Use and abuse of von Neumann measurement of the first kind

The classification of different quantum mechanic interpretations and solutions to the measurement problem is sometimes based on von Neumann analysis of the measurement process, in particular on his "measurements of the first kind". I will attempt to show that the mathematical foundation of the classification is incorrect. But what are measurements of the first kind?

Suppose we have a measurement device \(|M\rangle\) with three states:

• ready for measurement: \(|M_0\rangle\)
• detecting outcome A: \(|M_A\rangle\)
• detecting outcome B: \(|M_B\rangle\)

Also suppose we have a quantum system \(|\psi\rangle\) which can collapse on the two outcomes: \(|\psi_A\rangle\) and \(|\psi_B\rangle\). von Neumann demands the following time evolution which could be obtained by some interaction between the measurement device and the quantum system:

\(|\psi\rangle\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)

At first sight, this is perfectly reasonable. For example in the case of a Stern-Gerlach experiment, the interaction Hamiltonian corresponding to an inhomogenous magnetic field is:

\(H = -i\sigma_3 \frac{\partial}{\partial z}\)

and on plugging this into the Schrodinger equation it produces the desired effect (and I'll skip the computation details for brevity sake). So what could be wrong with using measurement of the first kind arguments into the classification of quantum mechanics interpretations?

If we try to use Schrodinger equation into the analysis of the measurement process, we need to do it in general, not only in particular cases where there is no issue. For example, what should \(|\psi_A\rangle\otimes |M_0\rangle \) evolve into? We know:

\(|\psi_A\rangle \otimes |M_0\rangle \rightarrow |\psi_A\rangle \otimes |M_A\rangle\)

 

because a repeated experiment yields the same outcome. But is this what really happens when we have an interaction Hamitonian able to generate: \(|\psi\rangle\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)? I'll show that this is impossible. The quantum axiom of repeated measurements yielding the same outcome is incompatible with von Neuman measurements of the first kind.

Before proving this, let's see why we care. If we have that pure unitary evolution satisfy:

(1)    \(|\psi_A\rangle \otimes |M_0 \rangle \rightarrow |\psi_A \rangle \otimes |M_A \rangle\)

(2)    \(|\psi_B\rangle \otimes |M_0 \rangle \rightarrow |\psi_B \rangle \otimes |M_B \rangle\)

then by superposition we get:

(3)    \((c_A|\psi_A\rangle + c_B|\psi_B\rangle)\otimes |M_0\rangle \rightarrow c_A|\psi_A\rangle \otimes |M_A\rangle + c_B|\psi_B\rangle \otimes |M_B\rangle\)

Now from (1), (2), and (3) the measurement problem can be presented as a ``trilemma'':

• the wave-function of a system is complete,
• the wave-function always evolves in accord with a linear dynamical equation,
• measurements have determinate outcomes.

where any two items contradict the third one. For example Bohmian interpretation violates the first item, spontaneous collapse the second item, and many worlds violates the third one. 

Now back to Eqs. 1 and 2 we notice something odd: the lack of a backreaction from the interaction Hamiltonian on the measurement device. Can this really happen? Here comes the algebraic formalism to the rescue. In this formalism, time evolution is represented by a product \(\alpha\) which up to various factors of 2, \(\hbar\) and \(i\) is the commutator. We also have a product \(\sigma\) which is the Jordan product of Hermitean operators and again up to various factors it is the anticommutator. For a bipartite system like the one we consider: "quantum system-measurement device", the product \(\alpha\) is:

\((A_1 \otimes A_2) \alpha_{12} (B_1 \otimes B_2) = (A_1 \alpha B_1 )\otimes (A_2 \sigma B_2) + (A_1 \sigma B_1 )\otimes (A_2 \alpha B_2)\)

where \(A_1, B_1\) are operators in the Hilbert space of system 1 (the quantum system), and \(A_2, B_2\) are operators in the Hilbert space of system 2 (the measurement device). If the interaction Hamiltonian is \(h_{12}=h_1\otimes h_2\), the time evolution in the Heisenberg picture for the operators in system one is as follows:

\((\dot{A_1} \otimes I) = (h_1\otimes h_2) \alpha_{12} (A_1 \otimes I) = (h_1 \alpha A_1) \otimes (h_2 \sigma I) + (h_1 \sigma A_1) \otimes (h_2 \alpha I)\)

Because the product \(\alpha\) is skew-symmetric we have \((h_2 \alpha I) = 0\) which demands:

\(\dot{A_1} = (h_1 \alpha A_1) h_2 \)

Since \((h_1 \alpha A_1 )\) cannot be zero because the quantum system does evolve in time, the only possibility to get \(A_1\) to be constant is for \(h_2\) to be zero. But then there is no coupling with the measurement device whatsoever.

So we seem to be stuck explaining repeated measurements. Repeated measurements do generate the same outcome (this is an experimental fact) and  \(|\psi_A\rangle \otimes |M_0 \rangle \rightarrow |\psi_A \rangle \otimes |M_A \rangle\) does happen. The only way out of this mathematical problem is to realize that the transition represents not a unitary time evolution, but a change in representation. But wait a minute: changes in representation are about Hilbert spaces and they do not happen inside a Hilbert space! And since they do not happen inside a Hilbert space, then there is no superposition and therefore:

Eq. 1 + Eq. 2 DOES NOT IMPLY Eq. 3

The way the usual story goes in explaining the measurement problem is:

the wave-function of a system is complete

+

the wave-function always evolves in accord with a linear dynamical equation

+

Eq. 1, Eq. 2 (repeated measurement axiom)

+

Eq. 3 (by superposition)

IMPLIES 

measurements DO NOT have determinate outcomes (because Eq. 3)

Because Eq. 1 + Eq. 2 DOES NOT IMPLY Eq. 3 the "trilemma" is bogus and  

• the wave-function of a system is complete
• the wave-function always evolves in accord with a linear dynamical equation
• measurements have determinate outcomes

are all compatible with each other. The trilemma is an artifact of using only one Hilbert space representation. If we insist on a single Hilbert space representation then Asher Peres (can we call him the forefather of QBism?) was right on demanding that quantum mechanics should not be applicable to itself. 

But now we have a solid foundation for Peres' handwaving: one universal Hilbert space representation (or to the extreme a nonsensical wavefunction of the universe) cannot explain everything. Why? Because this is incompatible with identical results for repeated measurements. There are many Hilbert space representations linked by the Grotendieck construction. 

The measurement process is the lifting of the representation degeneracy through equivalence breaking.

I will stop the measurement problem series here with the main conclusion stated on the line above. There is still one open problem before claiming success: derive Born rule in the Grotendieck Cartesian pair approach. This is a big work in progress. 

Next time at this blog I will look into classical emulations of quantum mechanics. Very interesting stuff which hold some big expectations on realizing a quantum computer without the fear of decoherence. This is all real, not a crackpot's dream of beating Bell's theorem, and it is actually built and working in the lab. But to what degree can a classical system realize pure quantum effects? Please stay tuned.