quantum mechanics - A "Hermitian" operator with imaginary eigenvalues

Let $${\bf H}=\hat{x}^3\hat{p}+\hat{p}\hat{x}^3$$ where $\hat{p}=-id/dx$. Clearly ${\bf H}^{\dagger}={\bf H}$, because ${\bf H}={\bf T} + {\bf T}^{\dagger}$, where ${\bf T}=\hat{x}^3\hat{p}$. In this sense ${\bf H}$ is "formally" self-adjoint.

It turns out that

$$ \psi_{\lambda}(x)=\frac{1}{|x|^{3/2}}e^{-\lambda/4x^2} \in L^2[-\infty,\infty]$$ is an eigenfunction of ${\bf H}$. In fact

$${\bf H}\psi_{\lambda}(x)=-i\lambda\psi_{\lambda}(x).$$

If we go through the usual proof that Hermitian operators have real eigenvalues, we see that something has to go wrong in the following steps:

$$ \langle\psi_{\lambda}(x)|{\bf H}\psi_{\lambda}(x)\rangle= \langle{\bf H}^{\dagger}\psi_{\lambda}(x)|\psi_{\lambda}(x)\rangle=\langle{\bf H}\psi_{\lambda}(x)|\psi_{\lambda}(x)\rangle=\langle\psi_{\lambda}(x)|{\bf H}\psi_{\lambda}(x)\rangle^{*}.$$

Clearly ${\bf H}\psi_{\lambda}(x) = -i\lambda\psi_{\lambda}(x)\in L^{2}.$ So all these operations seem to be well-defined.

Our book suggests that we look at ${\bf T}\psi_{\lambda}(x)$. It turns out that ${\bf T}\psi_{\lambda}(x),{\bf T}^{\dagger}\psi_{\lambda}(x) \notin L^{2}$. So when we are writing $$\langle\psi_{\lambda}(x)|{\bf H}\psi_{\lambda}(x)\rangle$$

we are really writing

$$\langle\psi_{\lambda}(x)|({\bf T} + {\bf T}^{\dagger})\psi_{\lambda}(x)\rangle=\langle\psi_{\lambda}(x)|{\bf T}\psi_{\lambda}(x)\rangle+\langle\psi_{\lambda}(x)|{\bf T}^{\dagger}\psi_{\lambda}(x)\rangle$$

where these inner products are no longer defined. Does the fact that these two inner products are undefined lead to this seemingly (and "formally") self-adjoint operator having imaginary eigenvalues? And how?

Answer

Although Emilio's answer is insightful, I don't think it directly answers your question. I'll attempt to do that here. This answer proceeds in two parts:

1. 
   

   We'll show that the operator you try to write down is hermitian with appropriate domain, but that it is not self-adjoint and has no self-adjoint extensions.

   
   
   


2. 
   

   We'll show that your formal manipulations have errors.

   
   

Part 1.

We set $\hbar=1$ for convenience throughout, and let $S(\mathbb R)$ denote Schwartz space. Recall that the $L^2(\mathbb R)$ inner product is defined as follows: \begin{align} \langle \psi, \phi\rangle = \int_{-\infty}^{\infty} dx\, \psi^*(x) \phi(x) \end{align}

We are going to use the following definition which appears on page 138 of Reed and Simon's Methods of Modern Mathematical Physics Volume II (Fourier analysis, self-adjointness):

Definition. For a symmetric operator $A$, we define its deficiency indices by \begin{align} n_+(A) &= \dim\mathrm{ker}(iI-A^\dagger) \\ n_-(A) &= \dim\mathrm{ker}(iI+A^\dagger) \\ \end{align}

We are also going to need the following result which is part of a corollary on page 141 of Reed and Simon:

Lemma. Let $A$ be a closed hermitian operator with deficiency indices $n_+(A)$ and $n_-(A)$, then $A$ is self-adjoint if and only if $n_+(A) = 0 = n_-(A)$, and $A$ has at least one self-adjoint extension if and only of $n_+(A) = n_-(A)$.

With this lemma, we can prove the following claim. What we are going to prove here tells us that there is no way to define $H$ self-adjoint operator on some domain in $L^2(\mathbb R)$.

Claim. The operator $H$ with domain $D(A) = S(\mathbb R)$ defined by \begin{align} H\psi(x) = -ix^3\frac{d\psi}{dx}(x) -i \frac{d}{dx} (x^3\psi(x)) \end{align} is closed and hermitian but not self-adjoint. Furthermore, $H$ has no self-adjoint extensions on $L^2(\mathbb R)$.

