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I've been trying to understand the preferred-basis problem in QM, specifally in the Everettian intepretation.

To quote an answer to another question which discusses this:

In my opinion, the situation is almost exactly analogous to the question of whether an abstract formulation of classical mechanics (e.g. Lagrangian mechanics) is satisfactory in the absence of a clear link between the mathematical formalism and our experiences. I could write down the math of Lagrangian mechanics very compactly, but it would not feel like a satisfactory theory until I told you how to link it up with your experiences (e.g. this abstract real scalar x = the position coordinate of a baseball) and you could then use it to make predictions. Similarly, a unitarily evolving wavefunction of the universe is not useful for making predictions unless I give you the branch structure which identifies where you are in wavefunction as well as the possible future, measurement-dependent versions of you. I would claim that the Copenhagen cook book for making predictions that is presented in introductory QM books is a correct but incomplete link between the mathematical formalism of QM and our experiences; it only functions correctly when (1) the initial state of our branch and (2) the measurement basis are assumed (rather than derived).

Also it is said here:

It thus seems that quantum mechanics has nothing to say about which observable(s) of the system is (are) being recorded, via the formation of quantum correlations, by the apparatus. This can be stated in a general theorem (Auletta, 2000; Zurek, 1981): When quantum mechanics is applied to an isolated composite object consisting of a system S and an apparatus A, it cannot determine which observable of the system has been measured—in obvious contrast to our experience of the workings of measuring devices that seem to be “designed” to measure certain quantities.

I really don't understand the problem here. I think I understand how a measurement would occur in an Everettian interpretation, and I don't see an obvious problem with choosing a basis.
To use an example (related to the answer given here):

Any procedure in a quantum system can be described by a unitary operator $U$ (quantum evolution) and/or a projection operator $P$ (measurement). If you want to bring two isolated subsystems in a state $|\psi_1\rangle\otimes|\psi_2\rangle$ into an entangled state $|\psi\rangle$ you need to ask what type of unitary operator $U$ and/or projection operator $P$ you should use such that: $$ P\left(U\left(|\psi_1\rangle\otimes|\psi_2\rangle\right)\right)=|\psi\rangle $$

As an example, imagine two $1/2-$spin systems in an initial state $|\uparrow\rangle \otimes|\uparrow\rangle$, doing the following procedures:

  1. A measurement of $\vec{S}_1\cdot \vec{S}_2=\frac{1}{2}\left[(\vec{S}_1+\vec{S}_2)^2 - > \vec{S}_1^{2}-\vec{S}_2^{2}\right]=\frac{1}{2}(\vec{S}_1+\vec{S}_2)^2-\frac{3}{4}$.
  2. Or a evolution by an hamiltonian $H\propto\vec{S}_1\cdot \vec{S}_2$ by $\Delta t\neq T$, where $T$ is the period of precession.

you are going to get an entangled state.

More generally, any measurement of a global observable like $\vec{S}_1\cdot \vec{S}_2$ produces an entangled state.

For the $U$ operators, any hamiltonian that cannot be writen as $H\neq > H_1\otimes1 + 1\otimes H_2$ will produce entangled states for times different than the period of oscillation, if there is any. This means that is sufficient to have an interaction between this two subsystems and avoid intervals of time $\Delta t=T$, where $T$ is some period of the system.

So what is the problem with defining measurement operators as systems that perform an entangling operation between the device and value seeking to be measured?

As in this example, I want to measure $S_1$ using $S_2$. So I begin with some initial state $(A|\uparrow\rangle+B|\downarrow\rangle) \otimes|\uparrow\rangle$.

What stops me from just considering the Hamiltonian that creates a unitary operator that turns this into:

$(A|\uparrow\rangle+B|\downarrow\rangle) \otimes|\uparrow\rangle \rightarrow (|A|^2|\uparrow\rangle \otimes|\uparrow\rangle + |B|^2|\downarrow\rangle \otimes |\downarrow\rangle)$

If I construct this, then if my second spin is my "detector" that I check, then isn't my device performing a measurement of the first particle? This would therefore be a device that checks the spin of a particle in the $\{|\uparrow\rangle, |\downarrow\rangle \}$ basis.

