Quantum Measurements: some technical background

Quantum Measurements: some technical background [From the projection postulate to density matrices & (introduction to) von Neumann measurements] (AKA: the boring lecture) First: One more example I wanted to get to last time about odd sorts of measurements and the surprises in store... PHY 2206, lecture 2 (17.1.12)

" Quantum seeing in the dark " (AKA: The Elitzur-Vaidman bomb experiment) A. Elitzur, and L. Vaidman, Found. Phys. 23, 987 (1993) P.G. Kwiat, H. Weinfurter, and A. Zeilinger, Sci. Am. (Nov., 1996) Problem: Consider a collection of bombs so sensitive that a collision with any single particle (photon, electron, etc.) is guarranteed to trigger it. Suppose that certain of the bombs are defective, but differ in their behaviour in no way other than that they will not blow up when triggered. Is there any way to identify the working bombs (or some of them) without blowing them up?

" Quantum seeing in the dark " C (AKA: The Elitzur-Vaidman bomb experiment) A. Elitzur, and L. Vaidman, Found. Phys. 23, 987 (1993) P.G. Kwiat, H. Weinfurter, and A. Zeilinger, Sci. Am. (Nov., 1996) D BS2 Bomb absent: Only detector C fires BS1

The measurement postulate

...measurement outcomes... Collapse of the wavefunction Future measurements of A will of course agree with this a i Measurement State preparation

What are the effects of measurement? Suppose we have two pawns, one black and one white, and I put one in each hand we can write this state as something like = + Obviously, if I open my left hand and measure the colour of its pawn, I find either black or white, not both from that point on, I describe the left pawn as one colour eigenstate, or. Is the other pawn still in a state of uncertain colour? No obviously, its state has also been affected by this measurement.

More refined version: the projection postulate When a i is found, the state Of course, this is not normalized, so the final state is actually Finding the left pawn to be black leaves the system in a state where the right pawn is known to be white (unsurprisingly).

Two effects of measurement 1. One thing happens, as opposed to all other possibilities 2. Interference between the different possibilities becomes impossible. 100% 50% 50% What s the state of the particles before the final beam-splitter? det. 2 A det. 1 B If no bomb was present, half the particles are in path A and half are in path B. If the bomb measures which path each is in... then still, half are in path A and half are in path B. The measurement didn t change the probabilistic description of the state... but without the bomb, interference caused all particles to interfere at the final beam splitter and go left; with the bomb, there is no such interference. Measurement destroyed phase information, but left the probabilities unchanged. A + e iϕ B A OR B.

How does the bomb cause the other detector to fire? The probability is given by the absolute square of this inner product, which is 1/4 + 1/4 = 1/2 (because the orthogonality of "peace" and "BOOM" cause the cross-terms to vanish).

Sneaky fact... No one knows why one thing happens instead of many simultaneous things... in fact, no one knows whether this is true (cf. relative-state, many-worlds, many-minds interpretations). No collapse process has ever been observed i.e., no case where we would make the wrong predictions if we didn t assume collapse. Yet to make sense of probabilities, one typically assumes that by the time you measure something, it s one thing or another. (But how do you know that when I measured it, I wasn t still in a probabilistic state? Wigner s friend. ) We can try (a) to understand what measurements do to coherence and/or (b) to search for a real collapse process, supplementary to quantum mechanics as we know it.

Statistical Mixtures How should we describe a statistical mixture, i.e., half the time I ll send you state A and half the time I ll send you state B? Classically: P(x) = P(x B) P(B) + P(x A) P(A) +... QM: shall we say Ψ(x) = Ψ Β (x) P(B) + Ψ Α (x) P(A) +...? P(x A) P(x Ψ A + Ψ B ) P(x B)

We need a formalism for this... A simpler example: a spin-1/2. In +z >, <S x > = <S y > = 0. In z>, <S x > = <S y > = 0. So in a 50/50 mixture, we d have <S x > = <S y > = 0, but no { +z> + e iφ z> } / 2 satisfies this. (They are all spins in the xy plane, with angle φ.) Any QM wave function you write down which is half A and half B will exhibit some interference; no wave function can describe the state after a measurement of A & B, which we all know makes interference impossible... "Pure states" "Mixed states" individual QM wave functions probabilistic mixtures of QM states. (e.g., results of measurements) "Density Matrices"

