According to the Copenhagen interpretation, the process of measurement is described by the collapse of the wavefunction, which is a non-unitary process. The Hermiticity of a Hamiltonian guarantees that the time dependent wavefunction undergoes a unitary transformation, so the Schrödinger equation can not be applied to the measurement process, and the Born rule should be used instead. The Many Worlds Interpretation avoids this issue by removing the collapse of the wavefunction and holds that the measurement process can be explained by an ordinary interaction between two quantum systems. This results in a final entangled state between the system to be measured and the measurement device. As I understand it, the total state evolves according to the evolution operator \begin{equation} U=\lim_{t\rightarrow\infty}\mathcal{T}e^{-i/\hbar\int_{t_0}^t Hdt} \end{equation} where $H$ is the interaction Hamiltonian describing the measurement process.

I have tried deriving the unitary operator $U$ which describes the measurement, but I did not obtain a unitary operator. Let the total wavefunction be written as \begin{equation} |\Psi\rangle=|\psi\rangle\otimes|\phi\rangle\; , \end{equation} where $|\psi\rangle=\sum_n|\psi_n\rangle$ is the state of the measured system and $|\psi_n\rangle$ are the eigenstates of the measured observable. In this notation, state $|\phi_n\rangle$ describes the state of the measurement device which has measured state $|\psi_n\rangle$ of the system. The measurement operator thus acts upon the states as \begin{equation} U|\psi_n\rangle\otimes|\phi\rangle=|\psi_n\rangle\otimes|\phi_n\rangle\; , \quad \forall |\phi\rangle\; . \end{equation}

Using the above relation, the matrix elements of $U$ are given by: \begin{equation} U_{n_1,n_2;m_1,m_2}=\langle\psi_{n_1}|\otimes\langle\phi_{n_2}|U|\psi_{m_1}\rangle\otimes|\phi_{m_2}\rangle=\langle\psi_{n_1}|\otimes\langle\phi_{n_2}|\psi_{m_1}\rangle\otimes|\phi_{m_1}\rangle=\delta_{n_1,m_1}\delta_{n_2,m_1}\; . \end{equation}

A unitary operator satisfies that $UU^\dagger=1$, that is to say $(UU^\dagger)_{n_1,n_2;m_1,m_2}=\delta_{n_1,m_1}\delta_{n_2,m_2}$. I have tried computing this property and obtained the following result: \begin{equation} (UU^\dagger)_{n_1,n_2;m_1,m_2}=\sum_{a,b}U_{n_1,n_2;a,b}(U_{m_1,m_2;a,b})^*=\sum_{a,b}\delta_{n_1,a}\delta_{n_2,a}\delta_{m_1,a}\delta_{m_2,a}\\ =\sum_{b}\delta_{n_1,m_1}\delta_{n_2,m_2}\delta_{n_1,n_2}\neq\delta_{n_1,m_1}\delta_{n_2,m_2}\; . \end{equation}

What is the reason for this inconsistency? I do not know if I made any wrong assumption about the MWI or maybe I did not define $U$ correctly. All the explanations I have read about the MWI do not delve into the mathematics behind the measurement, is there any book that covers this topic mathematically?

Edit: After posting the question I have realised that this derivation is equivalent to the quantum cloning problem, which is forbidden by the no cloning theorem. Therefore, there must be something wrong with the assumptions made. Maybe there are multiple degenerate states $|\phi_{n,g}\rangle$ corresponding to the measurement of $|\psi_n\rangle$?


4 Answers 4


Probably the most authoritative source on the Everett Interpretation that goes into the mathematics is Everett's thesis.

The way I understand it (and I can't claim to be an expert on this), the reasoning goes something like this.

Two non-interacting systems can be represented in a single joint equation like this.

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

The matrix in the middle should be thought of as block-diagonal. Each block represents the (possibly infinite-dimensional) Hamiltonian of one of the systems. (Obviously, this is not a rigorous presentation...)

