Suppose that $\lvert \psi_n \rangle$ are the eigenvectors of a Hamiltonian, $\hat{H}$, which span some Hilbert space $\mathcal{H}$ and satisfy $$\hat{H}\lvert \psi_n \rangle = E_n \lvert \psi_n \rangle.$$

Since all Hilbert spaces are vector spaces and thus linear, I would expect that for any operator $$\hat{O} = \hat{A} + \hat{B},$$ we should have that $$\langle \psi_n \rvert \hat{O} \lvert \psi_n \rangle = \langle \psi_n \rvert \hat{A}+\hat{B} \lvert \psi_n \rangle = \langle \psi_n \rvert \hat{A} \lvert \psi_n \rangle + \langle \psi_n \rvert \hat{B} \lvert \psi_n \rangle.$$

Are there any pathological cases where this simple identity fails? For example, when the Hamiltonian contains some distribution like $\delta(x)$?

Proposed counter-example: this question is motivated by my calculations following this paper on the systems of two identical atoms with s-wave contact interaction in a harmonic trap. The authors provide an analytical solution for this problem which I will employ. All following calculations will be expressed in natural harmonic oscillator units.

After decoupling the center of mass and relative motion, the relative Hamiltonian is given by $$H_\text{rel}= -\frac{1}{2} \nabla_r^2 +\frac{1}{2}r^2 + \sqrt{2} \pi a \delta^3(\vec{r}) \frac{\partial}{\partial r} r = H_\text{osc} + V_\text{pseudo}.$$ The solution (for example) for the ground state with energy $E_0 = -\frac{1}{2}$, corresponding to $\nu = -1$, is $$\psi_0(r) = \frac{e^{\frac{r^2}{2}}}{2 \pi^{3/4} \sqrt{\ln{2}}}\big(\text{ExpIntegralE}(\frac{1}{2},r^2)\big), $$ with $\lvert \vec{r} \rvert = r =\lvert \frac{1}{\sqrt{2}}(\vec{r}_1 - \vec{r}_2) \rvert$ and $\text{ExpIntegralE}(x)$ is the exponential integral function in Mathematica. The scattering length, $a$ associated with this solution is $a = \sqrt{\frac{\pi}{2}}$

It is easy to verify that in fact $$H_{rel} \psi_0(r) = -\frac{1}{2} \psi_0(r),$$ provided that one correctly accounts that the term proportional to $\delta^3(\vec{r})$ is cancelled by a corresponding term coming from the Laplacian. Since $$\psi_0(r) \sim_{r \to 0} \frac{1}{r},$$ we have to treat the Laplacian in a distributional sense such that $$\nabla^2_r \psi_0(r) = \nabla^2_{r \neq 0}\psi_o(r) -4 \pi \delta^3(\vec{r}) \text{Res}(\psi_0(0)).$$ $\text{Res}(f(r))$ is the residue of $f$ at $r$.

We should straight-forwardly have that $$\langle \psi_0 \rvert \hat{H}_{rel} \lvert \psi_0 \rangle = \langle \psi_0 \rvert E_0 \lvert \psi_0 \rangle = E_0.$$ However, when I break apart the Hamiltonian operator, I find that I get additional non-zero terms coming from the terms proportional to $\delta^3(\vec{r})$.

In particular, I find that $$ 4 \pi \int_0^{\infty} \psi_0(r) (-\frac{1}{2} \nabla_{r \neq 0}^2 +\frac{1}{2}r^2) \psi_0(r) r^2 \ dr = E_0,$$ whereas the remaining terms give $$ \int \psi_0(r) \Big(-2 \pi \delta^3(\vec{r})\text{Res}(\psi_0(0)) + \sqrt{2} \pi \sqrt{\frac{\pi}{2}} \delta^3(\vec{r}) \frac{\partial}{\partial r}\big(r \psi_0(r)\big) \Big) \ d^3 r = \frac{\pi}{4 \ln{2}}. $$

I cannot justify ignoring these terms since they were crucial in proving that $\psi_0$ is a solution to the Schrodinger equation with the correct eigenvalue, nor can I find any unaccounted for counter-terms.