So where exactly is there a problem with choosing a basis? I don't get where there is an issue. What else am I missing here?

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If there were a unitary $U$ that turns $$\tag{1} (A|\uparrow\rangle+B|\downarrow\rangle) \otimes|\uparrow\rangle=A\,|\uparrow\rangle \otimes|\uparrow\rangle+B\,|\downarrow\rangle \otimes|\uparrow\rangle $$ into $$ |A|^2|\uparrow\rangle \otimes|\uparrow\rangle + |B|^2|\downarrow\rangle \otimes |\downarrow\rangle $$ then it would map two orthogonal states to the same product state which cannot be: \begin{align} &U:\frac{|\uparrow\rangle \otimes|\uparrow\rangle+|\downarrow\rangle \otimes|\uparrow\rangle}{\sqrt{2}}\mapsto \frac{|\uparrow\rangle \otimes|\downarrow\rangle+|\downarrow\rangle \otimes|\downarrow\rangle}{2}\,,&(\text{ take }A=B=\frac{1}{\sqrt{2}})\,,\\ &U:\frac{|\uparrow\rangle \otimes|\uparrow\rangle-|\downarrow\rangle \otimes|\uparrow\rangle}{\sqrt{2}}\mapsto \frac{|\uparrow\rangle \otimes|\downarrow\rangle+|\downarrow\rangle \otimes|\downarrow\rangle}{2}\,,&(\text{ take }A=\frac{1}{\sqrt{2}},B=-\frac{1}{\sqrt{2}})\,. \end{align} If instead you are asking if there is a unitary $U$ that maps the state (1) to $$ A\,|\uparrow\rangle \otimes|\uparrow\rangle + B\,|\downarrow\rangle \otimes |\downarrow\rangle $$ the answer is yes. To see this note that $$ |\uparrow\rangle \otimes|\uparrow\rangle,\quad|\uparrow\rangle \otimes|\downarrow\rangle,\quad|\downarrow\rangle \otimes|\uparrow\rangle,\quad|\downarrow\rangle \otimes|\downarrow\rangle $$ is an orthonormal basis of the tensor product (the combined system describing the state and the measurement device).

The unitary $U$ we are looking for must map the first basis state $|\uparrow\rangle \otimes|\uparrow\rangle$ to itself, and the third basis state $|\downarrow\rangle \otimes|\uparrow\rangle$ to the fourth basis state $|\downarrow\rangle \otimes|\downarrow\rangle$.

To make the definition of $U$ complete represented it by the matrix $$ \left(\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{matrix}\right)\,. $$ That is it is the map that permutes the third and the fourth basis state.

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  • $\begingroup$ You're correctly pointing out a mistake in OP's example, but I think this should be a comment because it doesn't answer the overall question. $\endgroup$
    – user87745
    Mar 7, 2022 at 11:18
  • $\begingroup$ Agreed. Because of the long formulas I found it too long for a comment. $\endgroup$
    – Kurt G.
    Mar 7, 2022 at 12:01
  • $\begingroup$ @KurtG., thanks. So is this related to the issue or just a mistake on my part? That is, is the issue that you can't construct a unitary operator that causes entanglement between a measuring device and a measurable? $\endgroup$ Mar 7, 2022 at 12:03
  • $\begingroup$ @StevenSagona No, you can construct the kind of time-evolution operators you were hoping to construct -- that's not the issue. For example, see expression $2c$ here. $\endgroup$
    – user87745
    Mar 7, 2022 at 12:56
  • $\begingroup$ @DvijD.C., I wrote it as a second question here to not overcomplicate things. But is it possible to do a simpler version than the expression you linked? For example, would $(A|\uparrow\rangle+B|\downarrow\rangle) \otimes|\uparrow\rangle \rightarrow (A|\uparrow\rangle \otimes|\uparrow\rangle + B|\downarrow\rangle \otimes |\downarrow\rangle)$ be sufficient? $\endgroup$ Mar 7, 2022 at 13:27

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