This is a loose follow up to this question: Interpreting Argyres' spectrum of spontaneously broken SUSY QM.

In SUSY QM, the Hamiltonian can be cast as a 2x2 matrix $$ H = \frac{1}{2}p^2 + \frac{1}{2}W(q)^2 + \frac{1}{2}W'(q)\sigma_3= \begin{pmatrix}H_+&0\\0&H_-\end{pmatrix} $$ where $H_\pm = \frac{1}{2}p^2 + V_\pm$ are the spin-up and spin-down Hamiltonians and $V_\pm=\frac{1}{2}W(q)^2 \pm \frac{1}{2}W'(q)$ are their potentials (do I understand this correctly?).

Eigenfunctions of this system is always of the form $\begin{pmatrix}\phi_+\\0\end{pmatrix}$ and $\begin{pmatrix}0\\\phi_-\end{pmatrix}$. Question: In light of statements below, Is this always true?

On page 87 of Mariño's notes, he illustrates SUSY breaking with the superpotential $$W(q) = \lambda q^2 -\mu^2\,,$$ and the corresponding potentials are plotted:

Potentials plus and minus

Now, it is claimed that instantons allow transitions from the left vacuum to the right vacuum:

$$\langle q_+, \uparrow, T=-\infty |\hat\psi_+(t)| q_-, \downarrow, T=+\infty \rangle = \text{non-zero}$$

We need a $\psi_+$ operator to soak up a fermionic zero mode, and flips the spin from $+$ to $-$ or vice-versa. The mathematical computations is absolutely clear to me. But now I'm running into a frustrating contradiction:

If the Hamiltonian is diagonal, how can the system possibly flip the spin as it tunnels from one vacuum to the other? Remember, stationary states can always be put in the form $\phi_+$ or $\phi_-$ of definite spin. Flipping spin can only happen if the Hamiltonian has an off-diagonal term as in $\begin{pmatrix}H_+&H_x\\H_x&H_-\end{pmatrix}$. But in SUSY-QM it doesn't!

Is there some effective Hamiltonian that allows this?!, (with the off-diagonal elements acting as some kind of 't Hooft effective vertex?)? If so, standard treatments (e.g. numerical ones) to solve the QM system as an eigenvalue problem would totally miss this effect because they would start from a diagonal Hamiltonian! What's the resolution to this contradiction?


1 Answer 1


I hope that the readers will be indulgent towards this not-exactly-answer because I can't write in the comment section yet (so, I would be very grateful if you will not dislike it just for that)

Maybe I'm missing something but I find your question very strange. That's because you don't talk about transition amplitude with spin flipping. It would be $\langle q_+, \uparrow, T=-\infty | q_-, \downarrow, T=+\infty \rangle$ without $\hat\psi_+(t)$. Now it would be really surprising if this didn't vanish for the reasons you mentioned in your question.

But the $\hat\psi_+(t)$ operator flips the spin of the state. So it seems to me that in fact you are talking about $\langle q_+, \uparrow, T=-\infty | q_-, \uparrow, T=+\infty \rangle$. But in this case I would find nontrivial if it were vanishing, not the other way. And in this case the diagonality of the hamiltonian doesn't play a role because the states belong to the same spin subspace.

  • $\begingroup$ Ok I understand your answer -- but I am now even more confused! If I go back to action with the spinors $S=\int dt \big(\frac{1}{2}\dot{x}^2 - V(x) + i\psi^* \dot{\psi} - \psi^* \psi W''(x)\big)$, what transition amplitude (with particular attention to the $\psi$ degrees of freedom) am I really computing when I do instantons? (You know -- when I put in $x(\tau) \sim a \tanh (\omega(\tau-\tau_0)/2)$ ??) $\endgroup$
    – QuantumDot
    Commented May 20, 2014 at 19:06
  • $\begingroup$ Honestly, I thought that in the standard computation, we are computing zero-fermion to zero-fermion amplitude (corresponding to spin-down to spin-down amplitude). But apparently this is not true since according to Marino's notes its really $\langle \uparrow |\downarrow \rangle$. So how do I know what amplitude I was calculating in the first place? $\endgroup$
    – QuantumDot
    Commented May 20, 2014 at 19:11

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