quantum mechanics - Entanglement and coherence

I have come across a wonderful review of entanglement by Chris Drost in his answer to this post. One part that left me puzzled was: (This post is merely an attempt to understand a portion of Chris' answer, unfortunately I do not have enough reputation to ask this as a comment in his post, so I figured a new post wouldn't be a terrible idea as this is a rather important conceptual question for all beginners.)

Obviously, the product-states have a "quantum coherence" to both qubits: doing our double-slit experiment means that we see an interference pattern. Shockingly, entanglement weakens and sometimes eliminates this interference pattern. For example, the state $\sqrt{\frac 12} |00\rangle + \sqrt{\frac 12}|11\rangle$ describes an entangled state. If you pass the first qubit of this through the double-slit experiment, normal rules of quantum mechanics give the distribution $\frac 12 |f_0(x)|^2 + \frac 12 |f_1(x)|^2:$ classically overlapping bell curves!

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   Unfortunately I fail to see how by entangling two particles they lose their coherence. But When I have a particle $A$ in a superposition state $\psi_A = a|0\rangle + b|1\rangle$ and entangle it to another system $B,$ in state $\psi_B,$ my first particle still remains in a superposition, and its measurement is still random, is it not?

   
   


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   So why do we say that entanglement destroys coherence? It would be great if one could elaborately show this for the simplest entangled pairs! Is the point maybe that if $B$ is measured first, only then $A$ loses its coherence? (assume here a complete correlation).

   
   


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   Small digression if I may: if it is true that entangelemnt destroys coherence, does the converse mean that the concept of decoherence is tightly related to entanglement of a small system with its environment? Or in other words would decoherence happen at all without entanglement?

   
   

Answer

Okay, this is getting even more into depth, which is great stuff! I heartily recommend anyone who is this dedicated take a few courses on the subject, if you haven't already.

Here's the most basic formulation of quantum mechanics which adequately shows all of these properties, called the density-matrix or state-matrix formulation. Take a wavefunction $|\psi\rangle$ and identify the state-matrix $\rho = |\psi\rangle\langle\psi|$ with this state. The state matrix has all of the same information as the wavefunction but evolves according to the product rule,

$$i\hbar ~\frac {\partial\rho}{\partial t} = \hat H \rho - \rho \hat H.$$

As always, we predict expectation values of experiments by associating to their numerical parameters a Hermitian operator $\hat A.$ Now, instead of calculating this as the usual $\langle A \rangle = \langle\psi|\hat A|\psi\rangle$ we insert some orthonormal basis $I = \sum_i |i\rangle\langle i|$ into the middle of this expression as

$$\langle A \rangle = \sum_i \langle\psi|\hat A|i\rangle\langle i|\psi\rangle = \sum_i \langle i|\psi\rangle\langle\psi|\hat A|i\rangle = \sum_i \langle i|\rho ~ \hat A|i\rangle = \operatorname{Tr} \rho \hat A.$$ All expectation values are therefore traces of these matrix products. We can also insert a further identity inside these two between them, to find $\langle i | \rho | j\rangle = \rho_{ij},\;\langle j | \hat A | i\rangle = A_{ji},$ and so we have a matrix expression $\langle A \rangle = \sum_{ij} \rho_{ij} A_{ji},$ if you like. Any discrete basis of the Hilbert space will work even if it has no particular tie to our Hamiltonian.

Now suppose we have an observable which only impacts one subsystem of the whole system. Here we simply convert the basis to one that spans both subsystems, $|i, j\rangle$ and our observable has the form $\hat A \otimes I$ in terms of its effect on the respective systems. Our expression for the expected value is therefore:

$$\operatorname{Tr} \rho (\hat A \otimes I) = \sum_{ij} \langle i,j|\rho (\hat A \otimes I) |i, j\rangle$$ Inserting another identity $I = \sum_{mn} |m,n\rangle\langle m,n|$ we can look carefully to the second term: $$\langle A\rangle = \sum_{ij~mn} \langle i,j|\rho |m, n\rangle\langle m,n|(\hat A \otimes I) |i, j\rangle = \sum_{ij~mn} \langle i,j|\rho |m, n \rangle A_{mi} \delta_{nj}.$$ We therefore find that there is an expression for something which acts precisely as an effective substate-matrix $\tilde \rho$ for the subsystem: it reproduces all of the expectation values that you see above for any operator which only works in the substate. That substate matrix is: $$\tilde\rho_{ij} = \sum_{n} \langle i,n| \rho |j, n\rangle,$$ whence $\langle A \rangle = \sum_{ij} \tilde\rho_{ij} A_{ji}.$

We call the process which generates the substate matrix "tracing out" the rest of the superstate, because it has the same structure as a partial trace.