We now suppose that for a short period of time, the two sub-systems are allowed to interact. Interaction creates off-diagonal cross-terms in the Hamiltonian matrix in which the state of one system affects the other. (Because it's a Hamiltonian, when you integrate and exponentiate the results have to remain unitary - this is the 'unitary operator' I think you're looking for.)

$-i\hbar\frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & \epsilon_{12} \\ \epsilon_{21} & H_2}\pmatrix{\psi_1 \\ \psi_2}$

This interaction matrix can be diagonalised as $U^{-1}DU$ where $U$ is a constant unitary matrix of eigenstates, and $D$ is a diagonal (or at least block-diagonal) matrix. $D$ might have lots of eigenvalues (maybe infinitely many) but I'll show two to illustrate the idea.

$-i\hbar\frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=U^{-1}\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$

Move the $U$ across to the other side.

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

Now change variables to $\pmatrix{\phi_1 \\ \phi_2}=U\pmatrix{\psi_1 \\ \psi_2}$ to get:

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

This is again the equation for two non-interacting systems, but this time the two 'systems' in question are correlated linear combinations of the two original systems. Each represents a distinct outcome of the measurement; a separate 'world'.

(The division into eigenstates/worlds don't all have to be related to the states of the observed system. A lot of the choices will be to do with various quantum choices for events going on inside the observer, too. So it may be that for a two-choice quantum observable, what we get is two groups of $D_i$s, one group all corresponding to spin up, the other group to spin down. Again, I'm skipping past many details.)

If we are in the $\phi_1$ world, the state in the original basis will look like:

$U^{-1}\pmatrix{\phi_1 \\ 0}=\pmatrix{\psi_{11} \\ \psi_{21}}$

That's the observer in a state $\psi_{11}$ observing system 2 in state $\psi_{21}$. Similarly if we are in the $\phi_2$ world. The state of observer and observed are correlated in each 'world'.

That's my take on the idea. But for an authoritative answer, see Everett.


Haven't you just rederived the fact that collapse is not unitary?

People who subscribe to Copenhagen and Many Worlds use the exact same projectors to model measurements. But as the name suggests, what changes is their interpretation. Many Worlds holds that the projectors we use are only non-unitary because our approximation of declaring the system size to be finite was too severe. If we had kept track of everything, the operation that looked like a projection, when restricted to a subspace, would've actually acted as a unitary operator on the universe's whole state.

To actually find this operator (without saying equally cheap things like "evolving via the unvierse's Hamiltonian"), this point of view is completely non-constructive. There are as many operators performing a measurement as there are ways to measure something. In fact, any interaction which produces entanglement can be seen as one thing measuring another.


  1. How do we keep track of everything? The short answer is, we can't. This is the "Schroedigner equation for $10^{23}$ particles" thing that every textbook says is too hard. Keeping track of everything means knowing enough about the initial state of the universe to evolve it forward past the time of the measurement (which was inevitably guaranteed to happen by the deterministic nature of many worlds). If we could do this, we would know which value the wavefunction assigns to all of the alternate versions of ourselves who observe different things.

  2. If by "measurement device", you really mean everything involved in the measurement, then the system plus device space is large enough. But most people don't do this. They say a Geiger counter clicking if an atom decays is a measurement. But they say our brains registering an auditory sensation if the counter clicks is something else. It is not something else. The chain of A being measured by B which is measured by C never ends. We have to assume it ends (say at D) in order to restrict the ABCD system onto ABC and therefore get a chance of having the process appear non-unitary.

  3. Does this apply to Copenhagen as well? Yes. Entanglement and decoherence are part of quantum mechanics no matter what and therefore so is the idea that unitary evolution can look like projection when you don't keep track of everything. This is why I think Many Worlds is a more minimal set of assumptions. Copenhagen asserts that, in addition to this apparent collapse, there is a fundamental collapse as well.

