Exactness of SWKB for Shape Invariant Potentials

For Class I, $f_{1}=\mu$, a constant. In this case, $W(x,a)=\mu\,a+f_{2}(x)+u(a)$. We can regroup terms by defining $\tilde{u}(a)\equiv u(a)+\mu~{}a$, so that $W(x,a)=f_{2}(x)+\tilde{u}(a)$. Renaming $\tilde{u}$ back to $u$, eq.(9) yields $f_{2}~{}\dot{u}-f_{2}^{\prime}=-\frac{1}{2}\dot{g}-\dot{u}~{}u$, where dots denote derivatives with respect to $a$ and primes denote derivatives with respect to $x$. The RHS is independent of $x$. Since $f_{2}$ does not depend on $a$ and cannot be a constant, each side of the equation is equal to a constant, which we call $\epsilon$. Consequently $\dot{u}=\alpha$, so $u=\alpha~{}a+\beta$ for constants $\alpha$ and $\beta$. Finally, we can regroup terms one more time, such that $\tilde{f_{2}}(x)\equiv f_{2}(x)+\beta$ and $\tilde{\epsilon}\equiv\epsilon+\alpha\beta,$ and then rename $\tilde{f_{2}}$ back to $f_{2}$ and $\tilde{\epsilon}$ back to $\epsilon$.

Therefore, Class I superpotentials can be written as $W(x,a)=f_{2}(x)+\alpha~{}a$, where $\alpha f_{2}-f_{2}^{\prime}=\epsilon$ for constants $\alpha$ and $\epsilon$.

For Class II, $f_{2}=\mu$, a constant. In this case, $W(x,a)=a~{}f_{1}(x)+\mu+u(a)$. We regroup terms such that $u(a)+\mu\rightarrow u(a)$, so that $W(x,a)=a~{}f_{1}(x)+u(a)$. Then Eq.(9) requires $a\left(f_{1}^{2}-f_{1}^{\prime}\right)+f_{1}\left(u+a\dot{u}\right)=-\frac{1}{2}\dot{g}-\dot{u}~{}u$. Since $f_{1}$ is not constant and the right hand side is $x$-independent, this requires $u+a\dot{u}=a~{}\alpha$, so $u=\alpha a/2+B/a$ for constants $B$ and $\alpha$. Similar to Class I, we can shift $f_{1}$ by the constant $\alpha/2$ to absorb the $\alpha a/2$ term into $af_{1}$.

Therefore, Class II superpotentials can be written as $W(x,a)=af_{1}(x)+B/a$, where $f_{1}^{2}-f_{1}^{\prime}=\lambda$ for constants $B$ and $\lambda$.

For Class III, neither $f_{1}$ nor $f_{2}$ is constant. We first note that if $f_{2}$ is of the form $f_{2}(x)=\nu f_{1}(x)+\mu$ for any constants $\mu$ and $\nu$, then with a redefinition of $a\to a+\nu$, $W$ can be considered equivalent to a Class II superpotential in which $f_{2}$ is a constant. Similarly, if $u(a)$ is linear in $a$, this is equivalent to the case $u(a)=0$ by regrouping terms.

With these assumptions, we substitute the form of $W$ from Eq.(11) in Eq. (9) and get

$\displaystyle-a\left(f_{1}^{2}-f_{1}^{\prime}\right)+\left(f_{2}^{\prime}-f_{1}f_{2}\right)-f_{1}\left(u+a\dot{u}\right)-\dot{u}f_{2}=g/2+\dot{u}u~{}.$

(12)

The first two terms in this equation are respectively linear in $a$ and independent of $a$. Therefore, if there is any nonlinearity in $a$ on the right hand side of the equation, it could only come from the third and fourth terms of the left hand side. However, the right hand side of this equation is independent of $x$; since $f_{1}$ and $f_{2}$ are not constant and are linearly independent, the $x$-dependence in the third term cannot be canceled by the fourth term, and vice versa. Consequently, the coefficients of $f_{1}$ and $f_{2}$ in terms three and four must each be linear functions of $a$.

Linearity in $a$ of the $\dot{u}f_{2}$ term in Eq. (12) implies $u=\mu a^{2}/2+\nu a+\gamma$. Then $u+a\dot{u}=3\mu a^{2}/2+2\nu a+\gamma$, which is linear in $a$ only if $\mu=0$. Thus $u$ itself is linear in $a$, so we can set $u=0$. We then have $-a\left(f_{1}^{2}-f_{1}^{\prime}\right)+\left(f_{2}^{\prime}-f_{1}f_{2}\right)=g/2+\dot{u}u$. Since the right-hand-side is independent of $x$, this requires $f_{1}^{2}-f_{1}^{\prime}=\lambda$ and $f_{1}f_{2}-f_{2}^{\prime}=\varepsilon$ for constants $\lambda$ and $\varepsilon$.

