Supersymmetry. Consider the two operators A = i p/√2m + W(x) and A^† = -i p/√2m + W(x), for some function W(x). These may be multiplied in either order to construct two Hamiltonians: H1 ≡ A^†A = p^2/2m + V1(x) and H2 ≡ AA^† = p^2/2m + V2(x); V1 and V2 are called supersymmetric partner potentials.

Question

Supersymmetry. Consider the two operators

[latex] \hat{A}=i \frac{\hat{p}}{\sqrt{2 m}}+W(x) \quad ext { and } \quad \hat{A}^{\dagger}=-i \frac{\hat{p}}{\sqrt{2 m}}+W(x) [/latex], (3.116)

for some function W(x). These may be multiplied in either order to construct two Hamiltonians:

[latex] \hat{H}_{1} \equiv \hat{A}^{\dagger} \hat{A}=\frac{\hat{p}^{2}}{2 m}+V_{1}(x) \quad ext { and } \quad \hat{H}_{2} \equiv \hat{A} \hat{A}^{\dagger}=\frac{\hat{p}^{2}}{2 m}+V_{2}(x) [/latex]; (3.117)

[latex]V_1 [/latex] and [latex]V_2 [/latex] are called supersymmetric partner potentials. The energies and eigenstates of [latex] \hat{H}_{1} ext { and } \hat{H}_{2} [/latex] are related in interesting ways.

(a) Find the potentials [latex]V_1(x) [/latex] and [latex]V_2(x) [/latex], in terms of the superpotential, W(x).

(b) Show that if [latex] \psi_{n}^{(1)} [/latex] is an eigenstate of [latex] \hat{H}_{1} [/latex] with eigenvalue [latex] E_{n}^{(1)}, ext { then } \hat{A} \psi_{n}^{(1)} [/latex] is an eigenstate of [latex] \hat{H}_{2} [/latex] with eigenvalue Similarly, show that if [latex] \psi_{n}^{(2)}(x) [/latex] is an eigenstate [latex] \hat{H}_{2} [/latex] with eigenvalue [latex] E_{n}^{(2)}, ext { then } \hat{A}^{\dagger} \psi_{n}^{(2)} [/latex] is an eigenstate of [latex] \hat{H}_{1} [/latex] with the same eigenvalue. The two Hamiltonians therefore have essentially identical spectra.

(c) One ordinarily chooses W(x) such that the ground state of [latex] \hat{H}_{1} [/latex] satisfies

[latex] \hat{A} \psi_{0}^{(1)}(x)=0 [/latex] (3.118).

and hence [latex] E_{0}^{(1)}=0 [/latex]. Use this to find the superpotential W(x), in terms of the ground state wave function, [latex] \psi_{0}^{(1)}(x) [/latex]. (The fact that [latex] \hat{A} [/latex] annihilates [latex] \psi_{0}^{(1)} ext { means that } \hat{H}_{2} [/latex] actually has one less eigenstate than [latex] \hat{H}_{1} [/latex], and is missing the eigenvalue [latex] E_{0}^{(1)} [/latex].

(d) Consider the Dirac delta function well,

[latex] V_{1}(x)=\frac{m \alpha^{2}}{2 \hbar^{2}}-\alpha \delta(x) [/latex] (3.119).

(the constant term, [latex] m \alpha^{2} / 2 \hbar^{2} [/latex], is included so that [latex] E_{0}^{(1)}=0 [/latex]). It has a single bound state (Equation 2.132)

[latex] \psi(x)=\frac{\sqrt{m \alpha}}{\hbar} e^{-m \alpha|x| / \hbar^{2}} ; \quad E=-\frac{m \alpha^{2}}{2 \hbar^{2}} [/latex] (2.132).

[latex] \psi_{0}^{(1)}(x)=\frac{\sqrt{m \alpha}}{\hbar} \exp \left[-\frac{m \alpha}{\hbar^{2}}|x| ight] [/latex] (3.120).

Use the results of parts (a) and (c), and Problem 2.23(b), to determine the superpotential W(x) and the partner potential [latex] V_2(x) [/latex] . This partner potential is one that you will likely recognize, and while it has no bound states, the supersymmetry between these two systems explains the fact that their reflection and transmission coefficients are identical (see the last paragraph of Section 2.5.2).

Accepted Answer

(a)

[latex] \hat{H}_{1}=\hat{A}^{\dagger} \hat{A}=\left(-i \frac{\hat{p}}{\sqrt{2 m}}+W ight)\left(i \frac{\hat{p}}{\sqrt{2 m}}+W ight)=\frac{\hat{p}^{2}}{2 m}+\frac{i}{\sqrt{2 m}}(W \hat{p}-\hat{p} W)+W^{2}=\frac{\hat{p}^{2}}{2 m}+V_{1} [/latex].

