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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?

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Neat! This is really a pure math question, so someone might want move it to math.SE, but I like it. :) –  Michael Brown Oct 17 '13 at 4:27
    
Can you show explicitly that your $\psi$ is indeed an eigenvalue of your hamiltonian? –  BMS Oct 17 '13 at 5:59
    
@BMS Like, you want me to type it out? It's a quick calculation. You just have to be careful when calculating $d/dx |x|^{-3/2} = (-3/2){\rm sgn}(x)|x|^{-5/2}$. –  santa claus Oct 17 '13 at 6:07
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This problem is Example 3 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, arXiv:quant-ph/9907069, see p. 6, 39, 45-47. –  Qmechanic Oct 17 '13 at 7:22
    
@Qmechanic That's an interesting list of questions worth reading –  hwlau Oct 17 '13 at 8:15
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3 Answers 3

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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.

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Just a comment on your nice answer. Actually to apply the technology of deficiency indices it is not necessary to require that $H$ is closed, symmetric (not only Hermitean) is enough. Symmetric means Hermitean with dense domain. If $H$ is symmetric, $A$ is a self adjoint extension of $H$ if and only if it is a self adjoint extension of the closure $\overline{H}$ that is closed and symmetric. Moreover $n_\pm(H)= n_\pm(\overline{H})$, so dealing with $H$ itself as you did (and $H$ is not closed) is completely sufficient. –  V. Moretti Feb 27 at 7:29
    
@V.Moretti Thank you very much for that clarification; that makes me feel better about the answer. Since I, and I assume others, not most certainly not nearly as familiar with functional analysis are you are, would be mind pointing to a reference or two which might discuss the observation you have made in some detail in case you happen to know any? –  joshphysics Feb 27 at 20:22
    
Hi Josh. Unfortunately it seems that the literature is divided into two sets. A set of textbooks like Reed-Simon, Rudin, Prugovecki etc assumes that the operator must be symmetric and closed, another set like Dunford-Schwartz, Teschl, etc. assumes only symmetry. I never found a book discussing the interplay in details, so I produced my own proofs (actually very elementary) and I am inserting in the second edition of my book. Please send me an email to my institutional address (moretti@science.unitn.it) Tomorrow, I will send copy of the proofs. –  V. Moretti Feb 27 at 21:40
    
@V.Moretti Ah I see, that's unfortunate. I will email you; thank you! –  joshphysics Feb 27 at 21:42
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The problem with this hamiltonian is that there is a difference between symmetric/Hermitian operators and self-adjoint operators. It looks like a nit-picky mathematician's poking holes into everything, but it is in fact important:

In general, the domains of $\hat{A}$ and $\hat{A}^\dagger$ do not coincide. If $\hat{A}=\hat{A}^\dagger$ on $D(\hat{A})$, then $D(\hat{A})\subseteq D(\hat{A}^\dagger)$ holds and $\hat{A}$ is called symmetric or Hermitian. If, in addition, $D(\hat{A}^\dagger)=D(\hat{A})$, then $\hat{A}$ is called self-adjoint.

The important existence and reality theorems for eigenvalues and eigenvectors are usually only for self-adjoint operators. This is made clear in page 13 of your textbook. While your operator is indeed symmetric, it is unlikely to be self-adjoint.

More specifically, $H$ is densely defined and therefore has an adjoint $H^\dagger$, which is an operator on some domain $D(H^\dagger)$ which satisfies $\langle\phi|H\psi\rangle=\langle H^\dagger\phi|\psi\rangle$ for all $\psi\in D(H)$ and $\phi\in D(H^\dagger)$. Your real job is characterizing the domains of both operators and seeing if they coincide, or figuring out whether $H$ can be extended to a larger domain such that the adjoint's domain will coincide with the original domain. None of that is particularly easy.


The thing is, though, that these mathematical troubles very rarely come on their own and are usually accompanied by trouble in the corresponding classical problem. This is beautifully made clear, together with a clear exposition of the necessary mathematical facts, in the paper

Classical symptoms of quantum illnesses. Chengjun Zhu and John R. Klauder. Am. J. Phys. 61 no. 7, 605 (1993).

The main point is that unless the classical problem has well-defined solutions for all time and for all initial conditions, you really have no business complaining about unexpected behaviour in the quantum counterpart.

For your model, the classical hamiltonian $H=\frac12(q^3p+pq^3)=q^3 p$ produces the Hamilton equations $$ \left\{ \begin{array}{} \dot p=-\frac{\partial H}{\partial q}=&3q^2 p,\\ \dot q=\phantom{-}\frac{\partial H}{\partial p}=&q^3. \end{array} \right. $$ These are fairly easy to solve, and the solutions are not well behaved: $$ \left\{ \begin{array}{} \frac{1}{2q^2}&=t_0-t,\\ \frac{p_0^2}{p^2}&=(t_0-t)^3, \end{array} \right. $$ where $t_0$ and $p_0$ are constants of integration. Note, in particular, that there are no (real) solutions after a certain time $t_0=t_\text{in}+\frac{1}{2q(t_\text{in})^2}$. How, then, are you expecting reasonable physics out of the quantized version of this?