  • $\begingroup$ The "natural harmonic oscillator units" is vague. It would be much better to say exactly what conventions are used. Also, it makes no sense to say that you're working in any particular system of units because equations are not written "in units". A unit is a measure like 'meter' or 'second'. Equations are usually written in terms of abstract quantities where units don't exist. Perhaps you mean to comment on the dimensions of the various quantities in the equation? If so, please explicitly say how the variables are defined. $\endgroup$
    – DanielSank
    Commented Nov 9, 2015 at 4:22
  • 1
    $\begingroup$ @DanielSank My convention is the same as the relevant paper, namely that all quantities which ordinarily have dimension of energy should be though of as being divided by $\hbar \omega$ and those ordinarily with dimensions of length divided by $\sqrt{\hbar/m \omega}$. $\endgroup$ Commented Nov 9, 2015 at 5:03
  • $\begingroup$ @DanielSank And I don't mean to to be confrontational, but this isn't the first time I've gotten this objection here about saying 'in units...' Every graduate student and professor I have spoken with understands the intended meaning of this phrase; it even appears in the published literature (for example, the paper I linked). Is it the same as the ordinary meaning of 'units'? No, as you point out. But, saying it makes no sense is pedantic, in the same way that prescriptivists are pedantic when they say it makes no sense to use 'literally' figuratively. $\endgroup$ Commented Nov 9, 2015 at 5:11
  • $\begingroup$ Don't worry about confrontational tone. It makes sense that folks who work in any particular field understand common terminology even if it's vague. I don't see that as a reason not to improve it. I was always super confused about this "in units of..." terminology as a student and I know at least one other professional physicist who always finds it really confusing. In my own notes I write things like "$H$ means $H/\hbar$" etc. $\endgroup$
    – DanielSank
    Commented Nov 9, 2015 at 5:27
  • 1
    $\begingroup$ The "usual" way would be to let all operators act on Schwartz functions, and seek distributional solutions in the dual Schwartz space. However, you seem to be saying that $\psi_0(r)$ blows up at $r=0$, which would mean it isn't a Schwartz function, so this probably doesn't work here. The reason I'm talking about domains is that contradictions such as the one between your different calculations here often arise because operators were applied to stuff they are not actually defined on. $\endgroup$
    – ACuriousMind
    Commented Nov 9, 2015 at 19:35

1 Answer 1


Your last integral is $$ J = \int{d^3r \;\psi_0(r) \left[-2\pi\delta(\vec{r})Res(\psi_0(0)) + \sqrt{2}\pi a\delta(\vec{r})\frac{\partial}{\partial r}(r\psi_0(r))\right]} =\\ = 4\pi \int_0^\infty{dr\; r^2 \psi_0(r) \left[-2\pi\frac{\delta(r)}{2\pi r^2}Res(\psi_0(0)) + \sqrt{2}\pi a\frac{\delta(r)}{2\pi r^2}\frac{\partial}{\partial r}(r\psi_0(r))\right]} \\ = 4\pi\int_0^\infty{dr\;\psi_0(r)\left[-\delta(r)Res(\psi_0(0)) + \frac{ a}{\sqrt{2}}\delta(r)\frac{\partial}{\partial r}(r\psi_0(r))\right]} = \\ = 4\pi \int_0^\infty{dr\;\delta(r)\psi_0(r)\left[-Res(\psi_0(0)) + \frac{ a}{\sqrt{2}}\frac{\partial}{\partial r}(r\psi_0(r))\right]} $$ where I used $\delta(\vec{r}) = \frac{\delta(r)}{2\pi r^2}$ (see here). For $\psi_0(r) \approx \frac{\alpha}{r}$ when $r\rightarrow 0$ the last form above becomes $$ J = 4\pi \int_0^\infty{dr\;\delta(r)\left(\frac{\alpha}{r}\right)\left[-Res(\psi_0(0)) + \frac{ a}{\sqrt{2}}\frac{\partial}{\partial r}\alpha\right]} = \\ = - 4\pi \alpha \;Res(\psi_0(0)) \int_0^\infty{dr\;\frac{\delta(r)}{r}} = \\ = 4\pi \alpha \;Res(\psi_0(0)) \int_0^\infty{dr\;\delta'(r)} = 4\pi \alpha \;Res(\psi_0(0)) \frac{d}{dr}(1) = 0 $$ The first expression on the last line uses again a delta-identity, this time $\delta(r) = -r \delta'(r)$. The result obviously holds in distributional sense. This leaves $$ \langle \psi_0|\hat{H}_0|\psi_0\rangle = E_0 $$ as it should. I might have missed a factor of $\frac{1}{2}$ in the last integral over $\delta'$, but it wouldn't matter.


