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 $|\psi_n\rangle$?
Best Answer
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.