Quantum Measurement Process

NOTE: This page is still a work in progress and I have not yet produced the kind of simple example I am looking for

It is often alleged that “collapse” of a superposition to a statistical mixture violates the unitarity of quantum evolution. But while this is true of an isolated system in a pure state, the measurement process necessarily involves interaction of the observed system with some larger measurement apparatus which is in turn interacting with the world at large (including with the brain of the observer) and so is certainly not necessarily in a pure state from the perspective of the observer.

The purpose of this note is to provide a simplified model to illustrate (mainly just for paedagogical purposes as if it works this should all be trivial) how the unitary evolution of a pair of interacting systems can have their combined state start as the tensor product of a pure state of one (the system under observation) with a mixed state of the other (the measuring apparatus) and end up as a statistical mixture of tensor products (one for each value of the observable being measured and with weights corresponding to the squared moduli of the coefficients of the eigenstates in the initial pure superposition state of the observed system).

For simplicity we’ll consider just a two-valued observable with values +1 or -1 having eigenvectors $\ket{\psi_{+}}$ and $\ket{\psi_{-}}$ respectively and more general pure states of the observed system having the vector form $\ket{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}$ with $\alpha_{+}^2+\alpha_{-}^2 = 1$ which corresponds to the rank one density matrix $\ket{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}\bra{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}$.

For (a grossly simplified model of) the measurement apparatus we’ll assume not a pure state but a statistical mixture of two possible pure states represented by the density matrix $M = 0.5\ket{A}\bra{A}+0.5\ket{B}\bra{B}$

The combined system then has states in the tensor product of the two Hilbert spaces with the initial, pre-measurement state just being the operator tensor product of density matrices:

$$\rho=\ket{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}\bra{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}\otimes M$$

$$=(|\alpha_{+}|^2\ket{\psi_{+}}\bra{\psi_{+}}\otimes (\ket{A}\bra{A}+\ket{B}\bra{B})/2$$

$$+\alpha_{-}^*\alpha_{+}\ket{\psi_{-}}\bra{\psi_{+}}\otimes (\ket{A}\bra{A}+\ket{B}\bra{B})/2$$

$$+\alpha_{+}^*\alpha_{-}\ket{\psi_{+}}\bra{\psi_{-}} \otimes (\ket{A}\bra{A}+\ket{B}\bra{B})/2$$

$$+|\alpha_{-}|^2 \ket{\psi_{-}}\bra{\psi_{-}} )\otimes (\ket{A}\bra{A}+\ket{B}\bra{B})/2$$

$$=(|\alpha_{+}|^2(\ket{\psi_{+}A}\bra{\psi_{+}A}+\ket{\psi_{+}B}\bra{\psi_{+}B})/2$$

$$+\alpha_{-}^*\alpha_{+}(\ket{\psi_{-}A}\bra{\psi_{+}A}+\ket{\psi_{-}B}\bra{\psi_{+}B})/2$$

$$+\alpha_{+}^*\alpha_{-}(\ket{\psi_{+}A}\bra{\psi_{-}A}+\ket{\psi_{+}B}\bra{\psi_{-}B})/2$$

$$+|\alpha_{-}|^2(\ket{\psi_{-}A}\bra{\psi_{-}A}+\ket{\psi_{-}B}\bra{\psi_{-}B})/2$$

If we apply a unitary “evolution” operator U to the combined system we get

$$U\rho U^*=U(\ket{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}\bra{\alpha_{+}\psi_{+}+\alpha_{-}\psi_{-}}\otimes M)U^*$$

$$=|\alpha_{+}|^2(U\ket{\psi_{+}A}\bra{\psi_{+}A}U^*+U\ket{\psi_{+}B}\bra{\psi_{+}B}U^*)/2$$

$$+\alpha_{-}^*\alpha_{+}(U\ket{\psi_{-}A}\bra{\psi_{+}A}U^*+U\ket{\psi_{-}B}\bra{\psi_{+}B}U^*)/2$$

$$+\alpha_{+}^*\alpha_{-}(U\ket{\psi_{+}A}\bra{\psi_{-}A}U^*+U\ket{\psi_{+}B}\bra{\psi_{-}B}U^*)/2$$

$$+|\alpha_{-}|^2(U\ket{\psi_{-}A}\bra{\psi_{-}A}U^*+U\ket{\psi_{-}B}\bra{\psi_{-}B}U^*)/2$$

And the key is now to see if, with an appropriate choice of U, we can make the cross terms vanish.

