The Measurement Problem

The time-evolution of a quantum state characterized by a wavefunction $\Psi$ is given by the (linear) Schrödinger Equation

\[\mathrm{i} \hbar \frac{\partial}{\partial t} \Psi(\boldsymbol{r}, t) = H \Psi(\boldsymbol{r}, t) = - \frac{\hbar^{2}}{2m} \boldsymbol{\nabla}^{2} \Psi(\boldsymbol{r}, t) + V \Psi(\boldsymbol{r},t),\]

where $V$ encodes the interaction or potential between particles described by the wavefunction $\Psi(\boldsymbol{r},t)$.

Note that the Schrödinger Equation is a linear differential equation.

In bra-ket notation, one can write the Schrödinger Equation as

\[\mathrm{i} \hbar \frac{\partial}{\partial t} \vert \Psi(\boldsymbol{r}, t) \rangle = \hat{H} \vert \Psi(\boldsymbol{r}, t) \rangle.\]

Further, the quantum state can be expressed as a superposition:

\(\vert \Psi(\boldsymbol{r}, t) \rangle = \alpha_{1} \vert \psi_{1} \rangle + \alpha_{2} \vert \psi_{2} \rangle.\) Since the Schrödinger Equation is linear, we obtain

\[\mathrm{i} \hbar \alpha_{1} \frac{\partial}{\partial t} \vert \psi_{1} \rangle + \mathrm{i} \hbar \alpha_{2} \frac{\partial}{\partial t}\vert \psi_{2}\rangle = \alpha_{1} \hat{H} \vert \psi_{1} \rangle + \alpha_{2} \hat{H} \vert \psi_{2} \rangle.\]

The problem is: we never observe superpositions. We only observe one quantum state.

One might say: “… but the wavefunction collapses into a single quantum state by a measurement!”
But what is the difference between a ‘‘measurement’’ and the interaction of a bunch of particles (measurement apparatus) with a bunch of other particles (detected particles), which interaction is already described by $V$ in the Schrödinger Equation?
Every interaction is already contained in the Hamiltonian (by $V$) of the Schrödinger Equation!

Therefore, we have 3 options:

1. The Schrödinger Equation is wrong, or at least, quantum states do not evolve according to a linear differential equation as the Schrödinger Equation.
   $\Longrightarrow$ Objective-collapse theories, Ghirardi–Rimini–Weber theory
2. The quantum system is not described by $\Psi$ alone.
   $\Longrightarrow$ “hidden-variable theory”, e.g. Bohmian Mechanics
3. Who said that everything that we observe is everything that exists?! 😏
   $\Longrightarrow$ Many–Worlds interpretation

Although I am not completely satisfied with those options, I would side with deterministic approaches like Bohmian Mechanics.

One might say: “… but Bell’s Theorem proves that there is no hidden–variable theory!”
And this, dear friend, is unfortunately one of the biggest misconceptions circulating even in the highest academic circles in physics. This is wrong.

The conclusion of the empirical data and Bell’s Theorem is the following:
Every theory – with or without “hidden-variables” – which aims to reproduce the same measurement statistics as “ordinary” quantum mechanics and simultaneously assumes “statistical independence/measurement independence”, contains some kind of non-locality.

The choice is not between locality and determinism, since only superdeterminism (denial of “statistical independence/measurement independence”) can save locality.
The choice is between locality and “statistical independence/measurement independence”.

(L) locality condition:

\[P(A \vert a, b, \lambda) = P(A \vert a, \lambda) \\ P(B \vert a, b, \lambda) = P(B \vert b, \lambda)\]

(SI) statistical independence/measurement independence condition:

\[P(\lambda \vert a, b) = P(\lambda)\]

• $A/B$: measurement outcome of Alice/Bob

• $a/b$: measurement setting of Alice/Bob

• $\lambda$: ‘‘hidden variable’’ (although it is also possible that $\lambda = \Psi$, has nothing to do with a ‘‘hidden variable’’ and therefore applies to all theories – regardless of whether they assume $\Psi$ describes the entire system or not)

Everyone is pointing with their fingers on Bohmian Mechanics since it is explicitly non-local and therefore “incompatible with relativity”, while ignoring that every theory that is not (super)deterministic has to be non-local. This is the essence of Bell’s Theorem.