  4. The quantum process is the only thing there is in Many Worlds. It's as bare bones as it gets. If a positively polarized photon enters the same crystal as another photon in a superposition, they might get entangled. Initial and final states therefore look like \begin{align} \left | \psi_i \right > = \left | + \right > (\left | + \right > + \left | - \right >) / \sqrt{2} \\ \left | \psi_f \right > = (\left | ++ \right > + \left | -- \right >) / \sqrt{2}. \end{align} States of the crystal's atoms (which can interact in infinitely many ways) have been neglected which is why the above evolution looks non-unitary. Most critics of Many Worlds agree with all of this so far. But now imagine that a cat we last saw alive (+ state) is placed in Schroedinger's box where it goes into a superposition (between + and -). Opening the box can again be done in infinitely many ways and it will lead to a fully correlated state where the cat and our eyes are either both + or both -. This scenario is a measurement and it is isomorphic to the "photons in, photons out" process which colloquially never gets called a measurement.

  • $\begingroup$ How do we "kept track of everything" then? When you say that declaring the system size to be finite is too severe, are you refering to the system subspace, the measurement device subspace or the total "universal" state? In addition, if Many Worlds is just an interpretation, would not this argument also be valid for the Copenhagen interpretation? Finally, I do not see any issue with any interaction producing entanglement being seen as one thing measuring another. It is just a definition for a quantum process. $\endgroup$ Commented Jul 27, 2022 at 7:54
  • $\begingroup$ One thing I don't understand : Isn't it an assumption of MWI that there exists a physical phenomenon called "apparent collapse"? The assumption that there exist some physical systems, like humans, which can "find" themselves in a wavefunction branch, like in our subjective experience. Is this assumption not equivalent to the "measurement causes collapse" assumption of Copenhagen? $\endgroup$
    – Ryder Rude
    Commented Nov 29, 2022 at 12:05
  • $\begingroup$ The apparent collapse I'm talking about arises because an entangled state looks more classical when acted upon by a partial trace. This is inevitable in quantum mechanics regardless of interpretation. But the idea that people have a subjective experience of what their own state is like? MWI of course does not help here but that's because no science has ever been able to clarify what it means for people to "find" themselves. If there were hope of this, that would count in Copenhagen's favour which gives a smaller "space" for people to find themselves in. But I doubt the hard problem has hope. $\endgroup$ Commented Nov 29, 2022 at 12:39

The perfect measurement operator for a system whose states are on an $N$-dimensional Hilbert space onto another system with an $N$-dimensional Hilbert space can be written as

$$U=\sum_{ab}|\psi_a\rangle\langle\psi_a||\phi_{b\oplus a}\rangle\langle\phi_b|.$$

where $\oplus$ stands for addition modulo $N$. This operator produces the result: $$U|\psi_a\rangle|\phi_0\rangle=|\psi_a\rangle|\phi_a\rangle.$$

So when the measurement instrument is prepared in the relative state $|\phi_0\rangle$ its state after the measurement is $|\phi_a\rangle$ when the relative state of the measured system is $|\psi_a\rangle$.

We can write $$U^\dagger=\sum_{cd}|\psi_c\rangle\langle\psi_c||\phi_d\rangle\langle\phi_{d\oplus c}|.$$ The product is $$UU^\dagger=\sum_{abcd}|\psi_a\rangle\langle\psi_a||\psi_c\rangle\langle\psi_c||\phi_{b\oplus a}\rangle\langle\phi_b||\phi_d\rangle\langle\phi_{d\oplus c}|\\ =\sum_{abcd}\delta_{ac}\delta_{bd}|\psi_a\rangle\langle\psi_a||\phi_{b\oplus a}\rangle\langle\phi_{b\oplus a}|=I.$$

So the description of a perfect measurement in the MWI is consistent. There aren't really any textbooks using the MWI. There are some decent books on the maths of quantum theory such as "Lectures on Quantum Theory: Mathematical and Structural Foundations" by C. J. Isham. There are also books about the Everett interpretation such as "The Emergent Multiverse" by David Wallace.

