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In non relativistic quantum mechanic, we are dealing with a problem involving a particle in one dimensional space, and it has been given the potential and it reads:

$$V(x)~=~A'(x)^2-\frac{\hbar}{\sqrt{2m}}A''(x).\tag{1}$$

The kinetic terms is the usual $p^2/2m$. How can I show that the Hamiltonian of the system is semi-definite positive?

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  • $\begingroup$ What are $A'(x)$ and $A''(x)$? Are they derivatives of a function $A(x)$ or just two separate functions? $\endgroup$ – Gonenc Jun 19 '15 at 22:22
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    $\begingroup$ re: question tags - is susy or operators really relevant here? $\endgroup$ – Kyle Oman Jun 19 '15 at 22:50
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    $\begingroup$ @Qmechanic Are you sure you meant to add the "supersymmetry" (or for that matter, "operators") tag here? $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 20 '15 at 4:15
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jun 20 '15 at 15:15
  • $\begingroup$ Related question on Math.SE: math.stackexchange.com/q/1337079/11127 $\endgroup$ – Qmechanic Jun 24 '15 at 23:54
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Hint: Think of way to write the Hamiltonian as

$$\hat{H}~=~ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} +V(x)~=~\hat{B}^{\dagger}\hat{B} ~\geq~0 $$

for some first-order differential operator

$$\hat{B}~=~a(x)\frac{d}{dx} +b(x) ,$$

with suitable functions $a(x)$ and $b(x)$. Here the potential $V(x)$ is given by formula (1). Can you see what the operator $\hat{B}$ should/must be?

By the way, the function $A(x)$ is known as a superpotential, cf. e.g. Ref. 1, which also explains the connection to SUSY QM.

References:

  1. F. Cooper, A. Khare, and U. Sukhatme, Supersymmetry and Quantum Mechanics, Phys. Rept. 251 (1995) 267, arXiv:hep-th/9405029; Chapter 2.
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  • $\begingroup$ Thanks for the answer. Why the function $A(x)$ is known as a superpotential? And also, i don't understand how supersymmetry enter in this problem. However i just try to write the Hamiltonan as you suggest but i couldn't because there are 2 additional terms. $\endgroup$ – Andrea89 Jun 20 '15 at 10:41
  • $\begingroup$ Ok, i'm reading the article. $\endgroup$ – Andrea89 Jun 20 '15 at 10:49
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jun 20 '15 at 10:52
  • $\begingroup$ Ok, i have another question, maybe caused by my ignorance. Why the dagger of $\hat{B}$ take a minus in front of the derivative terms ? $\endgroup$ – Andrea89 Jun 20 '15 at 11:13
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    $\begingroup$ This is basically the same reason that the momentum operator $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ is a Hermitian operator. $\endgroup$ – Qmechanic Jun 20 '15 at 11:33

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