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

[JD_PM]

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.

 

[nrqed]

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 $a's$?

 

[Gaussian97]

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 $\rho_0=-1$, $\rho_1=\rho_2=\rho_3=1$

 

[JD_PM]

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

That's right.

 

[JD_PM]

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?

 

[JD_PM]

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.

 

[JD_PM]

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

 

[strangerep]

Try googling for "harmonic oscillator normalization video".

As for the math, try reviewing how the standard Riemann integral is defined, and then generalize to the 3D case.

 

[JD_PM]

Try googling for "harmonic oscillator normalization video".

Alright, so the idea now is to show how we normalize the harmonic oscillator in 1D and then see how can we generalize it for the 3D case (Source: introduction to Quantum Mechanics by D. Griffiths, second edition; note I name the equations as he does).

We know that

$$a^{\dagger} \psi_{n} = c_n \psi_{n + 1}, \ \ \ \ a \psi_n = d_n \psi_{n - 1} \ \ \ \ (2.63)$$

In words: we know that the annihilation/creation operators are proportional to $\psi_{n - 1} / \psi_{n + 1}$; where $c_n$ and $d_n$ are proportionality factors.

Given the following theorem:

$$\int_{-\infty}^{\infty} f^* (a^{\dagger} g) dx = \int_{-\infty}^{\infty} (a f)^* g dx \ \ \ \ (2.64.a)$$

$$\int_{-\infty}^{\infty} f^* (a g) dx = \int_{-\infty}^{\infty} (a^{\dagger} f)^* g dx \ \ \ \ (2.64.b)$$

Applying it to our problem we get

$$\int_{-\infty}^{\infty} (a^{\dagger} \psi_n)^* (a^{\dagger} \psi_n) dx = \int_{-\infty}^{\infty} (a a^{\dagger} \psi_n)^* \psi_n dx \ \ \ \ (2.64.c)$$

$$\int_{-\infty}^{\infty} (a \psi_n)^* (a \psi_n) dx = \int_{-\infty}^{\infty} (a^{\dagger} a \psi_n)^* \psi_n dx \ \ \ \ (2.64.d)$$

We also know that the Schrodinger equation for the 1D harmonic oscillator is

$$\hbar \omega \Big( a^{\dagger} a + \frac{1}{2} \Big) \psi = E \psi \ \ \ \ (2.57.a)$$

$$\hbar \omega \Big( a a^{\dagger} - \frac{1}{2} \Big) \psi = E \psi \ \ \ \ (2.57.b)$$

The excited states of the 1D harmonic oscillator are given by

$$\psi_n = A_n (a^{\dagger})^n \psi_0, \ \ \ \ E_n = \Big(n + \frac{1}{2} \Big) \hbar \omega \ \ \ \ (2.61)$$

By using equations $(2.57.a)$, $(2.57.b)$ and $(2.61)$ we get

$$a^{\dagger} a \psi_n = n \psi_n, \ \ \ \ a a^{\dagger} \psi_n = (n +1) \psi_n \ \ \ \ (2.65)$$

Combining $(2.63)$, $(2.64.a)$, $(2.64.b)$ and $(2.65)$ yields

$$\int_{-\infty}^{\infty} (a^{\dagger} \psi_n)^* (a^{\dagger} \psi_n) dx = |c_n|^2 \int_{-\infty}^{\infty} |\psi_{n +1}|^2 dx = (n+1) \int_{-\infty}^{\infty} |\psi_{n}|^2 dx$$

$$\int_{-\infty}^{\infty} (a \psi_n)^* (a \psi_n) dx = |d_n|^2 \int_{-\infty}^{\infty} |\psi_{n -1}|^2 dx = n \int_{-\infty}^{\infty} |\psi_{n}|^2 dx$$

Thus it follows that $|c_n|^2 = n+1$ and $|d_n|^2 = n$

Once we have found the coefficients we can simply plug them into $(2.63)$ to get

$$a^{\dagger} \psi_{n} = \sqrt{n+1} \psi_{n + 1}, \ \ \ \ a \psi_n = \sqrt{n} \psi_{n - 1} \ \ \ \ (2.63)$$

 

[JD_PM]

Thus it is clear that the normalization factor associated to the annihilation 1-D-HO operator is $\frac{1}{\sqrt{n}}$ and the normalization factor associated to the creation 1-D-HO operator is $\frac{1}{\sqrt{n + 1}}$.

However I still do not see how to generalize it to 3D to show a).

I will keep thinking about it.

 

[strangerep]

I sense you're making this harder than it needs to be.

You had
$$A^{\mu} _+ = \sum_{r = 0}^3 \sum_{\vec k \in \frac{2 \pi}{L} Z} \sqrt{\frac{\hbar c^2}{2V \omega_{\vec k}}} \epsilon_r^{\mu} (\vec k) a_r (\vec k) e^{-i \vec k \cdot \vec x}$$ and $$a_r(\vec k) \rightarrow \tilde{a_r}(\vec k) = \sqrt{\frac{V}{(2\pi)^3}} a_r (\vec k), \ \ \ \ \tilde {a^{\dagger}_r}(\vec k) = (\tilde{a_r}(\vec k))^{\dagger} ~.$$ Use the 2nd eqn to express $A^{\mu} _+$ into terms of the new operator $\widetilde a_r(\vec k)$. (Do similarly for $A^{\mu} _-$.)

Then use the sum-to-integral trick you already found, i.e., $$\frac{1}{V} \sum_{k} \rightarrow\frac{1}{(2 \pi)^3} \int d^3 \vec k ~,$$which is applicable when you pass to infinite volume.

For the commutator $[a_r(\vec k), a_s(\vec k')] = \rho_r \delta_{r,s} \delta_{\vec k, \vec k'}$, the discrete $\delta_{\vec k, \vec k'}$ becomes a 3D Dirac delta fn $\delta^3(\vec k- \vec k')$, and you're almost done.