# How Can I Find The Commutator of this of these $2$ operators (quantum)?

$$\hat a = \frac{m\omega \hat x + i\hat p_x}{\sqrt{m\omega \hbar}}$$

Where $$\hat x$$ is the position operator which is just $$x$$. And $$\hat p_x$$ is the momentum in the x-direction operator, which is $$-i\hbar \frac{\partial}{\partial x}$$.

And $$m,\omega$$ are respectively the mass of a particle in a harmonic potential, and its oscillation frequency.

I need to workout what the commutator $$[\hat a,\hat a^\dagger]$$ would be equal to.

I know the commutator of $$[a,b]=ab - ba$$

But I don't know how to complex transpose $$\hat a$$ in this situation, also it has a partial derivative - which makes me even more confused.

Could anyone shed some light on how I can approach this question and what steps I should take to get the commutator value?

I look forward to hearing your responses.

• Hi yoyo247, I've added MathJax to your post. For the future, you can find a basic MathJax tutorial here. Feb 22 at 14:02
• Appreciate it @J.Murray ! Feb 22 at 14:17

Note that $$-i\hbar\, \partial_x$$ is not the momentum operator, but the momentum operator evaluated in the position-representation, i.e. we have that $$\langle x|\hat{p} = -i\hbar\,\partial_x \langle x|$$. So you do not have to replace $$\hat{p}_x$$ with the partial derivative at all; just leave it as $$\hat{p}_x$$. Also note that both the position and the momentum operator are hermitian such that $$\hat{x}^\dagger = \hat{x}$$ and $$\hat{p}^\dagger = \hat{p}$$.

Regarding your first question: You need basic properties of the adjoint of an operator. For example, you could use the fact that for two operators, $$\hat{A}$$ and $$\hat{B}$$, the adjoint of their sum is given as the sum of their adjoints: $$\left(\hat{A}+ \hat{B}\right)^\dagger = \hat{A}^\dagger + \hat{B}^\dagger$$. Additionally, you can make use of the following property: $$\left(\lambda\, \hat{A}\right)^\dagger = \bar{\lambda}\, \hat{A}^\dagger$$, where $$\lambda \in \mathbb{C}$$.

You can apply these rules to find $$\hat{a}^\dagger$$. To evaluate the commutator between $$\hat{a}$$ and $$\hat{a}^\dagger$$, you further need the canonical commutation relation $$[\hat{x},\hat{p}] = i\hbar$$, which was derived in the answer by J. Murray above.

I hope this helps.

• Thanks - although I don't fully understand ⟨𝑥|𝑝̂ =−𝑖ℏ∂𝑥⟨𝑥| - would you be able to explain this and what it means? Feb 22 at 14:40
• $\left(\hat p|x\rangle \right)^\dagger=\langle x|\hat p\rightarrow -i\hbar\partial_x\langle x|$ in the position basis. @yoyo247 Feb 22 at 14:44

The trick here is always to supply a test function $$f$$ for the operators to act on. From there, $$[\hat a,\hat a^\dagger] f = \hat a\big( \hat a^\dagger f) - \hat a^\dagger\big( \hat a f)$$

As an example for the position and momentum operators, $$[\hat x,\hat p]f = \hat x\big(\hat p f) - \hat p(\hat x f) = x\big(-i\hbar f'(x)\big) +i\hbar \frac{d}{dx}\big(x f(x)\big)$$ $$= -i\hbar xf'(x) + i\hbar f(x) + i\hbar xf'(x) = i\hbar f(x)$$ $$\implies [\hat x,\hat p] = i\hbar \mathbb I$$ where $$\mathbb I$$ is the identity operator.

• This is very useful - I really appreciate the way you show that. Feb 22 at 14:36

$$\newcommand\dag\dagger$$ You can proceed as follows (as my professor did when he explained the topic), defining two ausiliar operators:

$$\hat{x}'= \sqrt{\frac{m \omega}{\hbar}} \hat{x}$$,

$$\hat{p}'= \frac{1}{\sqrt{m \omega \hbar}} \hat{p}$$

So rewriting both $$a$$ and $$a^\dag$$ in function of $$\hat{x}', \hat{p}'$$ they are

$$a=\frac{\hat{x}'+i\hat{p}'}{\sqrt 2}$$, $$a^\dag=\frac{\hat{x}'-i\hat{p}'}{\sqrt 2}$$

Then $$[a,a^\dag]=\frac{1}{2}[\hat{x}'+i\hat{p}',\hat{x}'-i\hat{p}'] \\=\frac{1}{2}\bigl([\hat{x}',\hat{x}']-[\hat{x}',i\hat{p}']+[i\hat{p}',\hat{x}']-[i\hat{p}',i\hat{p}']\bigr)\\ =\frac{1}{2}\bigl(-i[\hat{x}',\hat{p}']+i[\hat{p}',\hat{x}']\bigr)$$ Where the following were used: $$[A,A]=0$$ and $$[iA,B]=i[A,B]=[A,iB]$$ (the latest: $$[iA,B]=iAB-BiA=iAB-iBA=i[A,B]$$).

Now considering that since $$[\hat{x},\hat{p}]=i \hbar$$ then $$[\hat{x}',\hat{p}']=\sqrt{\frac{m \omega}{\hbar}}\cdot \frac{1}{\sqrt{m \omega \hbar}}[\hat{x},\hat{p}]=\frac{1}{\hbar}i \hbar=i$$ sobstituing in the previouses and remembering that $$[A,B]=-[B,A]$$,:

$$=\frac{1}{2}\bigl( -i^2+i(-i)\bigr)=\frac{1}{2}2=1$$

Then $$[a, a^\dag]=1$$

• This is a brilliant method - although how did you know [𝑖𝐴,𝐵]=𝑖[𝐴,𝐵] ? Otherwise I understand. Feb 22 at 14:39
• This method is the one that was presented as demonstration by my QM professor. I edit and add this information. $[iA,B]=iAB-BiA=iAB-iBA=i[A,B]$ Feb 22 at 14:42