# Fermionic ladder operators [closed]

After reading Dirac's method for finding the eigen energies of a harmonic oscillator by means of ladder operators and commutation relations, I tried making some exercises on them. First I did a system of N uncoupled oscillators, and I got there alright by following the oscillator example. Now I'm really struggling with this example of 'Fermionic ladder operators'; I don't seem to get anywhere following the same route...

Consider the system with Hamiltonian $$H=\sum_{i=1}^N(\frac{\hbar}{2}(b_ib_i^\dagger - b_i^\dagger b_i)$$ with the anti-commutation relations $[b_j,b_i^\dagger]_+=b_jb_i^\dagger + b_i^\dagger b_j = \delta_{ij}$ and $[b_j,b_i]_+=[b_j^\dagger,b_i^\dagger]_+=0$. Now I need to show that the eigenvalues of $N_j^b = b_j^\dagger b_j$ can only be 0 and 1. I have tried calculation the commutator $[H, b_j]$ (which led me to the answer for the bosonic operators, after a long calculation...). This gave me $[H, b_i]= \hbar b_i$, but it doesn't seem to lead me anywhere.... I have also tried writing $$N_j^b\vert n \rangle= \lambda \vert n \rangle \Longleftrightarrow (1-N_j^{b\dagger})\vert n \rangle = (1-\lambda)\vert n \rangle,$$ which seems promising, but really isn't...

## closed as off-topic by ACuriousMind♦, Gert, CuriousOne, Kyle Kanos, AccidentalFourierTransformApr 1 '16 at 18:17

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Gert, CuriousOne, Kyle Kanos, AccidentalFourierTransform
If this question can be reworded to fit the rules in the help center, please edit the question.

• What are $N^b$ and $N^{b^\dagger}$ ? This is not standard notations. – Adam Mar 28 '16 at 8:19
• @Adam b is just an index indicating that we're talking about the operator $b_j$, so $N^{b\dagger}=(b_j^\dagger b_j)^\dagger$. – Michael Angelo Mar 28 '16 at 9:23
• @AccidentalFourierTransform In which way? – Michael Angelo Apr 1 '16 at 18:35
• @AccidentalFourierTransform woops, typo, I'll edit. – Michael Angelo Apr 1 '16 at 18:37

## 1 Answer

Well, if $b$ and $b^{\dagger}$ are raising and lowering operators I know how you can argue why $N^{b}$ has eigenvalues 0 or 1 without any algebra, although I am a little unsure about the nature of $b$ and $b^{\dagger}$, maybe you should expand you question a little but.

Basically, when you apply $N_j^b$ to a ket, it first lowers the level by one the raises it by one, you end up with the state you started with. If those operators do not introduce any constant factor that is dependent on the state, then $N_j^b$ has eigenvalue of 1. Unless you apply it to ground state. You can not go lower than ground state of course, so lowering ground state would give you 0. So when applied to ground state $N_j^b$ has eigenvalue of 0.

I found it, there you go! The only thing is that I will be using a slightly different notation, I will use $f$ $f^\dagger$ for raising and lowering operators. $$N^2= (f^\dagger f)^2 = f^\dagger f f^\dagger f = f^\dagger(1-f^\dagger f) f=f^\dagger f = [f^\dagger, f]_+f^\dagger f= N$$ There I have used some of the expressions you gave several times for the last two steps. We found that $N^2 = N$. That means that eigenvalues also have to satisfy this equation i.e. $\lambda ^2 = \lambda$, only solutions are 0 or 1.

• Yes. For the uncoupled harmonic oscillator, the N matrix was just a diagonal matrix with the normalisation constant. For fermions, there is also a maximum level, and if you act on that with the raising operator you should also get zero. So I see why the eigenvalues should be zero and one (because I know we're talking about spin 1/2 particles), I just don't see how this follows from the Hamiltonian and commutation relations alone. Maybe you're right and the question is incomplete..? – Michael Angelo Mar 28 '16 at 8:38
• Oh god. You should have seen my 3 page long clusterf**k calculation :-D. Thanks a lot! – Michael Angelo Mar 28 '16 at 9:19
• I remembered I too could not figure it out until my tutor showed me the trick. BTW I remembered this is somehow related to Pauli exclusion principle and demonstrates that there can be no more that one particle in a single particle excited state and to each state corresponds an oscillator which is either excited once or not excited at all which is kind of neat. – Ilya Lapan Mar 28 '16 at 9:20