Therefore, Class III superpotentials can be written as $W(x,a)=af_{1}(x)+f_{2}(x)$, where $f_{1}^{2}-f_{1}^{\prime}=\lambda$ and $f_{1}f_{2}-f_{2}^{\prime}=\varepsilon$ for constants $\lambda$ and $\varepsilon$.

In this case, $W(x,a)=f_{2}(x)$, where $W^{\prime}(x)=f_{2}^{\prime}(x)=-\varepsilon$. From Eq. 9 $dg/da=-2\varepsilon$. To avoid level crossing, $dg/da$ must be positive. We define $\omega\equiv-2\varepsilon>0$, so $dg/da=\omega$, $E_{n}=n\omega\hbar$, therefore $\partial E_{n}/\partial\hbar=n\omega$. We now solve Eq. 16 with these values:

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{n\omega}{2}\int_{x_{1}}^{x_{2}}\frac{{\rm d}x}{\sqrt{E_{n}-W^{2}(x)}},$$

where $x_{1}$ and $x_{2}$ are given by solutions to $E_{n}-W^{2}=0$. We change the integration variable to obtain

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{n\omega}{2}\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{2dW}{\omega\sqrt{E_{n}-W^{2}(x)}}=n\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{\sqrt{E_{n}-W^{2}(x)}}=n\pi.$$

In this case, $W(x,a)=f_{2}(x)+\alpha\,a$, where $\alpha f_{2}-f_{2}^{\prime}=\epsilon$ for nonzero $\alpha$. With a redefinition $a+\varepsilon/\alpha\rightarrow a$, we can set $\varepsilon=0$ and $f_{2}-\epsilon/\alpha\rightarrow f_{2}$. Thus, we have

$\displaystyle W(x,a)=\alpha\,a+f_{2}(x)\quad\mbox{and}\quad\frac{\partial W}{\partial x}=f_{2}^{\prime}(x)=\alpha f_{2}(x)=\alpha\,(W-\alpha\,a)~{}.$

(17)

Since $f_{2}^{\prime}=\alpha f_{2}$, $f_{2}$ cannot be zero at any point, or it would be zero everywhere. Hence $W^{\prime}$ must have a definite sign, which for unbroken supersymmetry must be positive. This implies that $\alpha f_{2}>0$, so $\alpha$ and $f_{2}$ must have the same sign. Without loss of generality ¹¹1Here we have used the fact that the symmetry operations $W\rightarrow-W$ and $x\rightarrow-x$ discussed in Sec. II, do not change the value of the integral of Eq. (13). , we assume $\alpha<0$; consequently, $f_{2}<0$. From Eq. (9), we have $dg/da=-2\alpha^{2}\,a$. Because $dg/da>0$ we must have $a<0$ (and $a+n\hbar<0$ for all bound states), thus $W<\alpha\,a$. By integrating $g$, we get $E_{n}=g(a+n\hbar)-g(a)=\alpha^{2}\,a^{2}-\alpha^{2}(a+n\hbar)^{2}$. Then

$\displaystyle\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{1}{2}\frac{\partial E_{n}}{\partial\hbar}\int_{x_{1}}^{x_{2}}\frac{{\rm d}x}{\sqrt{E_{n}-W^{2}(x,a)}}=\frac{1}{2\alpha}\frac{\partial E_{n}}{\partial\hbar}\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{{\rm d}W}{(W-\alpha\,a)\sqrt{E_{n}-W^{2}(x,a)}}.$

We carry out the integration in the complex $W$ plane, as shown in Fig. 1.

The contour includes a pole at $W_{1}=\alpha\,a$, and a cut from ${-\sqrt{E_{n}}}$ to ${\sqrt{E_{n}}}$. The partial derivative $\frac{\partial I(a,n,\hbar)}{\partial\hbar}$ is then given by

	$\displaystyle\frac{\partial I(a,n,\hbar)}{\partial\hbar}$	$\displaystyle=$	$\displaystyle\frac{1}{4\alpha}\frac{\partial E_{n}}{\partial\hbar}\oint\frac{{\rm d}W}{(W-\alpha\,a)\sqrt{E_{n}-W^{2}(x,a)}}$		(18)
		$\displaystyle=$	$\displaystyle\frac{1}{4\alpha}\frac{\partial E_{n}}{\partial\hbar}\frac{2\pi\,i}{\sqrt{E_{n}-\alpha^{2}\,a^{2}}}=\frac{1}{4\alpha}\frac{\partial E_{n}}{\partial\hbar}\frac{2\pi\,i}{\sqrt{-\alpha^{2}(a+n\hbar)^{2}}}$
		$\displaystyle=$	$\displaystyle\frac{n\pi}{2\alpha^{2}}\,\frac{-2\alpha^{2}(a+n\hbar)}{-(a+n\hbar)}=n\pi~{},$

where we substituted ${\partial E_{n}}/{\partial\hbar}={-2\alpha^{2}(a+n\hbar)n}$.

