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

The Schrodinger picture allows for an evolving state, which evolves through a unitary, reversible evolution (Schrodinger'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!

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4 Answers 4

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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.

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Could you comment on why the previous version was wrong? Isn't $ |\Psi(t_2)\rangle =U(t_2,t_c)P_c U(t_c,t_1) |\Psi(t_1)\rangle$? –  Malabarba Aug 12 '13 at 0:25
    
I think you are right, I was confused because I thought I was projecting the time evolution operator down to the collapsed subspace, but you don't want to do that. Thank you for pointing this out! –  Andrew Aug 12 '13 at 1:36

I think you've misconstrued the Schrodinger picture. The Schrodinger and Heisenberg pictures are physical theories that make testable predictions, are strictly mathematically equivalent to each other, and are unitary. Neither theory says anything about wavefunction collapse.

Collapse (of the wavefunction or of an operator) is a feature of a particular interpretation of quantum mechanics (the Copenhagen interpretation, CI). The CI, like other interpretations of quantum mechanics, is not a physical theory and doesn't make testable predictions. The CI is not related to any particular picture of quantum mechanics.

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Thanks for this perspective on it. So, for example, in an interpretation like Everett's Many-Worlds the collapse is nonexistent and therefore we don't need a Heisenberg picture of the collapse. Is that the idea? –  FrancisFlute Aug 11 '13 at 18:57
    
@FrancisFlute: Right. In MWI you can dispense with collapse, in either the Schrodinger picture or the Heisenberg picture. –  Ben Crowell Aug 11 '13 at 19:23
    
Regardless of whether state collapse is physical, if you're doing any actual calculations to predict probabilities using quantum mechanics, you have to do something like state collapse. That is, you have to use some non-unitary projection. Otherwise, all you have is a wave function (Schroedinger picture) or an operator (Heisenberg picture) which you can't relate to the actual outcome of an experiment. –  Peter Shor Aug 12 '13 at 2:23
    
@PeterShor: you have to do something like state collapse What you need is the Born rule. The Born rule is mandatory, and without it QM is useless. Wavefunction collapse is optional, exists only in CI, and is a matter of philosophical taste. To my taste wavefunction collapse is a silly cartoon that can't possibly be taken seriously, but you don't have to agree with me any more than you have to agree with me about my favorite jazz musician. –  Ben Crowell Aug 12 '13 at 4:32
    
@Ben: Applying the Born rule is also not unitary. Whether you apply "wavefunction collapse" or "the Born rule", you need to do something that isn't strictly unitary in order to make predictions. –  Peter Shor Aug 12 '13 at 4:36

As you state, in the Heisenberg picture the operators evolve, and therefore their eigenvectors and eigenvalues are also evolving. But the basic procedure how a (von Neumann) measurement is performed stays the same as in the Schroedinger picture. The probabilities to measure a particular value of the observable (which must be an eigenvalue of the operator) is given by the projection of the (now constant) state vector onto the (now evolving) eigenvector.

In summary, both pictures have both concepts. The difference is only which part of the projection, the eigenvector or the state, changes with time.

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In Schrodinger representation, the "pseudo-collapse" is :

$$\sum_i~ a_i|\psi_i(t)\rangle \rightarrow ~~|\psi_o(t)\rangle$$

In Heisenberg representation, the "pseudo-collapse" is :

$$\sum_i~ a_i|\psi_i\rangle \rightarrow ~~|\psi_o\rangle$$

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