \sqrt{ \frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx} = \sum_{\mathbf{k}} \sqrt{\frac{2 \omega_{\mathbf{k}} \hbar c^2}{V}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

So:
[tex]\omega_{\mathbf{k}} \phi(x)+\mathrm{i} \dot{\phi}(x) = \sum_{\mathbf{k}} \sqrt{\frac{2 \omega_{\mathbf{k}} \hbar c^2}{V}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}[/tex]

And then I get stuck, because I have no idea how to get rid of the summation sign. I'm not certain of what happens to the V either, though I suspect it has something to do with the integral in some way.

Does anyone have any hints?
 
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Answers and Replies

  • #2
fzero
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First differentiate [tex]\phi(x)[/tex] w.r.t. time:

[tex]\dot{\phi}(x) = -\mathrm{i} \omega_{\mathbf{k}} \phi^+(x) + \mathrm{i} \omega_{\mathbf{k}} \phi^-(x) = -\mathrm{i} \omega_{\mathbf{k}} (\phi^+(x)-\phi^-(x))[/tex]
You've pulled the factors of [tex]\omega_{\mathbf{k}}[/tex] out of the summation here, which is incorrect. You can only write that

[tex]\dot{\phi}(x) =\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} \left( -\mathrm{i} \omega_{\mathbf{k}} a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}
+ \mathrm{i} \omega_{\mathbf{k}} a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx}\right) .[/tex]

And then I get stuck, because I have no idea how to get rid of the summation sign. I'm not certain of what happens to the V either, though I suspect it has something to do with the integral in some way.
The integration is normalized by the volume.
 
  • #3
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I'm a little unsure about the term [tex]\omega_{\mathbf{k}} \phi(x)[/tex]. Can the omega here be treated as a constant and be brought into the summation? Say by denoting it with k' instead of k?

I'm just a little unsure of how to proceed. I'm tempted to try and construct the RHS of the solution, expand the fields in summation form, then evaluate it to give the LHS, but I'm a bit unsure of how to manipulate the summations, especially when it comes to applying that integral.
 
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  • #4
fzero
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I'm a little unsure about the term [tex]\omega_{\mathbf{k}} \phi(x)[/tex]. Can the omega here be treated as a constant and be brought into the summation? Say by denoting it with k' instead of k?

I'm just a little unsure of how to proceed. I'm tempted to try and construct the RHS of the solution, expand the fields in summation form, then evaluate it to give the LHS, but I'm a bit unsure of how to manipulate the summations, especially when it comes to applying that integral.
You're not going to have to do anything fishy. Just remember that

[tex]\frac{1}{V} \int d^3x e^{ikx} e^{i k' x} f(k) \sim \delta^3(\mathbf{k}+\mathbf{k}') f(k) [/tex]

up to some factor that you will probably want to check in your notes for. Examine the expressions that you have for [tex]\phi[/tex] and [tex]\dot{\phi}[/tex] and determine an appropriate integral to use to compute linear combinations of [tex]a(\mathbf{k})[/tex] and [tex]a^\dagger(\mathbf{k})[/tex].
 
  • #5
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I'm self-teaching from Mandl and Shaw, and haven't seen anything like that yet :-S. I'm basically attempting exercise 3.1, from chapter 3 on the Klein Gordon field.
 
  • #6
fzero
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I'm self-teaching from Mandl and Shaw, and haven't seen anything like that yet :-S. I'm basically attempting exercise 3.1, from chapter 3 on the Klein Gordon field.
In their notation

[tex]
\int d^3x e^{ikx} e^{-i k' x} = \delta_{\mathbf{k}\mathbf{k}'}.
[/tex]

I can't see every page on Google Books, so I don't know if they explicitly introduce this around pages 10-11, but they did use it to derive equation 3.14 from 3.12. This type of integral is standard for Fourier transforms and you would have seen the 1d version in any quantum mechanics course.
 
  • #7
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Except mine apparently. I did Fourier transforms but the course skipped over delta functions. Thanks for the help. I'll read around a bit more, have another stab at it and let you know if I run into any more problems.
 
  • #8
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Right, I've had another go at this. Here's my working so far.

First I'll just outline the Fourier transform I'll be using (in one dimension).

