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...

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• 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

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.