Graduate Commutation relations between HO operators | QFT; free scalar field

[JD_PM]

TL;DR

I want to understand how can I prove

$[N(\vec k'), a_r^{\dagger}(\vec k)] = \delta_{\vec k', \vec k} a_r^{\dagger} (\vec k) \ \ \ \ (1)$



$[N(\vec k'), a_r(\vec k)] = -\delta_{\vec k', \vec k} a_r(\vec k) \ \ \ \ (2)$

I am getting started in applying the quantization of the harmonic oscillator to the free scalar field.

After studying section 2.2. of Tong Lecture notes (I attach the PDF, which comes from 2.Canonical quantization here https://www.damtp.cam.ac.uk/user/tong/qft.html), I went through my notes.

My notes state that the harmonic oscillator operators $a_r(\vec k)$ and $a_r^{\dagger}(\vec k)$ (where we know that $a^{\dagger}$ is the hermitian conjugate of $a$) satisfy the following commutator relations

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

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

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

Then an hermitian operator is introduced : $N(\vec k)$

$$N(\vec k) = \sum_{r=0}^{3} \rho_r a_r^{\dagger}(\vec k) a_r(\vec k)$$

And then 'the following commutation relations follow'

$$[N(\vec k'), a_r^{\dagger}(\vec k)] = \delta_{\vec k', \vec k} a_r^{\dagger} (\vec k) \ \ \ \ (1)$$

$$[N(\vec k'), a_r(\vec k)] = -\delta_{\vec k', \vec k} a_r(\vec k) \ \ \ \ (2)$$

(1) and (2) come completely out of the blue to me and I would like to understand and see how to get them.

How can I prove (1) and (2)? Do I have to use the Fourier transform Method Tong uses in 2.2?

Thanks.

 

[strangerep]

My notes state that the harmonic oscillator operators $a_r(\vec k)$ and $a_r^{\dagger}(\vec k)$ (where we know that $a^{\dagger}$ is the hermitian conjugate of $a$) satisfy the following commutator relations $$[a_r(\vec k), a_s(\vec k')] = [a_r^{\dagger}(\vec k), a_s^{\dagger}(\vec k')] = 0$$ $$[a_r(\vec k), a_s(\vec k')] = \rho_r \delta_{r,s} \delta_{\vec k, \vec k'}$$

Did you forget a dagger in the last eqn above?

$$[N(\vec k'), a_r^{\dagger}(\vec k)] = \delta_{\vec k', \vec k} a_r^{\dagger} (\vec k) \ \ \ \ (1)$$$$[N(\vec k'), a_r(\vec k)] = -\delta_{\vec k', \vec k} a_r(\vec k) \ \ \ \ (2)$$How can I prove (1) and (2)? Do I have to use the Fourier transform Method Tong uses in 2.2?

No, Fourier transforms are unnecessary here. Just evaluate the commutator using the fact that commutators are (bi)linear ($[A,B+C]=[A,B]+[A,C]$, etc), and also obey the Leibniz product rule, i.e., $[A,BC] = [A,B]C + B[A,C]$.

 

[JD_PM]

Did you forget a dagger in the last eqn above?

Yes.

No, Fourier transforms are unnecessary here. Just evaluate the commutator using the fact that commutators are (bi)linear ($[A,B+C]=[A,B]+[A,C]$, etc), and also obey the Leibniz product rule, i.e., $[A,BC] = [A,B]C + B[A,C]$.

Alright, let's evaluate $(1)$. By using the Leibniz product rule and the definition $[A, B] = AB - BA$ I got

$$[N(\vec k'), a_r^{\dagger}(\vec k)] = [\rho_r a_r^{\dagger} (\vec k') a_r (\vec k'), a_r^{\dagger}(\vec k)] = \rho_r a_r^{\dagger} (\vec k') [a_r (\vec k'), a_r^{\dagger}(\vec k)] + [\rho_r a_r^{\dagger} (\vec k'), a_r^{\dagger}(\vec k)]a_r (\vec k') = \rho_r a_r^{\dagger} (\vec k') \Big( a_r^{\dagger}(\vec k') a_r^{\dagger}(\vec k) - a_r^{\dagger}(\vec k) a_r (\vec k') \Big) + \Big( \rho_r a_r^{\dagger} (\vec k') a_r^{\dagger} (\vec k') - a_r^{\dagger} (\vec k) \rho_r a_r^{\dagger} (\vec k')\Big) a_r (\vec k')$$

OK. Then I expanded the products but I do not see how to get it all equal to $\delta_{\vec k', \vec k} a_r^{\dagger} (\vec k)$

Am I on the right track?

