# No equal boson/fermion masses when supersymmetry is broken?

I am currently studying the paper by Witten on supersymmetry and Morse theory. In the introduction it is stated that when supersymmetry is not broken, i.e. $$Q|0\rangle=0$$, the Hilbert space contains bosons and fermions of equal mass. I thought initially that this was the case simply because $$[Q,H]=0$$ (take f.e. a boson $$|b\rangle$$ with energy E, then $$HQ|b\rangle=QH|b\rangle=EQ|b\rangle$$ so its fermion counterpart has energy E as well) but I don't see why you can not use this to argue that there are also equal mass bosons/fermions when the symmmetry is broken (which is of course not the case). My question thus comes down to the fact that I don't understand why the $$Q|0\rangle=0$$ requirement is necessary and how symmetry breaking leads to different masses. In the paper there is also this sentence

Now in any quantum field theory if a symmetry operator (an operator which commutes with the Hamiltonian) annihilates the vacuum state, then the one particle states furnish a representation of the symmetry. In the case of a supersymmetric theory, if a solution of (8) does exist, then the Hilbert space of the theory contains bosons and fermions of equal mass.

• Here you find the paper by Witten. – Who Is Drubbels May 29 '20 at 9:20
• Welcome to Physics SE! I edited the question to fix an OCR (Optical Character Recognition) error in the excerpt. I assume it was an OCR error because I got the same result (Hubert instead of Hilbert) when I tried copying-and-pasting the same excerpt. – Chiral Anomaly May 30 '20 at 1:07
• Are you asking why/if symmetry breaking necessarily leads to different masses? Or are you asking why the condition $Q|0\rangle=0$ is important for showing that unbroken symmetry leads to equal particle masses? The sequence $HQ|b\rangle=QH|b\rangle=EQ|b\rangle$ says that $|b\rangle$ and $Q|b\rangle$ have the same energy, but it doesn't imply that $Q|b\rangle$ is a single-particle state, not even if $|b\rangle$ is a single-particle state, unless $Q|0\rangle=0$. (The condition "single-particle" is defined with respect to the vacuum state, which has zero particles by definition.) – Chiral Anomaly May 30 '20 at 3:06
• @ChiralAnomaly First of all, thanks for responding! Now, I am actually asking both questions although I think my confusion in both cases is related and although the answer to the first one is probably trivial since (if supersymmetry is broken) there simply isn't a 1 to 1 connection between the bosons and fermions anymore. Can this reasoning, together with your answer, also be a solution to my second question? I.e. when $Q|b\rangle\neq0$ then $Q|b\rangle$ still generates a state with equal energy but since $Q$ is not a symmetry anymore this is not a single-particle fermion state. – Who Is Drubbels May 30 '20 at 11:13
• I mean $Q|0\rangle\neq0$ of course. – Who Is Drubbels May 30 '20 at 13:19

Consider a bosonic state $$|b\rangle$$ and a fermionic one $$|f\rangle$$ of masses $$m_{b}$$ and $$m_{f}$$ related by $$Q_{\alpha}|b\rangle = |f\rangle.$$
Recall that $$P^{\mu}P_{\mu}|b\rangle = m_{b}^{2}|b\rangle \ \ , \ \ P^{\mu}P_{\mu}|f\rangle = m_{f}^{2}|f\rangle .$$
Also notice that if supersymmetry is unbroken then $$[Q_{\alpha},P^{\mu}]=0,$$ implies
$$[Q_{\alpha},P^{\mu}P_{\mu}]=0,$$ then from the comparison of $$P^{\mu}P_{\mu}Q_{\alpha}|b\rangle=P^{\mu}P_{\mu}|f\rangle=m_{f}^{2}|f\rangle,$$ with $$Q_{\alpha} P^{\mu}P_{\mu}|b\rangle=m_{b}^{2}Q_{\alpha}|f\rangle=m_{b}^{2}|f\rangle,$$ follows that $$m_{b}=m_{f}.$$
Why the above computation fails for the case of unbroken supersymmetry? The problem basically is that $$[Q_{\alpha},P^{\mu}P_{\mu}]=0$$ is no longer true (the supersymmetry algebra does not hold anymore).