In quantum mechanics, any kind of measurement causes the quantum state to change to something we'll call "definite", but sometimes it's actually not definite, just a very small variance.

But according to Schrodinger equation $$i\hbar \partial |\Psi\rangle/\partial t=H|\Psi\rangle,$$ the only way a state can change is by a Hamiltonian, i.e. even thought the position of a particle is not determined, the wave function should be absolutely deterministic. So how does a measurement change the function, or in mathematical words, what the Hamiltonian describing measurement?


2 Answers 2


The following argument is very vague and hand-wavey - as is inevitable when discussing measurement apparatus with on the order of $10^{23}$ particles. It is intended only to illustrate the idea. We can combine the Schrodinger equations for two independent systems prior to measurement using a matrix equation as follows:

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & 0 \\ 0 & H_2}\pmatrix{\psi_1 \\ \psi_2}$$

$\psi_1$ and $\psi_2$ are the observer and observed system states respectively, and may have many dimensions; even infinitely many. The matrix $H$ should be thought of as a block diagonal matrix. The zero blocks top-right and bottom-left in $H$ express the fact that the two systems do not interact. Each can vary completely independently of the other.

Now we modify the equation to introduce a weak interaction between the systems. The off-diagonal terms are set to non-zero values, expressing how each system affects the other. This represents the measurement process, and the details of the constant $\epsilon$ values define what sort of measurement it is.

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & \epsilon_1 \\ \epsilon_2 & H_2}\pmatrix{\psi_1 \\ \psi_2}$$

Now the equations are no longer independent, they are coupled. There is a classical technique we can use to disentangle coupled linear differential equations, which is to diagonalise the Hamiltonian. We write $H$ as a product $U^TDU$ where $U$ is a unitary matrix of orthogonal eigenvectors, and $D$ is a diagonal matrix of eigenvalues.

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=U^T\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$$

We can shift the $U^T$ over to the left hand side, using the fact that $U^TU=I$.

$$i\hbar \frac{\partial}{\partial t}U\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$$

And now we change basis by renaming $U\psi=\phi$.

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\phi_1 \\ \phi_2}=\pmatrix{D_1 & 0 \\ 0 & D_2}\pmatrix{\phi_1 \\ \phi_2}$$

Now this is again a system of completely independent equations, each acting as an independent system that does not interact with any of the others. But each component of $\phi$ is a mixture of $\psi_1$ and $\psi_2$ states in linear combination. The observer/observed system states in each component are correlated. Each of these corresponds to one possible outcome of the quantum measurement. Each consists of the observed system in a particular state, and the observer in a corresponding state of having observed it. But because the off-diagonal terms in the interaction Hamiltonian are all zero in this basis, none of the correlated observer/observed states can interact with any of the others. They are orthogonal to one another. To each, it is as if none of the others existed.

When we use this technique to decouple linear equations in classical physics, the orthogonal components are called normal modes of vibration. 'Normal' because they are orthogonal to one another. We can solve each independent sub-equation separately, and then the general solution is a linear combination of the normal modes.

Because an observer in such a 'normal mode' state cannot interact with, and hence perceive, any of the other normal modes, she has no way to tell whether they are really there. Maybe they all disappear, and get zeroed out? The assumption that all but one of them disappears for some unknown and unspecified reason is what we call wavefunction collapse. The assumption that they don't, but are simply unobservable is called the Everett Interpretation (or Many Worlds Interpretation). Each normal mode acts like a separate 'world'.

The evolution is entirely deterministic: all possible measurement outcomes happen, every time. But from the point of view of each orthogonal component of the observer's superposition, the outcome appears random. They each see exactly one thing happen: they see the particle show up in exactly one place, on one definite state.

