# Is Quantum Mechanics (norm)-consistent?

I edited a few small comments to the question in order to make it perhaps more comprehensible.

Today I came across the following question in quantum mechanics.

In Quantum mechanics it is common to use different so-called pictures, which I thought were obviously equivalent, but I am not so sure anymore. So please let me introduce the framework: Let $X$ be a Hilbert space.

The Schrödinger dynamics teaches that in the Hilbert space norm $\left\lVert e^{itH}\psi-\psi \right\rVert_{X} \rightarrow 0$ as $t$ goes to zero.

For density operators one uses the nuclear norm (positive trace-class operators with unit trace) and has $$\left\lVert e^{itH}\rho e^{-itH}-\rho \right\rVert_{\text{nuclear}} \rightarrow 0.$$

I would like to understand whether these two converges agree in certain situations, i.e. assume that $\rho(\psi)=\langle \bullet, \psi \rangle \psi$ is the projection onto $\psi.$

Then one can ask whether uniform convergence is still equivalent:

$$\sup_{\left\lVert \psi \right\rVert=1} \left\lVert e^{itH}\psi-\psi \right\rVert_{X} \rightarrow 0 \Leftrightarrow \sup_{\left\lVert \psi \right\rVert=1} \left\lVert e^{itH}\rho e^{-itH}-\rho \right\rVert_{\text{nuclear}}\rightarrow 0,$$

i.e. both pictures should agree, is that true? Or does only one of the implications hold?

Clearly uniform convergence in the Schrödinger picture holds true iff $H$ is bounded. What about the other picture? Clearly continuity is sufficient, but is it also necessary?

• It's a bit confusing to state this as an equivalence: both statements are true, and this is fairly elementary. In the second difference, you're comparing two rank one projections, so it doesn't really matter whether you use the trace or the operator norm. – Christian Remling May 26 '18 at 0:45
• Also, your question seems a bit basic for this site; it would have been a better fit for math.stackexchange.com – Christian Remling May 26 '18 at 0:48
• @ChristianRemling I guess the question is whether for an arbitrary family of unitary operators $U_t$ and a one-dimensional projection $P=\langle\cdot,\psi\rangle\psi$ the statements $\|U_t\psi-\psi\|_H\to 0$ and $\|U_tPU^{-1}_t-P\|_{\text{nuclear}}\to 0$ are equivalent. The answer is "yes", of course and I agree that it is easy. – fedja May 26 '18 at 0:52
• @ChristianRemling Indeed, I overlooked this possibility (projection defined by $-\psi$ is the same as the one defined by $\psi$). Thanks for the correction! :-) – fedja May 26 '18 at 11:04
• sorry, I hope I made things clearer – Hilbertspace May 26 '18 at 11:45

The answer to your new question is still yes. As you already pointed out yourself, we only need to show that if $H$ is unbounded, then $\sup_{\psi}\|e^{itH}P_{\psi}e^{-itH}-P_{\psi}\|$ will not go to zero as $t\to 0$ (and as I mentioned in my comment, since we're comparing two rank one projections, it doesn't really matter which norm we use here).
By passing to a spectral representation of $H$, we can focus on the situation where $X=L^2(\mathbb R, \mu)$, and $H$ is multiplication by the variable. Since $H$ is assumed unbounded now, the support of $\mu$ is unbounded as well. Take a $\psi$ of the form $\psi(x) =\chi_A(x)/\mu(A)^{1/2}$, with $A$ of small size and localized near an $x\simeq X$, $|X|$ large.
If we now take $t=\pi/(2X)$, say, then $$\|e^{itH}\psi-\psi\|^2 = \frac{1}{\mu(A)}\int_A |e^{itx}-1|^2\, d\mu(x) \simeq |e^{i\pi/2}-1|^2 .$$ Since both $e^{itH}\psi$ and $\psi$ are unit vectors, the two projections onto these vectors can obviously not be close to one another.
• I do not fully understand your argument. The OP seems to ask: The semigroup $e^{itH}$ is uniformly continuous if $H$ is bounded. Is the semigroup $T(t)\rho=e^{itH}\rho e^{-itH}$ uniformly continuous in the nuclear norm on rank one-projections iff $H$ is bounded as well? However, your answer seems to show that $e^{itH}$ is uniformly continuous iff $H$ is bounded, no?-However I do not see how this relates to the initial question. Would you mind helping me with this? – Sascha May 26 '18 at 21:38
• Yes, I'm proving the $\Longleftarrow$ direction of the equivalence stated by the OP, the other direction already having been observed by the OP, and what I show is that if $H$ is unbounded, then $\sup_{\psi}\|e^{itH}P_{\psi}e^{-itH}-\psi\|\not\to 0$. – Christian Remling May 26 '18 at 21:42
• @Sascha: Maybe what's confusing you about my answer is that I also make use of the (trivial) fact that $e^{itH}P_{\psi}e^{-itH}$ projects onto $e^{itH}\psi$, so to have the two projections at some distance from one another, we need $e^{itH}\psi$, $\psi$ to span appreciably distinct subspaces. (It is almost the same argument that shows that $e^{itH}$ is operator norm continuous if and only if $H$ is bounded.) – Christian Remling May 26 '18 at 21:47