# Commutator of two operators

1. Mar 12, 2013

### Denver Dang

1. The problem statement, all variables and given/known data
Hello.

I am supposed to find the commutator between to operators, but I can't seem to make it add up.
The operators are given by:
$$\hat{A}=\alpha \left( {{{\hat{a}}}_{+}}+{{{\hat{a}}}_{-}} \right)$$
and
$$\hat{B}=i\beta \left( \hat{a}_{+}^{2}-\hat{a}_{-}^{2} \right),$$
where alpha and beta are real numbers, i being the irrational number, and a+ and a- are the ladder operators.

Now, I just have to find the commutator [A, B]

2. Relevant equations

3. The attempt at a solution

By attempt is given by the following

$$\left[ \hat{A},\,\hat{B} \right]=\hat{A}\hat{B}-\hat{B}\hat{A}=\alpha \left( {{{\hat{a}}}_{+}}+{{{\hat{a}}}_{-}} \right)i\beta \left( \hat{a}_{+}^{2}-\hat{a}_{-}^{2} \right)-i\beta \left( \hat{a}_{+}^{2}-\hat{a}_{-}^{2} \right)\alpha \left( {{{\hat{a}}}_{+}}+{{{\hat{a}}}_{-}} \right)$$
=i\alpha \beta \left[ \begin{align} & -{{{\hat{a}}}_{+}}{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}+{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}}-{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}+{{{\hat{a}}}_{-}}{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}} \\ & -{{{\hat{a}}}_{+}}{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}+{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}}-{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}+{{{\hat{a}}}_{-}}{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}} \\ \end{align} \right]
$$=2i\alpha \beta \left[ \left( -\hat{a}_{+}^{2}{{{\hat{a}}}_{-}}+{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}}-{{{\hat{a}}}_{-}}{{{\hat{a}}}_{+}}{{{\hat{a}}}_{-}}+\hat{a}{{_{-}^{2}}_{-}}{{{\hat{a}}}_{+}} \right) \right]$$
Now, according to the answer I have gotten from my teacher, it is supposed to be:
$$\left[ \hat{A},\hat{B} \right]=2i\alpha \beta \hat{A}$$

But I am kinda lost in how to end up with the operator A in the end, and even another alpha constant, since A operator is equal to alpha and some ladder operators.

So, what am I missing ? :)

Last edited: Mar 12, 2013
2. Mar 12, 2013

### tia89

First of all, do not use directly the definition of commutator... remember that there are properties of the commutator you can use to do things simpler:
1) Commutator is linear, i.e. $[A+B,C]=[A,C]+[B,C]$ and $[\alpha A,B]=\alpha[A,B]$
2) Multiplication is treated like $[AB,C]=A[B,C]+[A,C]B$

Using these two properties, you can simplify a lot what you wrote. Then use also the definitions of commutations of ladder operators $[a_+,a_+]=[a_-,a_-]=0$, $[a_-,a_+]=1=-[a_+,a_-]$.
Now you are done. Anyway you're right as for the $\alpha$, you get only one so you have to put it into the definition of $A$ and you will get as result $2i\beta A$

3. Mar 12, 2013

### Denver Dang

Ahhh yes.

Haven't thought of re-writing it that way. Thank you :)

4. Mar 13, 2013

### tia89

R: Commutator of two operators

Always try to use this method in exercises like that, it's often much simpler than just splitting all up :)
And anyway, you're welcome

5. Mar 13, 2013

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