# Working out harmonic oscillator operators at $L \rightarrow \infty$

## Homework Statement:

Given

$$A^{\mu} (x) = A^{\mu} _+ + A^{\mu} _-$$

Where

$$A^{\mu} _+ = \sum_{r = 0}^3 \sum_{\vec k \in \frac{2 \pi}{L} Z} \sqrt{\frac{\hbar c^2}{2V \omega_{\vec k}}} \epsilon_r^{\mu} (\vec k) a_r (\vec k) e^{-i \vec k \cdot \vec x}$$

$$A^{\mu} _- = \sum_{r = 0}^3 \sum_{\vec k \in \frac{2 \pi}{L} Z} \sqrt{\frac{\hbar c^2}{2V \omega_{\vec k}}} \epsilon_r^{\mu} (\vec k) a_r^{\dagger} (\vec k) e^{i \vec k \cdot \vec x}$$

$$V = L^3, \ \ \ \ \ \ \ \ \ \omega_{\vec k} = ck^0 = c|\vec k|$$

Where $a_r (\vec k)$ and $a_r^{\dagger} (\vec k)$ are the harmonic oscillator operators, which satisfy the following commutation relations

$$[a_r(\vec k), a_s(\vec k')] = [a_r^{\dagger}(\vec k), a_s^{\dagger}(\vec k')] = 0$$

$$[a_r(\vec k), a_s(\vec k')] = \rho_r \delta_{r,s} \delta_{\vec k, \vec k'}$$

a) Show that it is necessary to rescale the harmonic oscillator operators (as shown below) if we want to take the limit $L \rightarrow \infty$

$$a_r(\vec k) \rightarrow \tilde{a_r}(\vec k) = \sqrt{\frac{V}{(2\pi)^3}} a_r (\vec k), \ \ \ \ \tilde {a^{\dagger}_r}(\vec k) = (\tilde{a_r}(\vec k))^{\dagger}$$

b) Give the commutation relations satisfied by $\tilde{a_r}(\vec k)$ and $\tilde {a^{\dagger}_r}(\vec k)$ at $L \rightarrow \infty$.

c) Give the expressions for $A^{\mu} _+$ and $A^{\mu} _-$ at $L \rightarrow \infty$.

## Relevant Equations:

$$A^{\mu} (x) = A^{\mu} _+ + A^{\mu} _-$$
Let's go step by step

a)

We know that the harmonic oscillator operators are

$$a^{\dagger} = \frac{1}{\sqrt{2 \hbar m \omega}} ( -ip + m \omega q)$$

$$a= \frac{1}{\sqrt{2 \hbar m \omega}} (ip + m \omega q)$$

But these do not depend on $L$, so I guess these are not the expressions we want to work out...

My guess is that I should first find L-dependent expressions for the harmonic oscillator operators and then work out the limit.

I guess $A^{\mu} _+$ and $A^{\mu} _-$ are not the operators themselves.

But what are these expressions?

A hint would be much appreciated.

Thank you.

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nrqed
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Homework Statement:: Given

$$A^{\mu} (x) = A^{\mu} _+ + A^{\mu} _-$$

Where

$$A^{\mu} _+ = \sum_{r = 0}^3 \sum_{\vec k \in \frac{2 \pi}{L} Z} \sqrt{\frac{\hbar c^2}{2V \omega_{\vec k}}} \epsilon_r^{\mu} (\vec k) a_r (\vec k) e^{-i \vec k \cdot \vec x}$$

$$A^{\mu} _- = \sum_{r = 0}^3 \sum_{\vec k \in \frac{2 \pi}{L} Z} \sqrt{\frac{\hbar c^2}{2V \omega_{\vec k}}} \epsilon_r^{\mu} (\vec k) a_r^{\dagger} (\vec k) e^{i \vec k \cdot \vec x}$$