So [latex] V_{1}=W^{2}+\frac{i}{\sqrt{2 m}}[W, \hat{p}] [/latex]. To calculate the commutator, use a test function f(x):

[latex] [W, \hat{p}] f(x)=W\left(-i \hbar \frac{d f}{d x}+i \hbar \frac{d}{d x}(W f)=i \hbar\left(-W \frac{d f}{d x}+\frac{d W}{d x} f+W \frac{d f}{d x} ight)=i \hbar \frac{d W}{d x} f \Rightarrow[W, \hat{p}]=i \hbar \frac{d W}{d x} ight. [/latex].

[latex] V_{1}=W^{2}-\frac{\hbar}{\sqrt{2 m}} \frac{d W}{d x} [/latex].

[latex] \hat{H}_{2}=\hat{A} \hat{A}^{\dagger}=\left(i \frac{\hat{p}}{\sqrt{2 m}}+W ight)\left(-i \frac{\hat{p}}{\sqrt{2 m}}+W ight)=\frac{\hat{p}^{2}}{2 m}-\frac{i}{\sqrt{2 m}}(W \hat{p}-\hat{p} W)+W^{2}=\frac{\hat{p}^{2}}{2 m}+V_{2} [/latex].

So [latex] V_{2}=W^{2}-\frac{i}{\sqrt{2 m}}[W, \hat{p}] [/latex], and hence [latex] V_{2}=W^{2}+\frac{\hbar}{\sqrt{2 m}} \frac{d W}{d x} [/latex].

(b)

[latex] \hat{H}_{1} \psi_{n}^{(1)}=E_{n}^{(1)} \psi_{n}^{(1)} \Rightarrow \hat{A}^{\dagger} \hat{A} \psi_{n}^{(1)}=E_{n}^{(1)} \psi_{n}^{(1)} \Rightarrow \hat{H}_{2}\left(\hat{A} \psi_{n}^{(1)} ight)=\hat{A} \hat{A}^{\dagger} \hat{A} \psi_{n}^{(1)}=\hat{A} E_{n}^{(1)} \psi_{n}^{(1)}=E_{n}^{(1)}\left(\hat{A} \psi_{n}^{(1)} ight) [/latex].

[latex] \hat{H}_{2} \psi_{n}^{(2)}=E_{n}^{(2)} \psi_{n}^{(2)} \Rightarrow \hat{A} \hat{A}^{\dagger} \psi_{n}^{(2)}=E_{n}^{(2)} \psi_{n}^{(2)} \Rightarrow \hat{H}_{1}\left(\hat{A}^{\dagger} \psi_{n}^{(2)} ight)=\hat{A}^{\dagger} \hat{A} \hat{A}^{\dagger} \psi_{n}^{(2)}=\hat{A}^{\dagger} E_{n}^{(2)} \psi_{n}^{(2)}=E_{n}^{(2)}\left(\hat{A}^{\dagger} \psi_{n}^{(2)} ight) [/latex].

(c)

[latex] \hat{A} \psi_{0}^{(1)}(x)=0 \Rightarrow\left(\frac{\hbar}{\sqrt{2 m}} \frac{d}{d x}+W ight) \psi_{0}^{(1)}(x)=0 \Rightarrow W(x)=-\frac{\hbar}{\sqrt{2 m}} \frac{\left(d \psi_{0}^{(1)} / d x ight)}{\psi_{0}^{(1)}}=-\frac{\hbar}{\sqrt{2 m}} \frac{d}{d x}\left[\ln \left(\psi_{0}^{(1)} ight) ight] [/latex].

(d)

[latex] W(x)=-\frac{\hbar}{\sqrt{2 m}} \frac{d}{d x}\left[\ln \left(\frac{\sqrt{m \alpha}}{\hbar} ight)-\frac{m \alpha}{\hbar^{2}}|x| ight]=\frac{\hbar}{\sqrt{2 m}} \frac{m \alpha}{\hbar^{2}} \operatorname{sign}(x)= \sqrt{\frac{m}{2}} \frac{\alpha}{\hbar} \operatorname{sign}(x) [/latex].

[latex] V_{2}=W^{2}+\frac{\hbar}{\sqrt{2 m}} \frac{d W}{d x}=\frac{m \alpha^{2}}{2 \hbar^{2}}+\frac{\hbar}{\sqrt{2 m}} \sqrt{\frac{m}{2}} \frac{\alpha}{\hbar} \frac{d}{d x}(\operatorname{sign}(x))= \frac{m \alpha^{2}}{2 \hbar^{2}}+\alpha \delta(x) [/latex].

I used the fact that

[latex] \operatorname{sign}(x)=-1+2 heta(x)= \begin{cases}-1, & (x<0) \ +1, & (x>0)\end{cases} [/latex].

and [latex] d heta / d x=\delta(x) [/latex] (Problem 2.23(b)).

Question 3.47: Supersymmetry. Consider the two operators A = i p/√2m + W(x)...

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