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Supersymmetric quantum mechanics
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== Example: the harmonic oscillator == The Schrödinger equation for the harmonic oscillator takes the form : <math>H^{\rm HO} \psi_{n}(x) = \bigg(\frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+\frac{m \omega^{2}}{2}x^{2}\bigg) \psi_{n}(x) = E_{n}^{\rm HO} \psi_{n}(x),</math> where <math>\psi_{n}(x)</math> is the <math>n</math>th energy eigenstate of <math>H^\text{HO}</math> with energy <math>E_{n}^\text{HO}</math>. We want to find an expression for <math>E_{n}^{\rm HO}</math> in terms of <math>n</math>. We define the operators : <math>A = \frac{\hbar}{\sqrt{2m}}\frac{d}{dx}+W(x)</math> and : <math>A^{\dagger} = -\frac{\hbar}{\sqrt{2m}}\frac{d}{dx}+W(x),</math> where <math>W(x)</math>, which we need to choose, is called the superpotential of <math>H^{\rm HO}</math>. We also define the aforementioned partner Hamiltonians <math>H^{(1)}</math> and <math>H^{(2)}</math> as : <math>H^{(1)} = A^{\dagger} A = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} - \frac{\hbar}{\sqrt{2m}} W^{\prime}(x) + W^{2}(x)</math> : <math>H^{(2)} = A A^{\dagger} = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} + \frac{\hbar}{\sqrt{2m}} W^{\prime}(x) + W^{2}(x).</math> A zero energy ground state <math>\psi_{0}^{(1)}(x)</math> of <math>H^{(1)}</math> would satisfy the equation : <math>H^{(1)} \psi_{0}^{(1)}(x) = A^{\dagger} A \psi_{0}^{(1)}(x) = A^{\dagger} \bigg(\frac{\hbar}{\sqrt{2m}}\frac{d}{dx}+ W(x)\bigg) \psi_{0}^{(1)}(x) = 0.</math> Assuming that we know the ground state of the harmonic oscillator <math>\psi_{0}(x)</math>, we can solve for <math>W(x)</math> as : <math>W(x) = \frac{-\hbar}{\sqrt{2m}} \bigg(\frac{\psi_{0}^{\prime}(x)}{\psi_{0}(x)}\bigg) = x \sqrt{m \omega^{2}/2} </math> We then find that : <math>H^{(1)} = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} + \frac{m \omega^{2}}{2} x^{2} - \frac{\hbar \omega}{2} </math> : <math>H^{(2)} = \frac{-\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} + \frac{m \omega^{2}}{2} x^{2} + \frac{\hbar \omega}{2}.</math> We can now see that : <math>H^{(1)} = H^{(2)} - \hbar \omega = H^{\rm HO} - \frac{\hbar \omega}{2}.</math> This is a special case of shape invariance, discussed below. Taking without proof the introductory theorem mentioned above, it is apparent that the spectrum of <math>H^{(1)}</math> will start with <math>E_{0} = 0</math> and continue upwards in steps of <math>\hbar \omega.</math> The spectra of <math>H^{(2)}</math> and <math>H^{\rm HO}</math> will have the same even spacing, but will be shifted up by amounts <math>\hbar \omega</math> and <math>\hbar \omega / 2</math>, respectively. It follows that the spectrum of <math>H^{\rm HO}</math> is therefore the familiar <math>E_{n}^{\rm HO} = \hbar \omega (n + 1/2)</math>.
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