Proof. We will show that $A$ is hermitian and closed but that $n_-(H) = 1$ while $n_+(H) = 0$. The desired result then follows immediately from the lemma. To show that $H$ is hermitian, it suffices to show that $\langle \psi, H\phi\rangle = \langle H\psi, \phi\rangle$ for all $\phi,\psi\in D(H) = S(\mathbb R)$. We have \begin{align} \langle \psi, H\phi\rangle &= \int_{-\infty}^\infty dx\, \psi^*\left(-ix^3\frac{d\phi}{dx} -i \frac{d}{dx} (x^3\phi)\right) \\ &= -2i\psi^*x^3\phi\Big|_{-\infty}^\infty +i\int_{-\infty}^\infty dx\left(\frac{d}{dx}(\psi^*x^3)+\frac{d\psi^*}{dx}(x)x^3\right)\phi \\ &= \int_{-\infty}^\infty dx\left(-i\frac{d}{dx}(\psi(x)x^3)-i\frac{d\psi}{dx}x^3\right)^*\phi \\ &= \langle H\psi, \phi\rangle. \end{align} The boundary term in the second equality vanished because $\phi$ is rapidly decreasing. This operator is closed (admittedly this is actually something I haven't been able to prove). Because $H$ is hermitian, its adjoint $H^\dagger$ has the same action on elements of its domain as $H$ itself. Moreover, let $D'$ the set of all $\psi\in L^2(\mathbb R)$ for which $H\psi$ is well-defined and also an element of $L^2(\mathbb R)$. Then the computation be performed above to demonstrate hermiticity shows that if $\phi\in D'$, then $\langle\phi, H\psi\rangle = \langle H\phi, \psi\rangle$ for all $\psi\in D(H)$, so $D(H^\dagger) = D'$. In particular, this domain is larger than that of $H$ which is therefore not self-adjoint.

Now if $\psi\in \mathrm{ker}(iI - A^\dagger)$, then $\psi$ obeys the following differential equation: \begin{align} i\psi - \left(-ix^3\frac{d\psi}{dx} -i \frac{d}{dx} (x^3\psi)\right)=0 \end{align} This differential equation can be simplified to give \begin{align} (1+3x^2)\psi + 2x^3\frac{d\psi}{dx} =0 \end{align} for $x> 0$ and $x<0$, we can separate variables and integrate to solve this differential equation. The result is (I used mathematica for this) \begin{align} \psi_>(x) &= \frac{e^{1/(4x^2)}}{x^{3/2}} + c_> \\ \psi_<(x) &= \frac{e^{1/(4x^2)}}{(-x)^{3/2}} + c_< \end{align} These solutions both diverge at the origin, so our differential equation does not yield an $L^2(\mathbb R)$ solution. This gives $\mathrm{ker}(iI - A^\dagger) = \{0\}$ so $n_+(H) = 0$. On the other hand, if $\psi\in \mathrm{ker}(iI + A^\dagger)$ then \begin{align} i\psi + \left(-ix^3\frac{d\psi}{dx} -i \frac{d}{dx} (x^3\psi)\right)=0 \end{align} This differential equation can be simplified to give \begin{align} (1-3x^2)\psi - 2x^3\frac{d\psi}{dx} =0 \end{align} which admits the normalized solution \begin{align} \psi(x) =\left\{ \begin{array}{cc} \frac{1}{\sqrt{2}}\frac{1}{|x|^{3/2}}e^{-1/(4x^2)} &, x\neq 0 \\ 0 &, x=0 \end{array}\right. \end{align} In fact, this is (up to normalization) exactly the function you wrote down in the original question statement. This function is in $D(H^\dagger)$. It follows that $\mathrm{ker}(iI + A^\dagger) = \mathrm{span}\{\psi_1\}$ so that $n_-(H) = 1$, as desired $\blacksquare$.

Part 2.

As for you manipulations, even if you were to enlarge the domain of $H$ to include $\psi_\lambda$, they would be wrong. Notice, for example, that you got a nonzero boundary term in the following computation: \begin{align} \langle \psi_\lambda, H\psi_\lambda\rangle &= \int_{-\infty}^\infty dx\, \psi_\lambda^*\left(-ix^3\frac{d\psi_\lambda}{dx} -i \frac{d}{dx} (x^3\psi_\lambda)\right) \\ &= -2i\psi_\lambda^*x^3\psi_\lambda\Big|_{-\infty}^\infty +i\int_{-\infty}^\infty dx\left(\frac{d}{dx}(\psi_\lambda^*x^3)+\frac{d\psi_\lambda^*}{dx}(x)x^3\right)\psi_\lambda \\ &= -2i\,\mathrm{sgn}(x)e^{-\lambda/(2x^2)}\Big|_{-\infty}^\infty +i\int_{-\infty}^\infty dx\left(\frac{d}{dx}(\psi_\lambda^*x^3)+\frac{d\psi_\lambda^*}{dx}(x)x^3\right)\psi_\lambda \\ &= -4i +\int_{-\infty}^\infty dx\left(-i\frac{d}{dx}(\psi(x)x^3)-i\frac{d\psi}{dx}x^3\right)^*\psi_\lambda \\ &= -4i+\langle H\psi_\lambda, \psi_\lambda\rangle \end{align} which we could have seen more easily by simply noting that \begin{align} \langle\psi_\lambda, \psi_\lambda\rangle = \frac{2}{\lambda} \end{align} which gives \begin{align} \langle \psi_\lambda, H\psi_\lambda\rangle &= -i\lambda \langle\psi_\lambda, \psi_\lambda\rangle = -2i \\ \langle H\psi_\lambda, \psi_\lambda\rangle &= (-i\lambda)^* \langle\psi_\lambda, \psi_\lambda\rangle = 2i \end{align} In particular, both of these computations show that \begin{align} \langle \psi_\lambda, H\psi_\lambda\rangle \neq \langle H\psi_\lambda, \psi_\lambda\rangle \end{align} in contradiction with what you claim.