Let us calculate the state matrix for $a |0\rangle + b |1\rangle$. This is very simple: it is

$$\rho = aa^* |0\rangle\langle 0| + a b^* |0\rangle\langle 1| + b a^* |1\rangle\langle 0| + b b^* |1\rangle\langle 1|,$$ or, written as a bona-fide matrix, $$\rho = \begin{bmatrix} a a^* & a b^* \\ b a^* & b b^*\end{bmatrix}.$$

Now let us entangle it with another system. We will use the CNOT operation to entangle it with a constant $|0\rangle$, generating $a |00\rangle + b |11\rangle.$ When we perform the above recipe to this system we find ourselves looking at a completely different density matrix:

$$\tilde\rho = \begin{bmatrix} a a^* & 0 \\ 0 & b b^*\end{bmatrix}.$$ Now let me explain why I couldn't use wavefunctions to get this result: it is that this state matrix cannot be expressed as a wavefunction, unless either $a = 0$ or $b = 0.$ The previous matrix $\rho$ is actually as general as a single-particle wavefunction can be, and it has off-diagonal terms. This one does not, precisely because there is no way that the "tracing out" step can convert a $|00\rangle\langle 11|$ term to anything "internal" to the substate matrix. It lives outside of the substate matrix and can only be measured by measuring both parts of the global state and comparing them!

The simplest observable is $\hat A_1 = |1\rangle\langle 1|$, measuring the probability that a qubit is in state $|1\rangle.$ Now suppose that we don't do this directly, but first evolve the state with a unitary matrix. This will correspond to a photon going through a slit corresponding to the qubit and then traveling to a photomultiplier tube at position $y$, which will "click" (transition from $|0\rangle$ to $|1\rangle$ with amplitudes $f_{0,1}(y)$ when only one of these is open. So the unitary transformation is, for some $\alpha_{0,1}$ that don't matter,

$$|0\rangle \mapsto \alpha_0(y) |0\rangle + f_0(y) |1\rangle\\ |1\rangle \mapsto \alpha_1(y) |0\rangle + f_1(y) |1\rangle.$$ We measure the resulting qubit, which results in $$\langle A \rangle = \operatorname{Tr} (U\rho U^\dagger~ \hat A_1) = \operatorname{Tr} (\rho~U^\dagger \hat A_1 U).$$ The matrix $U^\dagger \hat A_1 U$ is therefore $$\begin{bmatrix}\alpha_0^*&f_0^*\\\alpha_1^*&f_1^*\end{bmatrix} \begin{bmatrix}0&0\\0&1\end{bmatrix} \begin{bmatrix}\alpha_0&\alpha_1\\f_0 & f_1\end{bmatrix} = \begin{bmatrix}f_0 f_0^*& f_1 f_0^*\\f_0 f_1^* & f_1 f_1^*\end{bmatrix}.$$ This is our double-slit observable matrix.

From this you have enough to calculate the two cases, which are

$$\begin{align} \operatorname{Tr} (\rho \hat A) =& a a^* f_0 f_0^* + a b^* f_0 f_1^* + a^* b f_0^* f_1 + b b^* f_1 f_1^* = |a f_0(y) + b f_1(y)|^2\\ \operatorname{Tr} (\tilde \rho \hat A) =& a a^* f_0 f_0^* + b b^* f_1 f_1^* = |a f_0(y)|^2 + |b f_1(y)|^2.\end{align}$$ In fact in general the latter probability matrix, with no off-diagonal terms, behaves like a classical probabilistic mixture of classical bits $0$ and $1.$ That is a very general result from the linearity of trace; in general if $\rho = \sum_i p_i \rho_i$ then $\operatorname{Tr}(\rho \hat A) = \sum_i p_i \operatorname{Tr}(\rho_i \hat A)$, so the system behaves like a classical-probability-mixture of the different constituent $\rho_i$. (Caution: this basis is generally not unique. If you work it out, $\rho = \frac 12 |0\rangle\langle 0| + \frac 12 |1\rangle\langle 1|$ is actually the same as $\rho = \frac 12 |+\rangle\langle +| + \frac 12 |-\rangle\langle -|.$ I am telling you this because I have heard people who do not know this argue that this explains how Quantum Mechanics "chooses" a basis for its decoherence, hence why the world looks classical rather than quantum at a macro-scale... it doesn't really resolve that problem at all!

So that is how to easily understand entanglement as destroying coherence: the more you're entangled, the more the orthogonality of the other system kills your off-diagonal terms, and the more your substate looks like a classical probability mixture, transferring the cool quantum effects to the system-as-a-whole.