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Suppose we try to apply supersymmetry in quantum mechanics to a particular potential. If you come up with two partner potentials, and two partner Hamiltonians, and then look at the energy of the ground state, is it true that at most one of them can be zero?

If they are both zero, does that mean that supersymmetry is just not preserved by the vacuum states of the partner Hamiltonians? Can we still apply supersymmetry to this system if the energy of both ground states are zero?

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up vote 5 down vote accepted

Note that you begin with a superpotential $W$ which gives you a generalization of the creation/anihilation operator as $A^{+} = -\frac{d}{dx} + W(x)$ and $A^{-}= +\frac{d}{dx} + W(x)$. You will get 2 hamiltonians $H_- = A^+ A^-$ and $H_+ = A^- A^+$. So you obtain 2 different potentials $V_{+}$ and $V_{-}$ for superpartners $V_{-} (x)= W(x)^2 - \frac{dW}{dx}$ and $V_{+} (x)= W(x)^2 + \frac{dW}{dx}$.

Then, suppose that $\psi^-$ is a solution of $A^-\psi^-(x) = 0$ with $\psi^-$ normalizable, you will find that $H_- \psi_- = A^+A^-\psi^- = 0$, so $\psi^-$ is an eigenstate of $H_- $, with eigenvalue 0.

You have $\psi^-(x) \sim e^{-\int^x_{x_0} W(x) dx}$ (from the definition of $A^-$)

With the same reasoning, you will find that a $\psi^+$ solution of $A^+\psi^+(x) = 0$ will give you $\psi^+(x) \sim e^{+\int^x_{x_0} W(x) dx}$ which would be an eigenstate of $H_+ $, with eigenvalue 0.

But you have a problem, you can see that you cannot have, at the same time $\psi^+$ and $\psi^-$ normalizable, because $\psi^+(x) \sim \frac{1}{\large \psi^-(x)}$.

So, one of the functions $\psi^+$ or $\psi^-$ has to vanish. So, you have, at most, one ground state with zero energy.

[EDIT]

The practical link with supersymmetry is as follows. We define a 2 -dimensional space, with $\psi = (\psi_-, \psi_+)$. One of the $\psi$ states is a bosonic state, the other is a fermionic state.

We define the supersymmetric generators:

$Q^- =\left[ \begin{array}{cccc} 0 & 0 \\ A^- & 0 \end{array} \right]$

$Q^+ =\left[ \begin{array}{cccc} 0 & A^+ \\ 0 & 0 \end{array} \right]$

The hamiltonian is $H = Q^-Q^+ + Q^+Q^-$

$H =\left[ \begin{array}{cccc} A^+A^- & 0 \\ 0 & A^-A^+ \end{array} \right] = \left[ \begin{array}{cccc} H_- & 0 \\ 0 & H_+ \end{array} \right]$

It can be easily seen that $(Q^-)^2=(Q^+)^2 = 0$ and $[H,Q^-] = [H,Q^+] = 0$

Unbroken supersymmetry corresponds to a ground state $|0>$ such as $<0|H|0> = 0$, and this imply $Q^+|0> = Q^-|0> = 0$ In this case, one of the functions $\psi^+$ or $\psi^-$ is normalizable, and the other vanishes. For instance, suppose $\psi^-$ is normalizable, then it means that $A^- \psi^- =0$

The case of (spontaneously) broken supersymmetry, is when $<0|H|0> \neq 0$, which implies $Q^+|0> \neq 0$ or $Q^-|0> \neq 0$. In this case, neither of the ground states $\psi^+$ nor $\psi^-$ are normalizable.

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Thank you, this is a great response. How does one choose which ground state vanishes (Are you even allowed to choose which ground state vanishes)? If, for instance, $W(x)$ is a polynomial, one of the ground states goes to zero at $+\infty$ and the other at $-\infty$. How does one handle this? It seems arbitrary to solve $A^+ \psi^+(x) = 0$ and $A^- \psi^-(x) = 0$, to obtain two exponentials for the ground state, and then set one to zero. –  Randy Jun 4 '13 at 20:15
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For instance, if you take $W(x) = x$, you see that the only possible normalizable eigenstate is $\psi^-(x) = e^{-x^2/2}$ –  Trimok Jun 4 '13 at 20:31
    
Notice, that in some cases, neither $\psi^+$ Nor $\psi^-$ are normalizable (it depends on W) –  Trimok Jun 4 '13 at 20:33
    
My reference : Supersymmetric Quantum Mechanics: An Introduction, by Asim Gangopadhyaya , Jeffry V Mallow , Constantin Rasinariu. (World Scientific) –  Trimok Jun 4 '13 at 20:44
    
Right, I guess my question is suppose that neither $\psi^+$ and $\psi^-$ are normalizable. For instance, consider $W(x) = x^2$. Does this mean neither ground state has zero energy since they are not normalizable, even though $H_+ \psi^+ = 0$ and $H_- \psi^- = 0$? –  Randy Jun 4 '13 at 20:48
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