For $\lambda=0$, since $f_{1}^{\prime}=f_{1}^{2}$, we see that if $f_{1}=0$, at one point, it must be zero at all points. Hence $f_{1}$ must have a definite sign everywhere. Without loss of generality, we choose $f_{1}<0$. Then, since $W=af_{1}(x)+B/a$ must change sign to preserve supersymmetry, we must have $B>0$. Hence, Eq. (20) becomes

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{anB^{2}}{(a+n\hbar)^{3}}\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{\left(W-B/a\right)^{2}\sqrt{E_{n}-W^{2}}}~{}~{},$$

(21)

which integrates to

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{anB^{2}}{(a+n\hbar)^{3}}\frac{a^{2}B\pi}{(B^{2}-a^{2}E_{n})^{3/2}}=n\pi~{},$$

(22)

where we used $E_{n}=\frac{B^{2}}{a^{2}}-\frac{B^{2}}{(a+n\hbar)^{2}}$.

This class splits into two cases: $\lambda>0$ and $\lambda<0$.

We will consider first the case when $\lambda>0$. Because

$$E_{n}=\frac{B^{2}}{a^{2}}-\frac{B^{2}}{(a+n\hbar)^{2}}+\lambda\left[\,a^{2}-(a+n\hbar)^{2}\right]~{},$$

(23)

Eq. (20) becomes

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{a}{2}\left[\frac{2B^{2}n}{(a+n\hbar)^{3}}-2n\lambda(a+n\hbar)\right]\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{(W-W_{1})(W-W_{2})\sqrt{E_{n}-W^{2}}}$$

(24)

where the simple poles $W_{1}=B/a+a\sqrt{\lambda}$ and $W_{2}=B/a-a\sqrt{\lambda}$ are both greater than $\sqrt{E_{n}}$ , and $B>\sqrt{\lambda}(a+n\hbar)^{2}$. To compute this integral we first observe that it can be written as a sum of two integrals which can be evaluated independently. Using Eq. (23), we obtain²²2The computation shown assumes $a>0$. The case $a<0$ yields the same result.

	$\displaystyle J$		$\displaystyle\equiv\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{(W-W_{1})(W-W_{2})\sqrt{E_{n}-W^{2}}}$
			$\displaystyle=\frac{1}{W_{1}-W_{2}}\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{(W-W_{1})\sqrt{E_{n}-W^{2}}}+\frac{1}{W_{2}-W_{1}}\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{(W-W_{2})\sqrt{E_{n}-W^{2}}}$
			$\displaystyle=-\frac{\pi(a+n\hbar)}{2a\sqrt{\lambda}\,\left(B+\sqrt{\lambda}(a+n\hbar)^{2}\right)}+\frac{\pi(a+n\hbar)}{2a\sqrt{\lambda}\left(B-\sqrt{\lambda}\,(a+n\hbar)^{2}\right)}=~{}\frac{\pi(a+n\hbar)^{3}}{aB^{2}-a\lambda(a+n\hbar)^{4}}~{}.$

Substituting $J$ back into Eq. (24) we arrive at

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=n\pi~{}.$$

(25)

Let us now consider the case when $\lambda=-\mu<0$. Then Eq. (20) becomes

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{a}{2}\left[\frac{2B^{2}n}{(a+n\hbar n)^{3}}+2n\mu(a+n\hbar)\right]\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{(W-U_{1})(W-U_{2})\sqrt{E_{n}-W^{2}}}~{},$$

(26)

where the simple poles are $U_{1}=B/a+ia\sqrt{\mu}$ and $U_{2}=B/a-ia\sqrt{\mu}$. Note that this is a real positive integral ³³3 The case $\lambda<0$ requires $W^{\prime}>0$ at every point, so the derivative $(W^{2})^{\prime}=2WW^{\prime}=0$ at only those points where $W=0$, and this happens only at one point $x_{0}$. At $x_{0}$ the second derivative is positive, hence $W^{2}$ has only one extremum, a minimum at $x_{0}$. This implies that the integral $\int_{x_{1}}^{x_{2}}\left[E_{n}-W^{2}(x,a)\right]^{-1/2}dx$ is real and positive, as the integrand is real and positive at every point in the domain. . The complex factorization in Eq. (26) was done in order to carry out the calculations in the complex $W$ plane, as illustrated in Fig. (2). We obtain for the integral in Eq. (26):