For example if U is defined by the conditions

$$U\bra{\psi_{+}A}=\bra{\psi_{+}(A+B)/\sqrt{2}},$$

$$U\bra{\psi_{+}B}=\bra{\psi_{+}(A-B)/\sqrt{2}},$$

$$U\bra{\psi_{-}A}=\bra{\psi_{-}(A-B)/\sqrt{2}},$$

$$U\bra{\psi_{-}B}=\bra{\psi_{-}(-A-B)/\sqrt{2}},$$

then we get

$$\alpha_{-}^*\alpha_{+}(U\ket{\psi_{-}A}\bra{\psi_{+}A}U^*+U\ket{\psi_{-}B}\bra{\psi_{+}B}U^*)/2$$

$$=\alpha_{-}^*\alpha_{+}(\ket{\psi_{-}(A-B)/\sqrt{2}}\bra{\psi_{+}(A+B)/\sqrt{2}}$$

$$+\ket{\psi_{-}(A-B)/\sqrt{2}}\bra{\psi_{+}(-A-B)/\sqrt{2}})/2$$

$$=\alpha_{-}^*\alpha_{+}(\ket{\psi_{-}(A-B)}\bra{\psi_{+}(A+B)}$$

$$+\ket{\psi_{-}(-A-B)}\bra{-\psi_{+}(A-B)})/4$$

$$=\alpha_{-}^*\alpha_{+}(\ket{\psi_{-}}\bra{\psi_{+}}\otimes(\ket{B-A}\bra{A+B}$$

$$+\ket{B+A}\bra{A-B})/4$$

$$=\alpha_{-}^*\alpha_{+}(\ket{\psi_{-}}\bra{\psi_{+}}\otimes(\ket{B}\bra{A}+\ket{B}\bra{B}-\ket{A}\bra{A}-\ket{A}\bra{B}$$

$$+\ket{B}\bra{A}-\ket{B}\bra{B}+\ket{A}\bra{A}-\ket{A}\bra{B})/4$$

$$=\alpha_{-}^*\alpha_{+}(\ket{\psi_{-}}\bra{\psi_{+}}\otimes(2\ket{B}\bra{A}-2\ket{A}\bra{B})/4$$

WHICH DOESN’T QUITE WORK

and similarly

$$\alpha_{+}^*\alpha_{-}(U\ket{\psi_{+}A}\bra{\psi_{-}A}U^*+U\ket{\psi_{+}B}\bra{\psi_{-}B}U^*)/2$$

$$=\alpha_{+}^*\alpha_{-}(\ket{\psi_{+}(A-B)/\sqrt{2}}\bra{\psi_{-}(A-B)/\sqrt{2}}$$

$$+\ket{-\psi_{+}(A-B)\sqrt{2}}\bra{\psi_{-}(A-B)\sqrt{2}})/2$$

$$=\alpha_{+}^*\alpha_{-}(\ket{\psi_{+}(A-B)}\bra{\psi_{-}(A-B)}$$

$$+\ket{-\psi_{+}(A-B)}\bra{\psi_{-}(A-B)})/4$$

But for the diagonal terms we get

$$|\alpha_{+}|^2(U\ket{\psi_{+}A}\bra{\psi_{+}A}U^*+U\ket{\psi_{+}B}\bra{\psi_{+}B}U^*)/2$$

$$=|\alpha_{+}|^2(\ket{\psi_{+}(A+B)/\sqrt{2}}\bra{\psi_{+}(A+B)/\sqrt{2}}$$

$$+\ket{\psi_{+}(A-B)/\sqrt{2}}\bra{\psi_{+}(A-B)/\sqrt{2}})/2$$

$$=|\alpha_{+}|^2\ket{\psi_{+}}\bra{\psi_{+}}(\ket{(A+B)}\bra{(A+B)}$$

$$+\ket{(A-B)}\bra{(A-B)})/4$$

$$=|\alpha_{+}|^2\ket{\psi_{+}}\bra{\psi_{+}}(\ket{A}\bra{A}+\ket{B}\bra{B})/2$$

and

$$|\alpha_{-}|^2(U\ket{\psi_{-}A}\bra{\psi_{-}A}U^*+U\ket{\psi_{-}B}\bra{\psi_{-}B}U^*)/2$$

$$=|\alpha_{-}|^2/2(\ket{\psi_{-}(M_{+}-M_{-})/\sqrt{2}}\bra{\psi_{-}(M_{+}-M_{-})/\sqrt{2}}$$

$$+\ket{\psi_{-}(M_{+}-M_{-})/\sqrt{2}}\bra{\psi_{-}(M_{+}-M_{-})/\sqrt{2}})$$

$$=|\alpha_{-}|^2\ket{\psi_{-}}\bra{\psi_{-}}(\ket{(M_{+}+M_{-})}\bra{(M_{+}+M_{-})}$$

$$+\ket{(M_{+}-M_{-})}\bra{(M_{+}-M_{-})})/4$$

$$=|\alpha_{-}|^2\ket{\psi_{-}}\bra{\psi_{-}}(\ket{M_{+}}\bra{M_{+}}+\ket{M_{-}}\bra{M_{-}})/2$$

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