 

[vanhees71]

Just use
$$[\hat{A} \hat{B},\hat{C}]=\hat{A} [\hat{B},\hat{C}] + [\hat{A},\hat{C}] \hat{B},$$
which you prove by writing out the commutators on both sides of the equation, and
$$\hat{N}(\vec{k}')=\hat{a}^{\dagger}(\vec{k}') \hat{a}(\vec{k}').$$

 

[JD_PM]

Just use
$$[\hat{A} \hat{B},\hat{C}]=\hat{A} [\hat{B},\hat{C}] + [\hat{A},\hat{C}] \hat{B},$$
which you prove by writing out the commutators on both sides of the equation, and
$$\hat{N}(\vec{k}')=\hat{a}^{\dagger}(\vec{k}') \hat{a}(\vec{k}').$$

Hi vanhees71

Alright so I have already applied such property:

$$[N(\vec k'), a_r^{\dagger}(\vec k)] = [\rho_r a_r^{\dagger} (\vec k') a_r (\vec k'), a_r^{\dagger}(\vec k)] = \rho_r a_r^{\dagger} (\vec k') [a_r (\vec k'), a_r^{\dagger}(\vec k)] + [\rho_r a_r^{\dagger} (\vec k'), a_r^{\dagger}(\vec k)]a_r (\vec k')$$

My issue is that I get stuck in evaluating crossed terms; what do I have to take into account once I apply $[\hat{A} \hat{B},\hat{C}]=\hat{A} [\hat{B},\hat{C}] + [\hat{A},\hat{C}] \hat{B},$ ?

Thank you

 

[strangerep]

Alright, let's evaluate $(1)$. By using the Leibniz product rule and the definition $[A, B] = AB - BA$ I got

$$[N(\vec k'), a_r^{\dagger}(\vec k)] = [\rho_r a_r^{\dagger} (\vec k') a_r (\vec k'), a_r^{\dagger}(\vec k)] = \rho_r a_r^{\dagger} (\vec k') [a_r (\vec k'), a_r^{\dagger}(\vec k)] + [\rho_r a_r^{\dagger} (\vec k'), a_r^{\dagger}(\vec k)]a_r (\vec k') = \rho_r a_r^{\dagger} (\vec k') \Big( a_r^{\dagger}(\vec k') a_r^{\dagger}(\vec k) - a_r^{\dagger}(\vec k) a_r (\vec k') \Big) + \Big( \rho_r a_r^{\dagger} (\vec k') a_r^{\dagger} (\vec k') - a_r^{\dagger} (\vec k) \rho_r a_r^{\dagger} (\vec k')\Big) a_r (\vec k')$$ OK. Then I expanded the products but I do not see how to get it all equal to $\delta_{\vec k', \vec k} a_r^{\dagger} (\vec k)$

Am I on the right track?

You seem to have managed to railroad yourself into a swamp.

It should have been easy. Use the known commutation relations between the c/a operators (which you already wrote down in your opening post).

 

[Haborix]

Some confusion might be alleviated if you are more careful about labels on operators. Use something other than $r$, like say $s$, for the operators forming the number operator. You will always get in a muddle when you use the same label for something summed over and something not summed over.

 

[JD_PM]

You seem to have managed to railroad yourself into a swamp.

Gosh, you are absolutely right! Unfortunately I am a specialist in overcomplicating things. I am working to fix that.

It should have been easy. Use the known commutation relations between the c/a operators (which you already wrote down in your opening post).

Actually it was only about using the known commutation relations.

We know that

$$[N(\vec k'), a_r^{\dagger}(\vec k)] = [\rho_r a_r^{\dagger} (\vec k') a_r (\vec k'), a_r^{\dagger}(\vec k)] = \rho_r a_r^{\dagger} (\vec k') [a_r (\vec k'), a_r^{\dagger}(\vec k)] + [\rho_r a_r^{\dagger} (\vec k'), a_r^{\dagger}(\vec k)]a_r (\vec k')$$

The thing is that we do not need to apply the definition $[A, B] = AB - BA$ but simply notice that $[a_r (\vec k'), a_r^{\dagger}(\vec k)] =\delta_{\vec k', \vec k}$ and $[a_r^{\dagger} (\vec k'), a_r^{\dagger}(\vec k)] = 0$

Thus we get

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

Applying the exact same logic to $(2)$ I get

$$[N(\vec k'), a_r(\vec k)] = -\delta_{\vec k', \vec k} a_r(\vec k')$$

Notice a slightly, slightly difference with the provided solutions; I get $a_r(\vec k')$ instead of $a_r(\vec k)$. I'd bet is a typo made by the person who did the exercise. What do you think?

Thanks.

 

[strangerep]

Notice a slightly, slightly difference with the provided solutions; I get $a_r(\vec k')$ instead of $a_r(\vec k)$. I'd bet is a typo made by the person who did the exercise. What do you think?

It's not a typo. Think carefully about the effect of the $\delta$ term on the RHS.

 

[JD_PM]

It's not a typo. Think carefully about the effect of the $\delta$ term on the RHS.

Do you mean that as we have the Kronecker delta we can either write $\vec k$ or $\vec k'$?

 

[strangerep]

Do you mean that as we have the Kronecker delta we can either write $\vec k$ or $\vec k'$?

Yes.