	$\displaystyle J\equiv$		$\displaystyle\int_{-\sqrt{E_{n}}}^{\sqrt{E_{n}}}\frac{dW}{(W-U_{1})(W-U_{2})\sqrt{E_{n}-W^{2}}}$
	$\displaystyle=$		$\displaystyle\frac{\pi(a+n\hbar)}{2a\sqrt{\mu}}\left(\frac{1}{\sqrt{\left(\sqrt{\mu}(a+n\hbar)^{2}-iB\right)^{2}}}-\frac{1}{\sqrt{\left(\sqrt{\mu}(a+n\hbar)^{2}+iB\right)^{2}}}\right)$
	$\displaystyle=$		$\displaystyle\frac{\pi(a+n\hbar)}{2a\sqrt{\mu}}\left(\frac{e_{1}}{{\sqrt{\mu}(a+n\hbar)^{2}-iB}}-\frac{e_{2}}{{\sqrt{\mu}(a+n\hbar)^{2}+iB}}\right)~{},$		(27)

where $e_{1},e_{2}=\pm 1$. When substituted into Eq. (26), this gives

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=\frac{1}{2}\pi n\left(e_{1}-e_{2}+\frac{iB(e_{1}+e_{2})}{\sqrt{\mu}(a+n\hbar)^{2}}\right)~{}.$$

(28)

Since this is a real positive integral, we must have $e_{1}+e_{2}=0$ and $e_{1}=1$. We arrive at

$$\frac{\partial I(a,n,\hbar)}{\partial\hbar}=n\pi~{}.$$

(29)

In this case, $f_{1}^{\prime}=f_{1}^{2}$, hence $f_{1}$ cannot be zero at any point. Also, $f_{2}^{\prime}=f_{1}\,f_{2}-\varepsilon$. The homogeneous equation for $f_{2}^{\prime}=f_{1}\,f_{2}-\varepsilon$ is solved by $f_{2}=\alpha f_{1}$. A particular solution is $f_{2}=\frac{1}{2}\varepsilon/f_{1}$. Thus, the superpotential takes the form $W=af_{1}+\alpha f_{1}+\frac{1}{2}\varepsilon/f_{1}=(a+\alpha)f_{1}+\frac{1}{2}\varepsilon/f_{1}\equiv af_{1}+\frac{1}{2}\varepsilon/f_{1}$, where we have redefined the parameter $a$. From Eq. 9, $dg/da=-2\varepsilon>0$, implies $\varepsilon<0$, which requires that $W^{\prime}=af_{1}^{2}-\varepsilon/2>0$, because $a>0$⁴⁴4Unbroken supersymmetry requires that $W=0$ for some value of $f_{1}$, which occurs when $f_{1}^{2}=|\varepsilon|/2a$, so $a>0$.. So

$\displaystyle\frac{\partial I(a,n,\hbar)}{\partial\hbar}$

$\displaystyle=$

$\displaystyle\frac{1}{2}\frac{\partial E_{n}}{\partial\hbar}\int_{x_{1}}^{x_{2}}\frac{dx}{\sqrt{E_{n}-\left(af_{1}+\frac{1}{2}\varepsilon/f_{1}\right)^{2}}}~{}.$

Changing the integration variable to $f_{1}$,

$\displaystyle\frac{\partial I(a,n,\hbar)}{\partial\hbar}$

$\displaystyle=$

$\displaystyle\frac{1}{2}\frac{\partial E_{n}}{\partial\hbar}\int_{f_{L}}^{f_{R}}\frac{df_{1}}{f_{1}^{2}\sqrt{E_{n}-\left(af_{1}+\frac{1}{2}\varepsilon/f_{1}\right)^{2}}},$

where ${f_{L}}$ and ${f_{R}}$ are the turning points on the $f_{1}$ axis, where the square root in the denominator is zero⁵⁵5Since $df_{1}=f_{1}^{2}\,dx$, the relative positioning of ${f_{L}}$ and ${f_{R}}$ remains the same as $x_{1}$ and $x_{2}$.. Moving to the complex $f_{1}$-plane,

	$\displaystyle\frac{\partial I(a,n,\hbar)}{\partial\hbar}$	$\displaystyle=$	$\displaystyle\frac{1}{2}\frac{1}{2}\frac{\partial E_{n}}{\partial\hbar}\oint\frac{df_{1}}{f_{1}^{2}\sqrt{E_{n}-\left(af_{1}+\frac{1}{2}\varepsilon/f_{1}\right)^{2}}}$
		$\displaystyle=$	$\displaystyle\frac{1}{4}{\frac{1}{2}\,(-2\varepsilon)\,}n\oint\frac{df_{1}^{2}}{f_{1}^{2}\sqrt{E_{n}f_{1}^{2}-\left(af_{1}^{2}+\frac{1}{2}\varepsilon\right)^{2}}}=-\frac{1}{4}\varepsilon n(2\pi i)\frac{2}{-i\,\epsilon}=n\pi.$