quantum mechanics - State collapse in the Heisenberg picture

I've been studying quantum mechanics and quantum field theory for a few years now and one question continues to bother me.

The Schrödinger picture allows for an evolving state, which evolves through a unitary, reversible evolution (Schrödinger’s equation, represented by the Hamiltonian operator) and an irreversible evolution (wave function collapse, represented by a projection operator).

The Heisenberg picture holds the states constant and evolves the operators instead. It provides an equivalent representation of the unitary evolution on operators, but I haven't yet seen an equivalent Heisenberg representation of wave function collapse. Is there any accepted explanation for how to represent state collapse in the Heisenberg picture?

Thanks!

Answer

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. In other words, the operator 'collapses' by picking up a projection piece that kills the unphysical part of the state.

Forget about pictures for a second, the physical thing is the full matrix element

\begin{equation} \langle \psi,t_1 | U(t_1,t_2) \mathcal{O}(t_2) U(t_2,t_1) | \psi,t_1 \rangle \end{equation}

Knowledge of the hamiltonian is buried inside of the time evolution operator $U$.

The Schrodinger picture amounts to grouping the $U$ with the state so that $|\psi(t)\rangle =U(t,t_*)|\psi(t_*)\rangle $, the Heisenberg picture amounts to grouping the $U$ with the operator so that $\mathcal{O}(t)=U(t,t_*)\mathcal{O}(t_*)U(t_*,t)$. This is clearly an artificial split and nothing can ever depend on your choice of picture: if you express things in terms of the full matrix element the difference between the pictures always amounts to a different way of grouping terms.

How do we describe collapse? There is some special time $t_c$, the collapse time, at which something non-unitary happens. We cannot use $U$ to evolve past $t_c$.

Or in other words, the relationship \begin{equation} U(t_2,t_1)=U(t_2,t_c)U(t_c,t_1) \end{equation} is no longer true for $t_2>t_c>t_1$. We need to include a projection operator, as you said in your question: \begin{equation} U(t_2,t_1)=U(t_2,t_c)N_c P_c U(t_c,t_1) \end{equation} where $P_c$ is the operator that projects us down onto the collapsed subspace, and where $N_c$ is a normalization factor so that the state is correctly normalized after collapse. The projection operator will by hermitian and satisfies $P_c^2=P_c$, although the full operator that is being applied at $t_c$, namely the combination $N_c P_c$, is not a projection operator.

So let's say we want to evaluate the physical matrix element, we have to include this projection operator

\begin{equation} \langle \psi,t_1 | U(t_1,t_c) N_c^* P_c U(t_c,t_2) \mathcal{O}(t_2) U(t_2,t_c) N_c P_c U(t_c,t_1) | \psi,t_1 \rangle \end{equation}

So again we have a choice of how we group things. We could group things in a Schrodinger way so that

\begin{array} \ |\psi(t_c+\epsilon)\rangle &=& N_c U(t_c+\epsilon,t_c)P_c U(t_c,t_1)|\psi,t_1\rangle \\ & =& N_c P_c |\psi,t_c\rangle + O(\epsilon) \end{array}

This is the 'state collapse.' At $t_c$ the state changes so that it is projected down onto a subspace.

Or, we could group things in a Heisenberg way, so that

\begin{array} \ \mathcal{O}(t_c+\epsilon) &=& U(t_c+\epsilon,t_c) N_c^* P_c U(t_c,t_1) \mathcal{O(t_1)} U(t_1,t_c)N_c P_c U(t_c,t_c+\epsilon)\\ &=& |N_c|^2 P_c \mathcal{O}(t_c) P_c + O(\epsilon) \end{array}

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.

EDIT # 1: I previously said $U(t_2,t_1)=U(t_2,t_c)P_c U(t_c,t_1)$, which is incorrect. The basic point still stands but the math was technically wrong.

Whoops! I was right the first time. Thanks to Bruce Connor for making me rethink through this point. 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. But that is not what we want here: the time evolution operator is special. The point is that you project down to the subspace (say a position eigenstate) at $t_c$, then you evolve normally from there. In particular you are allowed to evolve out of the subspace. For example, after we observe a particle at position $x$ the particle is allowed to evolve a probability to be at $x'$. You don't want to force the evolution to stay in the subspace, that's what the second $P_c$ would have done.

EDIT # 2: Sorry for all the edits, this is a little more subtle to get exactly right than I originally thought. 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.