Finally, as a coda, let me address Qmechanic's very interesting comment. It is true that for a given physical system, you will have a single hamiltonian and many other physical observables. How, then is one to make sense of this construction for an arbitrary observable? I would counter that any arbitrary observable can be considered as a hamiltonian, and that things like the spectrum do follow from properties like the corresponding time evolution.

Take some arbitrary self-adjoint, as-nice-as-necessary physical observable $\hat A$ in a physical system with state space $\mathcal H$. Even if it does not make physical sense, you can definitely postulate the flow associated with that observable, i.e. the curve $t\to|\psi(t)\rangle$ in $\mathcal H$ that obeys the Schrödinger-like equation $$i\partial_t|\psi(t)\rangle=\hat A|\psi(t)\rangle.$$ If you want to bring the spectrum into play, you can solve this equation by decomposing the flow into the observable's eigenbasis, so $|\psi(t)\rangle=\sum_n\psi_n e^{-ia_n t}|n\rangle$, where $\hat A|n\rangle=a_n|n\rangle$ and $\psi_n=\langle n|\psi(0)\rangle$. (Here one needs to assume that $|\psi(0)\rangle$ is "general" or "random" enough that all the $\psi_n$ are nonzero.

The question, though, is how can you extract the spectrum from the time evolution? The answer to that is to Fourier transform into the frequency domain: define $$|\tilde\psi(\omega)\rangle=\int_{-\infty}^\infty\text dt \, e^{i\omega t}|\psi(t)\rangle$$ and see what the eigenvector decomposition does to it: $$|\tilde\psi(\omega)\rangle =\int_{-\infty}^\infty\text dt \, e^{i\omega t}\sum_n\psi_n e^{-ia_n t}|n\rangle =\sum_n\psi_n|n\rangle \int_{-\infty}^\infty\text dt \, e^{i(\omega -a_n) t} =\sum_n\delta(\omega -a_n)\psi_n|n\rangle . $$ That is: the Fourier transform of the $\hat A$-induced flow is equal to a number of spikes at the eigenvalues of $\hat A$, and the eigenvectors are the coefficients.

To put it another way, this offers a means of obtaining the spectrum from the time evolution: solve it in some way, Fourier transform the solution, and then read the eigenvalues from the support of the transform and the eigenvectors from the value at those points. In fact, this is a useful technique and it has quite wide deployment in numerical solutions of certain classes of problems. (For more information, see David Tannor's excellent textbook Introduction to Quantum Mechanics: A Time-Dependent Perspective (Ebookee).)

... and, of course, if you tried to diagonalize in this way an operator with the kind of problems described above, you'd be heading straight for trouble. Surely it is unreasonable to ask the quantum flow to be well-behaved when the classical flow isn't!

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Just a small comment from the Devil's advocate: The Hamiltonian operator (which describes the time evolution of the system) could in principle be different from OP's operator ${\bf H}$ (which is then just a particular observable of the system). –  Qmechanic Oct 17 '13 at 16:08
    
@Qmechanic See response in the post. –  Emilio Pisanty Oct 17 '13 at 19:59
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@EmilioPisanty, interesting answer, especially the second paper. I was wondering if you could explain exactly how the domains $D({\bf H})$ and $D({\bf H^{\dagger}})$ differ. After all, it seems (prima facie) $D({\bf H})=D({\bf T})\cap D({\bf T^{\dagger}})=D({\bf H^{\dagger}})$. –  santa claus Oct 18 '13 at 1:24
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@EmilioPisanty furthermore, what is the erroneous step in the following calculation? $ \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^{*}.$ All inner products are presumably evaluated in the same Hilbert space. –  santa claus Oct 18 '13 at 1:26
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+1 very nice! Almost makes me wish I did functional analysis! (alas our maths department got gutted to a bare bones "mathematics for engineers" type curriculum; a few of us managed to scrape through before most of the gutting but that still left certain gaps for us :S) –  Michael Brown Oct 18 '13 at 5:59
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This problem is posed and solved by: François Gieres his article: Mathematical surprises and Dirac’s formalism in quantum mechanics .

It is given in example 3 page 6 and its solution is given in: (3a) page 39 and continued on pages 45 and 46 where the von Neumann’s criterion for self adjontiness is formulated and applied to the problem.

As explained in the article, the basic reason that the function $\psi_{\lambda}$ is not an eigenfunction of $\mathbf{H}$ is that it does not belong to the Schwartz space $\mathcal{S}(\mathbf{R})$ (of rapidly decreasing functions) which the mutual domain of definition of both $\hat{x}$ and $\hat{p}$.

In the second part it is shown that $\psi_{\lambda}$ is an eigenfunction of $\mathbf{H^{\dagger}}$ and there is no way to extend the domain of $\mathbf{H}$ to make it self adjoint.

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