2
$\begingroup$

My (superficial) understanding of the Many-Worlds interpretation of QM is that it accounts somewhat for the fundamentally probabilistic nature of QM... but to my mind it doesn't seem to capture non-commutativity.

In a toy version of quantum mechanics, we have a system described by a normalised state $\psi\in\mathsf{H}$ and we can make a measurement of $\psi$ with an orthogonal projection $p\in B(\mathsf{H})$. The outcome is zero or one, with probability $\langle \psi,p\psi\rangle$ we get one.

If we get one, the state collapses to $p\psi$ (or rather $\widehat{p\psi}:=p\psi/\|p\psi\|$).

Take another orthogonal projection $q$ such that $pq\neq qp$. We can measure the conditioned state $p\psi$ with $q$, and we will find, say, $q=0$ with probability: $$\langle \widehat{p\psi},(I_{\mathsf{H}}-q)\widehat{p\psi}\rangle.$$

In this case where $\overline{q}=I_{\mathsf{H}}-q$, the state will transistion to: $$\widehat{\overline{q}p\psi}:=\frac{\overline{q}p\psi}{\|\overline{q}p\psi\|}.$$

Now we can measure $\widehat{\overline{q}p\psi}$ with $p$. Classically we find $p=1$ with probability one, but in this toy quantum mechanics, and with these sequential measurements, the probability that we see $p=1$, then $q=0$, then $p=0$ is non-zero.

I am imagining what we are doing here is we have pebbles under shells, and the pebbles are black or white, and have a one or a zero on them (so classically the pebbles are either of type black-one, black-zero, white-one, white-zero) and $p$ might be the measurement that is $1$ for black, $0$ for white, and $q$ is the number. Classically, once we see $p=1$ that pebble will always be black... but in this toy version of quantum mechanics it can, in what I feel is a violation of causality, later measured to be white. Of course I am mixing here a mathematical toy with something that is physics, and there might be effects at play, such as evolution of the wavefunction between measurements that makes this not make any sense at all.

But to my mind, if such quantum behaviour were observable, it would look as odd in Many Worlds as it does in the world we think we live in.

I was thinking in my head of this idea that after a quantum measurement time itself collapses in a way, so that in the world that it collapsed to, $\widehat{p\psi}$, time and causality are not conserved, and events that happened in the past in the larger world $\psi$, did not in fact happen in the past in the collapsed world $\widehat{p\psi}$, but maybe are possible future events (or something like that).

As I am obviously struggling to formulate anything of sense above, let me ask a question:

Does the Many Worlds interpretation ever get attacked on this basis, that a state can be measured to give seemingly non-causal results, in sequential measurements as described above, say, $p=1$, then $q=0$, then $p=0$?

$\endgroup$

1 Answer 1

4
$\begingroup$

Sequential projections can rotate a state. There is nothing non-causal about this, since there is no way of measuring (projecting) without interfering with the state. The third measurement involves a fundamentally different state from the first measurement, in a way which is not captured by your pebble-example.

This is not related to MWI in any particular way. MWI is about how we interpret the probabilities involved, but has the same experimental predictions as e.g. Copenhagen interpretation.

I think your example is well captured by the sequential Stern Gerlach experiment. In short, you take a stream of electrons and divide them into spin up and spin down as measured along the z-axis. You discard the spin down electrons. You then take the resulting stream and divide them into spin up and spin down as measured along the x-axis. You discard the spin down electrons.

If you then again divide the electrons into spin up and spin down as measured along the z-axis, you again find spin down electrons! Despite them being filtered away in the first filter, they have returned due to your x-filter in the middle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.