However, there are many more orthogonal components than there are outcomes, because the measurement Hamiltonian includes all the quantum interactions going on inside the observer, too. There are on the order of Avogadro's number of particles involved in a brain and set of sensory organs, give or take a handful of powers of ten. So these $10^{23}+$ dimensions are divided roughly between the states of the observed system's outcomes in proportion to the squared magnitude of the wavefunction. Probability normalisation requires the magnitudes add up to 1, and the squared magnitude of a sum of orthogonal vectors is the sum of the squares of all the vectors' magnitudes - that's just a multidimensional version of Pythagoras' theorem. If the squared magnitude is divided between the many components roughly equally, then the Born rule (probability proportional to squared magnitudes) should arise naturally.

The other common objection to this interpretation is the choice of basis. The states of the system under observation may have degenerate eigenvalues, and no definite 'obvious' choice of basis. But the joint measurement Hamiltonian does not apply only to the system observed, but also to the observer, and it is highly unlikely that these degeneracies will remain unsplit. The nature of the measurement (what observable you are measuring) is defined by the off-diagonal perturbations to the joint Hamiltonian, which describes how the systems interact.

There are many aspects to the Everett interpretation that are not fully understood or explained, but a significant fraction of physicists consider it simpler, more elegant, and more explanatory than the alternatives.

However, the Everett Interpretation is not universally accepted. Many object to the multitude of unobservable entities as a violation of Occam's Razor. If we suppose these unobservable alternative components to the wavefunction simply disappear (with no explanation as to how or why they do so), the predictions about what we would observe are exactly the same. There is no way to experimentally distinguish these hypotheses, so the choice of which to believe isn't a scientific question.

The collapse of the wavefunction, in which all but the one component you're in are presumed to disappear, has plenty of philosophical problems of its own. It's non-deterministic, irreversible, and breaks with Schrodinger's equation. The collapse propagates faster-than-light and hence (in some reference frames) backwards in time, so it conflicts badly with our intuitions about causality. But since the parts that change with the collapse are all the unobservable alternatives disappearing, this has no observable consequences and cannot be used to transmit information. So, no worries, eh? The nature of the non-local interactions that maintain the quantum correlations between widely-separated measurements remain mysterious.

If you want a Hamiltonian to describe measurements, you need a no-collapse interpretation, because collapse necessarily violates the smooth unitary evolution. If you want to explain irreversible collapse, then I don't think a Hamiltonian approach will work.

  • 2
    $\begingroup$ How do you justify the interchange of $U$ and $\frac{\partial}{\partial t}$ when the "system" $H_1=H_1(t)$ is time varying? $\endgroup$
    – hyportnex
    Commented Aug 13, 2023 at 15:01
  • $\begingroup$ First of all, thanks for the good and instructive explanation. Could the difficulties in this description come, in part, from an attempt to use solely the language of wave functions, i.e., to remain in the realm of pure states? Could a switch to the density matrix language make life easier, at least partially? $\endgroup$ Commented Aug 13, 2023 at 21:46
  • $\begingroup$ @hyportnex if $U$ is time invariant then $\frac{\partial}{\partial t}$ should commute with $U$. If $H_1$ is time dependent then to be time-invariant $U$ would need to simultaneously diagonalize $H_1(t)$ for all $t$ (unlikely). But if you accept some extra environmental systems then you can approximate any time-dependent Hamiltonian by a time-independent one via Feynman's clock construction. $\endgroup$
    – Sam Jaques
    Commented Aug 14, 2023 at 19:12
  • $\begingroup$ @SamJacques I kinda understand what you are saying but is it even conceivable that for a time-variable Hamiltonian there be a time-independent diagonalizing unitary operator for all times, speaking of it just as a matter of principle? $\endgroup$
    – hyportnex
    Commented Aug 14, 2023 at 22:21
  • $\begingroup$ "What is the Hamiltonian describing measurement?" is maybe a nice way of trying to get people who understand QM but dont undersand MWI to get the idea behind MWI $\endgroup$ Commented Aug 16, 2023 at 17:48

Measurement is an irreversible process, while the time evolution given by the Schrödinger equation is reversible.

The Schrödinger equation gives you unitary time evolution. But wave function collapse is a projection and projection operators are not unitary.


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