  • $\begingroup$ Thank you for the derivation, this means that a measument can only be performed if the measurement instrument is prepared in the state $|\phi_0\rangle$, right? In my derivation I assumed an arbitrary state $|\phi\rangle$. Another question, since $U$ is unitary, it can be written as $U=e^{-i/\hbar H T}$. Does this mean that $H$ is the interation term describing a measurement performed over a time T? $\endgroup$ Commented Jul 28, 2022 at 7:45
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    $\begingroup$ Yes a measurement requires that the measuring instrument is prepared in a standard state. If the Hamiltonian is constant then the expression you have given is correct. If it isn't constant, things get a bit more complicated physics.stackexchange.com/a/355570/28512 $\endgroup$
    – alanf
    Commented Jul 28, 2022 at 8:20

What I'll describe has been formally explained both for qubits and more generally. Hopefully this answers your question(s); the last equation is the unitary operator you seek. It is not unique, though, but I hope its derivation will be helpful to you!

Suppose you want to measure some observable $A$ that has the spectral decomposition $$ A \, = \, \sum\limits_{m=0}^{M-1} \, a^{\,}_m \, \mathbb{P}^{\,}_m \, ,$$ where $\{a^{\,}_m\}$ are the $M$ unique eigenvalues of $A$ (and also the possible outcomes of measuring $A$).

If the eigenvalue $a^{\,}_m$ is nondegenerate (i.e., has exactly one associated eigenstate $\left| m \right\rangle$), then $\mathbb{P}^{\,}_m = \left| m \middle\rangle \hspace{-0.3mm} \middle\langle m \right|$ projects onto that eigenstate. If $a^{\,}_m$ is degenerate, then $\mathbb{P}^{\,}_m$ projects onto the space of eigenstates with eigenvalue $a^{\,}_m$. The projectors satisfy $$ \mathbb{P}^{\,}_m \, \mathbb{P}^{\,}_n \, = \, \delta^{\,}_{m,n} \mathbb{P}^{\,}_m~~~,~~~\sum\limits_{m=0}^{M-1} \, \mathbb{P}^{\,}_m \, = \mathbb{1}\,,~$$ where $\mathbb{1}$ is the identity operator.

The Stinespring Dilation Theorem tells us that quantum channels—including measurements—can be expressed via isometries. Measuring $A$ (above) in the initial physical state $\left| \psi \right\rangle$ is represented by $$ \left| \psi \right\rangle \, \to \, \left| \psi' \right\rangle \, = \, \sum\limits_{m=0}^{M-1} \, \left( \mathbb{P}^{\,}_m \left| \psi \right\rangle \right)^{\,}_{\rm phys} \otimes \left| m \right\rangle^{\,}_{\rm out}\,,$$ where the "phys" subscript labels the physical state of the system after measurement, and the "out" subscript labels the state of the measurement apparatus after measurement (i.e., $m$ denotes that the outcome $a^{\,}_m$ was observed upon measurement). The isometry is given by $$\mathbb{V}^{\,}_{A} \, = \, \sum\limits_{m=0}^{M-1} \, \mathbb{P}^{\,}_m \otimes \left| m \right\rangle \,,$$ and noting that the outcome states obey $\left\langle m \middle| n \right\rangle = \delta^{\,}_{m,n}$, you can confirm that this channel is isometric by deriving that $\left\langle \psi' \middle| \psi' \right\rangle =\left\langle \psi \middle| \psi \right\rangle$ is length preserving (an isometry is a length-preserving linear map from one Hilbert space to a possibly larger one, hence Stinespring "dilation").

Isometries can be embedded in unitaries quite easily, by dilating the initial Hilbert space. Physically, this just means we need to identify the "default" state of the measurement apparatus. This is actually quite physical: Most quantum measurements work by detecting emitted photons. The default outcome is no photon detected, and the "click" outcome is a photon detected.

