Note that this normalizes correctly since Tr(W A) = Σ a Tr(W P a) = Tr(W Σ a P a) = Tr(W) = 1. Thus, each quantum event produces a change of the form W → W A = Σ a P a W P a, where A is the corresponding quantity that is being measured by that event. W → W A ≡ Σ a Tr(W P a) × P a W P a/Tr(W P a) = Σ a P a W P a. If a mixed state W = p W₀ + q W₁ (p,q ≥ 0, p + q = 1) stands for "W₀ with probability p, W₁ with probability q" (recursively applied to W₀ and W₁ if they are also mixed states) then the entire reduction itself can be succinctly wrapped up as: There are two ways to treat this, depending on what you consider a mixed state to represent. Introducing the "trace" operator, defining it by the property Tr(A|ψ>, then the above reduction can be restated as W → P a W P a/Tr(W P a) with probability Tr(W P a). It states that this happens with probability |P a|ψ>| 2/ = /. the latter equality applying since P a † = P a = P a 2 for projection operators. When restated, the rule asserts that the state W = |ψ> becomes P a|ψ> = P a|ψ>. The Born rule says that the result of applying the measurement for quantity A is a quantum event that reduces the state |ψ> to the state P a|ψ> with probability |P a|ψ>| 2/ and that the value measured by the event is (a). It won't shed any light to consider them here, so I'll just keep to the simpler case of discrete spectra only. More general operators can be considered that have spectra that are continuous, or mixed continuous/discrete. Since the eigensubspaces span the entire space for |ψ>, then it is also assumed the projections add to 1: Σ a P a = 1. Example: W C = p |0> + √q exp(iφ) |1>)(√p is to reduce |ψ> to the sum of its a-eigenvector components only zeroing out all the other eigenvectors - it projects |ψ> down to the eigenspace of the value (a) for operator Â. More general mixed states can be contemplated that are continuous linear combinations of pure states, rather than discrete sums. A mixture of such states corresponding to mutually orthogonal vectors, with non-negative mixing coefficients that add up to 1 gives you a mixed state. The best way to handle that and remove the ambiguity is to refer, instead, to W = |ψ> as the state. Among other issues, it has both phase and normalization ambiguity: different non-zero rescalings and different phases yield the same state. To answer your question in the most direct way possible, it's first necessary to clarify that the wave function vector |ψ> is not the state, but is better regarded as a "square root" of the state. In quantum theory, in the Heisenberg Picture, those same equations hold for the quantized versions of the same - up to operator ordering ambiguity, while the state becomes a timeless denotation of an entire history, rather than that of a system and its progression in time. You aren't just projecting the state down to a subspace, you are projecting the state and then rescaling it so that it has the correct normalization.īetween the Schrödinger and Heisenberg picture, the latter is the one most directly connected to dynamics in the form you're used to seeing it in, with the dynamic equations governing the various quantities be they ordinary differential equations of motion or partial differential equations for fields. You don't want to force the evolution to stay in the subspace, that's what the second $P_c$ would have done.ĮDIT # 2: Sorry for all the edits, this is a little more subtle to get exactly right than I originally thought. For example, after we observe a particle at position $x$ the particle is allowed to evolve a probability to be at $x'$. In particular you are allowed to evolve out of the subspace. The point is that you project down to the subspace (say a position eigenstate) at $t_c$, then you evolve normally from there. But that is not what we want here: the time evolution operator is special. I was confused because I thought wanted the transformation rule $P_c U P_c$, which is how you would project the time evolution operator to the collapsed subspace. Thanks to Bruce Connor for making me rethink through this point. The basic point still stands but the math was technically wrong. This is "the operator being projected onto a subspace." The state is the same, but the operator now includes a projection piece that cancels out the part of the state that is no longer physical.ĮDIT # 1: I previously said $U(t_2,t_1)=U(t_2,t_c)P_c U(t_c,t_1)$, which is incorrect. In other words, the operator 'collapses' by picking up a projection piece that kills the unphysical part of the state.įorget about pictures for a second, the physical thing is the full matrix element Well, I think you said the answer yourself when you used the words "projection operator." In the Heisenberg picture the operators get projected down to a subspace at the time of the collapse.
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