$$V = L^3, \ \ \ \ \ \ \ \ \ \omega_{\vec k} = ck^0 = c|\vec k|$$

Where $a_r (\vec k)$ and $a_r^{\dagger} (\vec k)$ are the harmonic oscillator operators, which satisfy the following commutation relations

$$[a_r(\vec k), a_s(\vec k')] = [a_r^{\dagger}(\vec k), a_s^{\dagger}(\vec k')] = 0$$

$$[a_r(\vec k), a_s(\vec k')] = \rho_r \delta_{r,s} \delta_{\vec k, \vec k'}$$

a) Show that it is necessary to rescale the harmonic oscillator operators (as shown below) if we want to take the limit $L \rightarrow \infty$

$$a_r(\vec k) \rightarrow \tilde{a_r}(\vec k) = \sqrt{\frac{V}{(2\pi)^3}} a_r (\vec k), \ \ \ \ \tilde {a^{\dagger}_r}(\vec k) = (\tilde{a_r}(\vec k))^{\dagger}$$

b) Give the commutation relations satisfied by $\tilde{a_r}(\vec k)$ and $\tilde {a^{\dagger}_r}(\vec k)$ at $L \rightarrow \infty$.

c) Give the expressions for $A^{\mu} _+$ and $A^{\mu} _-$ at $L \rightarrow \infty$.
Relevant Equations:: $$A^{\mu} (x) = A^{\mu} _+ + A^{\mu} _-$$

Let's go step by step

a)

We know that the harmonic oscillator operators are

$$a^{\dagger} = \frac{1}{\sqrt{2 \hbar m \omega}} ( -ip + m \omega q)$$

$$a= \frac{1}{\sqrt{2 \hbar m \omega}} (ip + m \omega q)$$

But these do not depend on $L$, so I guess these are not the expressions we want to work out...

My guess is that I should first find L-dependent expressions for the harmonic oscillator operators and then work out the limit.

I guess $A^{\mu} _+$ and $A^{\mu} _-$ are not the operators themselves.

But what are these expressions?

A hint would be much appreciated.

Thank you.
What is the meaning of $\rho_r$ in the commutation relation of the $a's$?

What is the meaning of $\rho_r$ in the commutation relation of the $a's$?
I assume that is defined in the usual way, as $\rho_0=-1$, $\rho_1=\rho_2=\rho_3=1$

I assume that is defined in the usual way, as $\rho_0=-1$, $\rho_1=\rho_2=\rho_3=1$
That's right.

Any hint for a) would be appreciated.

Thanks :)

strangerep
You're told to work with $A^\mu(x)$, i.e., in position space. Moreover, the implication of $V$ is that it's a harmonic oscillator restricted to a finite cubic volume of sidelength $L$. That restricts the allowable frequencies (modes) based on boundary conditions at the walls. A general solution is a linear combination of these modes, but each mode has an appropriate normalization constant.

Now, regarding part (a), what happens as $L\to\infty$? Does such normalization still make sense?

That restricts the allowable frequencies (modes) based on boundary conditions at the walls.
Actually I think that the periodic boundary conditions should look like $x' = x + L$ , $y' = y + L$ and $z' = z + L$

Does such normalization still make sense?
I would say it does not. I'd say we have to rescale the oscillator operators so that $A^{\mu} _+$ and $A^{\mu} _-$ have physical sense (i.e. they do not blow up).

I came to such a conclusion due to some reading (QFT; Mandl and Shaw):

While using a finite normalization volume $V$, we should be summing over a group of allowed wave vectors $\vec k$, where $\vec k = \frac{2 \pi}{L}(n_1, n_2, n_3), \ \ \ \ n_1, n_2, n_3 = 0, \pm 1,...$

For $V \rightarrow \infty$ we have

$$\frac{1}{V} \sum_{k} \rightarrow\frac{1}{(2 \pi)^3} \int d^3 \vec k$$

The normalization volume $V$ must then drop out of all physically significant quantities, such as transition rates.

Alright I think I understand what we're doing at a). Still stuck in how to show the Math though...

strangerep