Statistical mixtures... Conclusion: no superposition state (that is, no quantum state at all) correctly describes a statistical mixture. What s wrong? We need P(x) = { P(x A) + P(x B) } / 2, but P(x) = Ψ(x) 2, which isn t linear in Ψ. We would like a description of the state ρ such that probabilities were linear in ρ; then only could we naturally describe mixtures of such states. Well, remember that P(i) = <i Ψ><Ψ i>... this isn t linear in Ψ>, but it is linear in Ψ><Ψ ; perhaps that projector is a more desirable description of the quantum state than the wave function.

A linear description of the q. state So, let us define ρ = Ψ><Ψ

Was that clear? Who can define the trace for me in Dirac notation?

And what about mixed states? So instead of a wave function Ψ>, we describe a pure state by the density matrix ρ = Ψ><Ψ ; and we replace the formula <A> = <Ψ A Ψ> with the equivalent <A> = Tr ρ A. The essential property of a statistical mixture is that all expectation values are just the weighted averages of those for the individual pure states. Our expression for expectation values is linear in the density matrix i.e., we can keep using that expression with mixed states, if we define the mixed-state density matrix itself as a weighted average.

Density matrices

Interpretation of matrix elements pure Diagonal elements = probabilities Off-diagonal elements = "coherences" (provide info. about relative phase)

Density matrices for mixed states Note: probabilities still 50/50, but no coherence.

Purity Tr ρ = 1 is required for total probability = 1. (Or realize that Tr ρ = Tr ρi = <I> = 1 trivially.) As a Hermitian matrix, ρ can be diagonalized. For a pure state, ρ is a projector onto Ψ, so one eigenstate is Ψ (with eigenvalue 1) and all others have eigenvalue 0. ρ 2 is thus exactly the same. I.e., Ψ><Ψ Ψ><Ψ = Ψ> 1 <Ψ = Ψ><Ψ For a pure state, Tr ρ 2 = 1. For an impure state, no diagonalization into a single projector exists, so Tr ρ = Σ P i = 1 but Tr ρ 2 = Σ P i 2 < 1. This quantity Tr ρ 2 is sometimes called the linear purity. Any mixed state -- even one made by combining nonorthogonal states -- can be written as a mixture of orthogonal states. Decompositions are not unique.

What happens if you don't look at part of your system? When you calculate expectation values, you trace over the system. If your operators depend only on a subsystem, then it makes no difference whether you trace over other systems before or after:

Decoherence arises from throwing away information Taking this trace over the environment retains only terms diagonal in the environment variables i.e., no cross-terms (coherences) remain if they refer to different states of the environment. (If there is any way even in principle to tell which of two paths was followed, then no interference may occur.) ρ s when env is ρ s when env is

... coherence lost There is still coherence between and, but if the environment is not part of your interferometer, you may as well consider it to have "collapsed" to or. This means there is no effective coherence if you look only at the system.

It doesn t matter what the environment measures! We ve been talking as though sometimes the environment sees +z and sometimes it sees -z, and then we mix the two. But what if some evil experimenter in the environment decided to measure ±x instead of ±z? It wouldn t change a thing. The trace is basis-independent; when I sum over all possible states of the environment, this means it doesn t matter whether I m ignoring the outcome of an x- measurement or a z-measurement. This is important! It means nothing I do on the environment (no rotation, in this instance) changes the probability distribution observed on the system.

Decoherence: the party line When a particle interacts with a measurement device, the two subsystems become entangled (no separable description). Coherence is still present, but only in the entire system; if there is enough information in the measurement device to tell which path your subsystem followed, then it is impossible to observe interference without looking at both parts of the system. The effective density matrix of your system (traced over states of the measuring apparatus) is that of a mixed state. Coherence is never truly lost, as unitary evolution preserves the purity of states. In principle, this measurement interaction is reversible. In practice, once the system interacts with the "environment", i.e., anything with too many degrees of freedom for us to handle, we cannot reverse it. Just as in classical statistical mechanics, it is the approximation of an open system which leads to effective irreversibility, and loss of information (increase of entropy). Loss of Information = Loss of Coherence