[tex]\int^\infty_{-\infty} \mathrm{e}^{\mathrm{i}a x} \mathrm{e}^{-\mathrm{i}w x}\mathrm{d}x=2\pi \delta(w-a)[/tex]
And the inverse:
[tex]\int^\infty_{-\infty} \mathrm{e}^{\mathrm{i}a w} \mathrm{e}^{\mathrm{i}w x}\mathrm{d}w=2\pi \delta(x+a)[/tex]

From before:

[tex]
\dot{\phi}(x) =-\mathrm{i} \sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2 \omega_{\mathbf{k}}}{2 V}} \left( a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}
- a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx}\right)
[/tex]

and

[tex]
\dot{\phi}(x) =\sum_{\mathbf{k}} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} \left( a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx}
+ a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx}\right)
[/tex]

I first integrate [tex]\mathrm{i} \dot{\phi}(x) \mathrm{e}^{\mathrm{i} k' x}[/tex]:

[tex]\sum_{\mathbf{k}} \int \mathrm{d}^3 \mathbf{x} \sqrt{\frac{\hbar c^2 \omega_{\mathbf{k}}}{2 V}} \left( a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx} \mathrm{e}^{\mathrm{i} k' x} - a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx} \mathrm{e}^{\mathrm{i} k' x}\right) = \pi \sum_{\mathbf{k}} \sqrt{\frac{2 \hbar c^2 \omega_{\mathbf{k}}}{V}} \left( a(\mathbf{k}) \delta(k-k') - a^\dagger (\mathbf{k}) \delta(k+k')\right)[/tex]

Then [tex]\omega_{\mathbf{k}} \phi (x)[/tex]:

[tex]\omega_{\mathbf{k}} \sum_{\mathbf{k}} \int \mathrm{d}^3 \mathbf{x} \sqrt{\frac{\hbar c^2}{2 V \omega_{\mathbf{k}}}} \left( a(\mathbf{k}) \mathrm{e}^{-\mathrm{i}kx} \mathrm{e}^{\mathrm{i} k' x} + a^\dagger (\mathbf{k}) \mathrm{e}^{\mathrm{i}kx} \mathrm{e}^{\mathrm{i} k' x}\right) = \pi \omega_{\mathbf{k}} \sum_{\mathbf{k}} \sqrt{\frac{2 \hbar c^2 }{V \omega_{\mathbf{k}}}} \left( a(\mathbf{k}) \delta(k-k') + a^\dagger (\mathbf{k}) \delta(k+k')\right)[/tex]

I'm a little confused about what to do now. Those delta functions could be troublesome, but on the other hand, they could be used to remove the creation operator terms in some way. There's also that omega in front of the latter of the two integrals, which is still confusing me, and I'm not sure how to get around that. I'm tempted to treat the summation as a Riemann sum in some way, with a limiting case, but still I'm not sure how I'd do this, as I'm not multiplying by a [tex]\Delta k[/tex] anywhere, but then again an integral would deal with the delta functions quite nicely.
 
  • #9
fzero
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They've chosen to treat the sum over momenta as a discrete sum, so you should use Kronecker deltas instead of Dirac deltas. I probably confused you, but I followed up with the correct formula later. The deltas therefore pick out one term in the sum.

However, you are summing over k in the expressions for [tex]\phi[/tex], so you should really be considering [tex]\omega_{\mathbf{k}'} \phi[/tex] since you're going to get a very confusing result when you evaluate the sums otherwise.

Also the factors of [tex]2\pi[/tex] don't seem to match the conventions of the book. It doesn't change the physical interpretation, but will give you problems matching numerical factors.
 
  • #10
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Awesome. That's really helpful! I feel I have enough information to progress now. Thanks a lot!

Actually one last thing. How would you represent the second of the Dirac deltas (with the plus in) in Kronecker delta format? Or would it be eliminated in the integral by default?

EDIT: actually how about [tex]\delta_{k(-k')}[/tex]?
 
  • #11
fzero
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Awesome. That's really helpful! I feel I have enough information to progress now. Thanks a lot!

Actually one last thing. How would you represent the second of the Dirac deltas (with the plus in) in Kronecker delta format? Or would it be eliminated in the integral by default?

EDIT: actually how about [tex]\delta_{k(-k')}[/tex]?
That's fine. Sometimes you'll see the notation [tex]\delta_{k+k',0}[/tex].
 
  • #12
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Okay, I've pretty much arrived at the answer, except I still need to remove a factor of V somehow. After I integrate I get this:

[tex]\omega_{\mathbf{k'}} \sum_{\mathbf{k}} \sqrt{\frac{2 \hbar c^2 }{V \omega_{\mathbf{k}}}} \left( a(\mathbf{k}) \delta_{kk'} + a^\dagger (\mathbf{k}) \delta_{k(-k')}\right)=\sqrt{\frac{2 \hbar c^2 \omega_{\mathbf{k'}}}{V}} a(\mathbf{k'})[/tex]

But I don't know what to do about that V, which according to the answer needs to be on top.
 
  • #13
fzero
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I thought I checked those factors, but I might have made a mistake. Their convention might be that

[tex]

\frac{1}{V} \int d^3x e^{ikx} e^{-i k' x} = \delta_{\mathbf{k}\mathbf{k}'},

[/tex]

which would be in better agreement with the formulas you were using that had [tex]2\pi[/tex].
 
  • #14
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Ah therein would lie part of my confusion: this isn't stated formally in the book. All sorted now.

Thanks for humouring me on this one.
 

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