Is my explanation valid? (commutation relations)

1. Apr 18, 2014

Xyius

I am doing a problem on the "super symmetric harmonic oscillator" Defined by..

$$\hat{H}=\hat{H}_b+\hat{H}_f= \hbar \omega \left( \hat{b}^{\dagger}\hat{b}+\hat{f}^{\dagger}\hat{f} \right)$$

I am given the operator..

$$\hat{Q}=\sqrt{\hbar \omega} \hat{b}^{\dagger} \hat{f}$$

and asked to show that $[\hat{H},\hat{Q}]=0$. It seems like just a matter of plugging everything in and evaluating, but I am getting commutators of the following forms..

$[\hat{b},\hat{f}]$
$[\hat{b},\hat{f}^{\dagger}]$
$[\hat{b}^{\dagger},\hat{f}]$
$[\hat{b}^{\dagger},\hat{f}^{\dagger}]$
$[\hat{b},\hat{b}^{\dagger}]$
$[\hat{f},\hat{f}^{\dagger}]$

And to me, it seems like all of these commutation relations would be equal to zero for the following reason.

The eigenkets of this system are $|n_b,n_f>$. Where $n_b$ is the number of bosons and $n_f$ is the number of fermions. So say for example we have $[\hat{b},\hat{f}]$, we have..

$$\hat{b}\hat{f}|n_b,n_f>=\hat{b}|n_b,n_f-1>=|n_b-1,n_f-1>$$

And

$$\hat{f}\hat{b}|n_b,n_f>=\hat{f}|n_b-1,n_f>=|n_b-1,n_f-1>$$

Thus the commutator is zero. All of these can be proved this way. My question is, is this a correct way of thinking?

2. Apr 18, 2014

dauto

All the b's comute with all the f's but
$[\hat{b},\hat{b}^{\dagger}]$, and
$[\hat{f},\hat{f}^{\dagger}]$
do not commute.

3. Apr 18, 2014

Fredrik

Staff Emeritus
I'm not familiar with this problem, but I have to ask if you're sure that that's the effect of the b and f operators on the number states? I'm thinking that in the usual harmonic oscillator problem, we had $a^\dagger a|n\rangle=n|n\rangle$, which means that $a^\dagger a$ is a number operator, but if your calculations are correct, then $b^\dagger b$ wouldn't be a number operator in this problem. Is it supposed to be?

4. Apr 18, 2014

Xyius

Why is that exactly? My explanation is this..

$\hat{b}\hat{b}^{\dagger}|0>=\hat{b}|1>=|0>$

But..

$\hat{b}^{\dagger}\hat{b}|0>=0$

Which are different. But is there a more general way of looking at this?

5. Apr 18, 2014

Xyius

Yes it is supposed to be the number operator. $\hat{b}^{\dagger}\hat{b}$ can be the boson number operator and $\hat{f}^{\dagger}\hat{f}$ can be the fermion number operator.

I found a very good paper about it here..

http://gaitskell.brown.edu/courses/SeniorThesis/2003_SeniorThesis/2003SeniorThesis_Wellman.pdf

edit: I am not sure if that is the effect of the operators. In my problem is says they are the raising and lowering operators of bosons and fermions respectively.

6. Apr 18, 2014

andrien

You can actually use a cheap way out by noting that those supersymmetric generator replace a fermion by boson of exact same energy and hence if you use that commutator and act on an energy eigenvector then.......(figure that out).you can use brute force though.Anyway It is a very fundamental commutator in supersymmetry.

7. Apr 18, 2014

Fredrik

Staff Emeritus
I think that only tells us that there's a function $F_b$ such that
$$\hat b|n_b,n_f\rangle=F_b(n_b,n_f)|n_b-1,n_f\rangle$$ and similarly for the other operators. If you want to calculate the commutators, you will need to figure out what those functions are.

8. Apr 19, 2014

dauto

Yes, the b's are the typical bosonic ladder operators satisfying
$\hat{b}|n_b>=\sqrt{n_b}|n_b-1>,$
$\hat{b}^{\dagger}|n_b>=\sqrt{n_b+1}|n_b+1>,$ and
$[\hat{b},\hat{b}^{\dagger}]=1,$
while the f's are fermionic ladder operators satisfying
$\hat{f}|n_f>=\sqrt{n_f}|n_f-1>,$
$\hat{f}^{\dagger}|n_f>=\sqrt{n_f+1}|n_f+1>,$ and
$\{\hat{f},\hat{f}^{\dagger}\}=1.$
Notice the anti commutators in the last line

9. Apr 21, 2014

Xyius

Thanks for the replies guys. I am now trying to prove that $[\hat{Q}_1,\hat{H}]=0$ where $\hat{Q}_1=\hat{Q}+\hat{Q}^{\dagger}$ Where $\hat{Q}$ $\hat{Q}^{\dagger}$ are defined in my first post. What I have tried was simply expanding them out. But I get confused when I get here.

$\hat{b}^{\dagger}\hat{f}\hat{b}^{\dagger}\hat{b}+\hat{b}^{\dagger} \hat{f}\hat{f}^{\dagger}\hat{f}+\hat{f}^{\dagger} \hat{b}\hat{b}^{\dagger} \hat{b}+\hat{f}^{\dagger} \hat{b} \hat{f}^{\dagger}\hat{f}+\hat{b}^{\dagger}\hat{b}\hat{b}^{\dagger} \hat{f}+\hat{f}^{\dagger} \hat{f}\hat{b}^{\dagger} \hat{f}+\hat{b}^{\dagger} \hat{b} \hat{f}^{\dagger} \hat{b}+\hat{f}^{\dagger} \hat{f}\hat{f}^{\dagger} \hat{b}$

I don't know where to go from here to make this equal to zero. I know it must have something to do with getting anticommutation relations with fermionic operators, but anytime I try something, I can get some terms to cancel, but other terms wont. Maybe I am just not seeing it....

10. Apr 21, 2014

Fredrik

Staff Emeritus
If {f,f}=1, then ff=1/2 (times the identity operator). So you should be able to do things like this: ff*f =f(-ff*+{f*,f}) = -f*/2. (Hats and daggers are hard to type, so I simplified the notation). Also, the f's commute with the b's, right? I would try to reorder the factors in each term, so that all terms have the factors in the same order, e.g b's to the left of the f's, and daggered operators to the right of the undaggered operators.

Does the Hamiltonian really look like the one you posted? As I said, I'm not familiar with this problem at all, but I would have expected something like $\sum_k b_k^\dagger b_k$ rather than just $b^\dagger b$.