# Partial trace over continuous degrees of freedom

Let us suppose we have some quantum system whose Hilbert space admits a bipartition $$\mathscr{H}\simeq \mathscr{H}_A\otimes \mathscr{H}_B$$. Let $$|n\rangle_A$$ be a basis of $$\mathscr{H}_A$$ and $$|m\rangle_B$$ be a basis of $$\mathscr{H}_B$$. Then $$|n,m\rangle_{A,B}$$ is a basis of the composite system.

A general mixed state $$\rho$$ can be written always as $$\rho=\sum \rho_{nmn'm'}|n,m\rangle_{AB}\langle n',m'|$$

And we can define the partial trace over $$\mathscr{H}_A$$ to be $$\operatorname{Tr}_A\rho=\sum \rho_{mm'}|m\rangle_B\langle m'|,\quad \rho_{mm'}=\sum_n \rho_{nmnm'}$$

This seems to work well if $$|n\rangle_A,|m\rangle_B$$ are discrete basis.

Now, in Quantum Mechanics we often work with "continuous bases". These are not really rigorous since the "position eigenstates" $$|x\rangle$$ are not well defined. Still, this is very useful.

So now suppose that for $$\mathscr{H}_A$$ we take such a $$|x\rangle_A$$ basis. Then a basis of $$\mathscr{H}$$ is $$|x,m\rangle_{A,B}$$ and the state is $$\rho=\sum \int \rho_{mm'}(x,x') |x,m\rangle_{A,B}\langle x',m'|$$

And we can in analogy take a partial trace $$\operatorname{Tr}_A\rho=\sum \rho_{mm'}|m\rangle_B\langle m'|,\quad \rho_{mm'}=\int \rho_{mm'}(x,x)dx$$

It seems everything is fine, but now take the pure state $$\rho = |z,\sigma\rangle\langle z,\sigma|$$ (for instance, a particle located at $$z$$ with spin $$S_z$$ being $$\sigma$$). If we evaluate $$\rho_{mm'}(x,x')=\delta_{m\sigma}\delta_{m'\sigma}\delta(x-z)\delta(x'-z)$$

and the partial trace involves $$(\delta(x-z))^2$$ which is ill-defined.

Still, there should certainly be a way of talking about partial traces when continuous degrees of freedom like position and momentum are involved.

How is that done correctly? How to avoid this issue of getting a delta function squared when taking a partial trace over continuous degrees of freedom?

• With your $\delta$ convention, you have the same problem with pure states when you try to normalize them. Intuitively, I'd say those should be square roots of $\delta$s (since the $\delta$ should appear in the probabilities, not the amplitudes). Nov 2 '19 at 23:50

I wouldn't say that the square of the delta function is ill-defined, there are ways to define functions of $$\delta$$, see e.g. this question. Rather, the problem is that it's the wrong result.
You are essentially looking for a formal way to write a $$\rho(x,y)$$ such that: $$\int dxdy \,\rho(x,y) |x\rangle\!\langle y|=|z\rangle\!\langle z|.\tag A$$ I guess you got the expression with the square of the Dirac delta by using $$\langle x|z\rangle=\delta(z-x)$$. This expression is, however, incorrect: integrating $$\langle x|z\rangle$$ over $$x$$ (or $$z$$) does not give you the identity. It is therefore more correct to use something like $$\langle x|z\rangle=\sqrt{\delta(x-z)},\tag B$$ so that $$\int dx |\langle x|z\rangle|^2=1$$, as it should be. There is no problem in defining this object if you think of distributions as "underdetermined functions" (that is, functions only defined via their integral properties): $$\sqrt{\delta(x)}$$ is a function whose square integrates to the identity and is zero outside of $$x=0$$. On a more rigorous level, you can make sense of these objects e.g. via holomorphic calculus, see this question.
Using (B), you get $$\rho(x,y)=\delta(z-x)^{1/2}\delta(z-y)^{1/2}$$. Integrating, you get $$\int dxdy\,\delta(z-x)^{1/2}\delta(z-y)^{1/2}|x\rangle\!\langle y|=|x\rangle\!\langle x|.$$ You can probably prove that this holds using specific properties of delta functions. A simple, if handwavy, argument is is based on understanding the delta functions as the limit of functions concentrated at $$x=0$$. Using rectangle functions for ease of calculation, you then have $$\int dxdy \, \delta(z-x)^{1/2}\delta(z-y)^{1/2}f(x,y) \simeq f(z,z) \Delta \frac{1}{\sqrt\Delta}\frac{1}{\sqrt\Delta}=f(z,z).$$ What I'm doing here is using the fact that $$\delta(x)=0$$ for $$x\neq0$$ to restrict the domain of integration to a "very small" region $$\Delta$$ around, in this case, $$z$$. If $$f$$ is continuous, then for $$\Delta$$ small enough I can always approximate $$f(x,y)\sim f(z,z)$$. Then, using rectangular approximations for the deltas, we have that both $$\delta$$ equal $$1/\sqrt\Delta$$ in this infinitesimal region. Because there are two such $$\delta$$, we get a factor $$(1/\sqrt\Delta)^2$$, which cancels out with the $$\Delta$$ coming out of the integration region.