For simplicity, let's suppose that the state $\left|0 \right\rangle$ is the default state of the detector. Let's also define the operator $$ X^m \, = \, \sum\limits_{k=0}^{M-1} \, \left| k + m \, \text{mod} \, M \middle\rangle \hspace{-0.3mm} \middle\langle k \right| \, ,$$ which is a cyclic "raising operator" for the $M$-state Hilbert space that stores the measurement outcome (with $m$ some integer). The unitary analogy of the isometric measurement equation I wrote above is $$ \left| \psi \right\rangle \otimes \left| 0 \right\rangle \, \to \, \mathsf{V}^{\,}_A \,\left| \psi \right\rangle \otimes \left| 0 \right\rangle \, = \, \sum\limits_{m=0}^{M-1} \, \left( \mathbb{P}^{\,}_m \left| \psi \right\rangle \right)^{\,}_{\rm phys} \otimes \left| m \right\rangle^{\,}_{\rm out}\,,$$ where the unitary channel that captures measurement of $A$ is simply $$ \mathsf{V}^{\,}_A \, = \, \sum\limits_{m=0}^{M-1} \,\mathbb{P}^{\,}_m \otimes X^m \, ,$$ where $\mathbb{P}^{\,}_m$ acts on the physical Hilbert space and $X^m$ shifts the outcome state from the default state $\left| 0 \right\rangle$ to the observed outcome state $\left| m \right\rangle$. You can confirm that this operator is unitary.

In the case of qubits, this is even more simple! The outcome qubit remains in the default state if the physical qubit is in the state 0, and flips if the physical qubit is in the state 1. This is nothing more than a CNOT gate where the physical qubit is the "control" bit and the outcome qubit is the "target" bit! This is true for measurement of Pauli "string" operators, whose eigenvalues are always $\pm 1$.

Anyway, the unitary operator $\mathsf{V}$ follows from the Stinespring Dilation Theorem and a basic assumption about how measurements work. But it is also exactly what the many-worlds interpretation describes. I am currently working to connect it to von Neumann's pointer Hamiltonian for measurements, which seem related to what you wrote down.

  • 1
    $\begingroup$ Thanks for the detailed derivation. When you introduce the operator $X$ I think that you should generalise it to $ X^m \, = \, \sum\limits_{k=0}^{M-1} \, \left| k + m \, \text{mod} \, M \middle\rangle \hspace{-0.3mm} \middle\langle k \right| $ so that the last equation makes sense. $\endgroup$ Commented Dec 2, 2022 at 8:45
  • $\begingroup$ Also, I find it a bit troublesome that the measuring device needs to be prepared in the same initial condition in every measurement. Doesn't the screen in the double slit experiment get excited with each detection? $\endgroup$ Commented Dec 2, 2022 at 8:49
  • $\begingroup$ @TheAverageHijano This is generally a requirement for unitary measurement. But from the experimental standpoint, this is not at all a challenge. Really, it's like performing the double slit experiment multiple times: you would "reset" the screen each time you did the experiment. Each outcome register contains only one measurement outcome, so it only gets reset in between experiments. It really reflects the state of the measurement apparatus. Hopefully that clarifies things? $\endgroup$ Commented Dec 2, 2022 at 21:02
  • $\begingroup$ Nearly all quantum measurements are performed using photon absorption. In the double-slit experiment you care about resolving where the photons land. Imagine if you only cared about whether a photon is absorbed. Then, you need one detector per photon you check for. If a photon is absorbed, the detect flips from the 0 state to the 1 state, which is accomplished by $X$. Otherwise, it remains 0. You can read off the outcomes from each detector at any point. If you want to use the same detector, you have to reset between measurements, and write the result down. $\endgroup$ Commented Dec 2, 2022 at 21:05

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