The above takes into account only the dominant contribution to $\psi_0$ for $r\rightarrow 0$ before proceeding with the integral. It has been objected that this is not a valid procedure since taking into account zero and higher order terms in $\psi_0$ would render the integral nonzero. For instance, using the identity $\delta(r) = -r \delta'(r)$ in the 4th line from the top above yields $$ -4\pi \int_0^\infty{dr\;\delta'(r)\left(r\psi_0(r)\right)\left[-Res(\psi_0(0)) + \frac{ a}{\sqrt{2}}\frac{\partial}{\partial r}(r\psi_0(r))\right]} = \\ = 4\pi \lim_{r\rightarrow 0}\frac{\partial}{\partial r}\left[\left(r\psi_0(r)\right)\left(-Res(\psi_0(0)) + \frac{ a}{\sqrt{2}}\frac{\partial}{\partial r}(r\psi_0(r))\right)\right] \neq 0 $$ since both $r\psi_0$ and $\frac{\partial}{\partial r}r\psi_0$ retain terms linear in $r$ as $r\rightarrow 0$. If this is indeed correct, then $\langle \psi_0|\hat{H}_0|\psi_0\rangle = E_0$ is compromised.

I don't think this is the case, for the following reason. The original equation $\hat{H}_0\psi_0 = E_0\psi_0$ is satisfied provided the laplacian is taken in the distributional sense and the singular pseudo-potential term is taken to mean $$\sqrt{2}\pi a\delta(\vec{r})\frac{\partial (r\psi_0)}{\partial r}(0)$$ This is just the standard interpretation of the singular term. If it is taken into account from the beginning when writing the integral for $\langle\psi_0|\hat{H}_0|\psi_0\rangle$, the integral $J$ discussed above would not appear in the first place and everything would check out trivially. So the terms that make $J$ non-trivial here are the same ones that are wiped out under the singularity in the original differential equation. In view of this, we can say that dismissing higher order contributions in $\psi_0$ when integrating the singularities functions in the same way: it wipes out irrelevant terms accompanying the singularity. On the other hand, including these terms in the integral is no longer as benign as carrying them along in the context of the differential equation. This is simply because they change the nature of the function on which the singular distribution acts. So basically the problem comes down to $$ \int{d^3r \;\psi_0(r) \left[-2\pi\delta(\vec{r})Res(\psi_0(0)) + \sqrt{2}\pi a\delta(\vec{r})\frac{\partial}{\partial r}(r\psi_0(r))\right]} \\ \neq \\ \int{d^3r \;\psi_0(r) \left[-2\pi\delta(\vec{r})Res(\psi_0(0)) + \sqrt{2}\pi a\delta(\vec{r})\frac{\partial(r\psi_0)}{\partial r}(0)\right]} $$

  • $\begingroup$ You've shown that taking the first term in the Laurent expansion of $\psi_0(r)$ about $r=0$ does not give any contribution. The problem is that you need to also include the constant terms, in particular $\psi_0(r) \propto \frac{1}{r} - \frac{\sqrt{2}}{a} + \mathcal{O}(r)$. The $\mathcal{O}(r)$ terms can be neglected, but the constant terms cannot since they give a non-zero contribution at $r=0$ $\endgroup$ Commented Nov 9, 2015 at 9:53
  • $\begingroup$ Actually, I found an even more grave error. In the 4th line, you distribute the wavefunction inside the parenthesis, but you only apply it to the first term. A factor of the wavefunction is missing from the left of the remaining 2 terms. $\endgroup$ Commented Nov 9, 2015 at 21:16
  • $\begingroup$ Further, I should say that I was wrong to suggest to neglect the $\mathcal{O}(r)$ terms. You need to carry these around because they cancel with some of the $\mathcal{O}(1/r)$ terms $\endgroup$ Commented Nov 9, 2015 at 21:31
  • $\begingroup$ The thing in the 4th line was a typo, but you had a point about the effect of higher order terms. Please see the note I added on the issue. $\endgroup$
    – udrv
    Commented Nov 10, 2015 at 8:46

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