# How to handle a delta-function divergence from an infinite-dimensional trace of a quantum operator with a general continuous parameter?

Say I am working in the standard Schrödinger-style Hilbert space that corresponds to square-integrable functions on 3-space. Some normal operator $$\hat{A}$$ therefore has an eigenbasis indexed by 3 general continuous parameters, $$|\lambda_1,\lambda_2,\lambda_3\rangle \equiv |\lambda\rangle$$, such that

$$$$\hat{A}|\lambda\rangle = \lambda_1 |\lambda\rangle .$$$$

Now I want to compute the trace of an operator in the same space, call it $$\hat{B}$$, which shares an eigenbasis. Naïvely, this would just be

$$$$\text{Tr}[\hat{B}] = \int d^3\lambda \; \langle\lambda|\hat{B} |\lambda\rangle.$$$$

The issue is that

$$\langle\lambda|\hat{B} |\lambda\rangle = \delta^{(3)}(\lambda-\lambda^\prime) B(\lambda),$$

so the trace has a delta-function divergence. Now, I'm aware from physical grounds in my problem that there ought to be a volume factor, and there is a conventional handwave I've seen in similar problems where one can replace a delta function in momentum with a volume-factor,

$$\delta^{(3)}(p)|_{p=0} \rightarrow V.$$

But this clearly doesn't work for a general parameter. This is clear from the fact that it leads to self-contradiction if you try and use the same basis but indexed by a continuous monotonic functions of the momenta, but for them the remedy is easy: you can just use the relation

$$\delta(f(p)) = \frac{\delta(p)}{|f'(p)|} ,$$

and then do the same replacement. This serves as a clear warning that if I hand-wave this problem away it could cause serious problems: the divergence may not be a mere overall scaling factor but a function of $$\lambda$$ which would completely change the integral. So I want to know, what can I do in the case when I am using some general parameter which labels states which are not also eigenstates of momenta? I've tried explicitly regularising the integral, by selecting a function to act as an envelope, $$f_\epsilon(x)$$, such as

$$$$f(x) = \frac{1}{(2\pi)^{3/2}} e^{-\epsilon x^2/2}, \quad \epsilon^{-3/2} = V,$$$$

then defining the trace as

$$$$\text{Tr}[\hat{B}] \equiv \int d^3 x\; \langle x|\hat{B}|x\rangle f_\epsilon(x)= \int d^3\lambda\; B(\lambda) \left[ \int d^3x \; \langle\lambda|x\rangle \langle x|\lambda\rangle f_\epsilon(x) \right],$$$$

where I've inserted two identity operators $$I = \int d^3 \lambda\; |\lambda\rangle\langle\lambda|$$ between the $$|x\rangle$$ vectors and the operator to get the second identity. I have explicit expressions for my eigenstates in the position-basis, the $$\langle x|\lambda\rangle$$, and I'm fairly sure this in principle could work, but: 1) I've found the integral over $$x$$ with the regularising envelope intractable - the eigenfunctions I'm working with are pretty ugly; and 2) it seems a step removed from the fundamental issue, which should in principle be independent of the vast range of regularising functions I could use and seems to be some sort of basic property of the relation between my basis and the position or momentum bases. Is there are general principle to be used here? I feel the correct function to be using would be some sort of determinant?

I'd be very grateful for any help!

• I don't understand what you mean by the volume factor $V$ not working for a "general parameter" --- can you give an example of the scenario you mean? Any calculation I've done, the volume factor has worked for me. After all it's just a regularization so anything physical shouldn't depend on how you regularize things anyways. Commented Nov 28, 2020 at 15:57
• So the specific example I'm working on is a bit esoteric - I'm working in Schwinger's "proper time" Hilbert space, for a constant applied electric field, so there are actually 4 continuous parameters. Three of them are just $\omega,k_y,k_z$ (with E-field in the x direction), but the fourth is $\lambda = \pm \sqrt{k^2_y+k^2_z \pm ieE}$, a sort of complex-valued mass parameter, which is needed to use the Sauter constant-field solutions as a complete orthogonal basis for bispinor profiles in spacetime. Commented Nov 28, 2020 at 16:40
• I don't think issues with the value being complex-valued should be directly relevant here though, the basic situation would be similar if you used energy as a parameter instead of one of the momenta in a Schrodinger system with a non-trivial applied potential, I think? So long as the system is such that the eigenstates aren't also plane waves, so the energy isn't simply a function of the momenta. Commented Nov 28, 2020 at 16:42
• The thing that makes me wary of the volume factor is that it would clearly get the wrong result if, for instance, I used $k^3$ as a parameter instead of $k$, as you'd get exactly the same integral but with an additional $1/(3k^2)$ from the different normalisation of the ket-vectors. Commented Nov 28, 2020 at 16:45
• Actually I'm wrong in that last comment, it would cancel from $d(k^3) = 3k^2dk$. Maybe I can just use volume... or it's at least a plausible enough hypothesis to proceed with it and see if I get the known result. Commented Nov 28, 2020 at 16:49

The issue is that your $$d^3\lambda$$ needs a density of states factor $$\rho(\lambda)$$ to match the $$\delta^3(\lambda-\lambda')$$: $$\sum_{\lambda_1,\lambda_2,\lambda_3}\to \int d^3 \!\lambda \,\rho(\lambda)$$ For plane waves this is standard and gives you your Volume factor, but If you are doing anything more rechereche you will need to compute it.
• I thought this might be the case. I persuaded myself it wasn't based on the fact that the density of states factor seems to cancel with a rescaling of normalisation of the eigenstates in the case of functions of momenta, e.g. $f=f(k)$, $df/f^\prime = dk$, $\sqrt{f^\prime} |f\rangle = |k\rangle$, so $\int dk \langle k | \hat{B} | k \rangle = \int df \langle f| \hat{B} | f \rangle$. Does this argument not extend to the general case? Commented Nov 27, 2020 at 21:32
• There is still the factor of $dk/2\pi$, and this would be different again if you wrote $dE$ as the measure with $E=k^2/2m$. Commented Nov 27, 2020 at 21:34
• I mean, I don't think it would be, right? I handled the $dk = mdE/k$, which would be different but would cancel with factors from redefining the ket vectors so that $\langle E | E^\prime \rangle = \delta(E-E^\prime)$, $\langle k | k^\prime \rangle = \delta(k-k^\prime)$. I think this is besides the point though! I suspect that there probably is a density of states factor, just from a different origin, so if you know of a very general expression for a density of states I'd be very interested? (I am doing something more rechereche.) Commented Nov 27, 2020 at 21:40