29
$\begingroup$

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!

$\endgroup$
18
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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$? $\endgroup$ – Malabarba Aug 12 '13 at 0:25
  • $\begingroup$ 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! $\endgroup$ – Andrew Aug 12 '13 at 1:36
  • $\begingroup$ @Andrew How does one calculate the value $N_c$ from the point of view of the Heisenberg picture? $\endgroup$ – Oscar Cunningham Sep 17 '15 at 10:42
  • $\begingroup$ This is not good since after a projection the operators no longer satisfy their defining relations. In particular, the projections of operators satisfying the CCR usually violate the CCR. $\endgroup$ – Arnold Neumaier Nov 18 '15 at 11:25
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – FrancisFlute Aug 11 '13 at 18:57
  • 3
    $\begingroup$ 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. $\endgroup$ – Peter Shor Aug 12 '13 at 2:23
  • 1
    $\begingroup$ @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. $\endgroup$ – Ben Crowell Aug 12 '13 at 4:32
  • 1
    $\begingroup$ @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. $\endgroup$ – Peter Shor Aug 12 '13 at 4:36
  • 1
    $\begingroup$ @PeterShor: The Born rule isn't a function from the Hilbert space to the Hilbert space, so the question of whether it's unitary is undefined. It's true that you need something other than the Schrodinger equation (which predicts unitary evolution). The Born rule is logically separate from the Schrodinger equation. The Born rule isn't an alternative to wavefunction collapse. Wavefunction collapse is an optional philosophical interpretation of the Born rule. $\endgroup$ – Ben Crowell Aug 12 '13 at 4:41
2
$\begingroup$

In the Heisenberg picture, the correct description of a dissipative process (of which the collapse is just the the simplest model) is through a quantum stochastic process. There is an extended literature on this. Properly designed, these processes preserve the commutation relations between key observables during the time evolution, which is an essential consistency requirement.

Note that the collapse is a very simplified description of a measurement process and that most measurements are not well described by it. For more realistic modeling one usually employs a density matrix (von Neumann) representation, and describes the evolution in terms of a Lindblad equation. For the analogue in the Heisenberg picture, see. e.g., a paper by Accardi et al., A stochastic golden rule and quantum Langevin equation for the low density limit.

$\endgroup$
1
$\begingroup$

Between 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. 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.

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. Among other issues, it has both phase and normalization ambiguity: different non-zero rescalings and different phases yield the same state. The best way to handle that and remove the ambiguity is to refer, instead, to W = |ψ><ψ|/<ψ|ψ> as the state.

These are pure states. 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. More general mixed states can be contemplated that are continuous linear combinations of pure states, rather than discrete sums. Example: WC = p |0><0| + q |1><1| is a classical bit (a mixed state), while WQ = (√p |0> + √q exp(iφ) |1>)(√p <0| + √q exp(iφ) <1|) = WC + √(pq) (exp(iφ) |0><1| + exp(-iφ) |1><0|) is a quantum bit, which is a pure state. The state ½ WC + ½ WQ would then be 50% pure; so purity can range from 0 to 100%.

A physical quantity A, when quantized, becomes an operator  that has a decomposition of the form  = Σa a Pa, into a linear combination of projection operators Pa and eigenvalues a. The effect of the projection Pa|ψ> 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. Since the eigensubspaces span the entire space for |ψ>, then it is also assumed the projections add to 1: Σa Pa = 1.

More general operators can be considered that have spectra that are continuous, or mixed continuous/discrete. It won't shed any light to consider them here, so I'll just keep to the simpler case of discrete spectra only.

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 Pa|ψ> with probability |Pa|ψ>|2/<ψ|ψ> and that the value measured by the event is (a).

When restated, the rule asserts that the state W = |ψ><ψ|/<ψ|ψ> becomes Pa|ψ><ψ|Pa/<ψ|PaPa|ψ> = Pa|ψ><ψ|Pa/<ψ|Pa|ψ> ... the latter equality applying since Pa = Pa = Pa2 for projection operators. It states that this happens with probability |Pa|ψ>|2/<ψ|ψ> = <ψ|Pa|ψ>/<ψ|ψ>.

Introducing the "trace" operator, defining it by the property Tr(A|ψ><ψ|B) = <ψ|BA|ψ>, then the above reduction can be restated as W → Pa W Pa/Tr(W Pa) with probability Tr(W Pa).

There are two ways to treat this, depending on what you consider a mixed state to represent. 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: W → WA ≡ Σa Tr(W Pa) × Pa W Pa/Tr(W Pa) = Σa Pa W Pa.

Thus, each quantum event produces a change of the form W → WA = Σa Pa W Pa, where A is the corresponding quantity that is being measured by that event.

Note that this normalizes correctly since Tr(WA) = Σa Tr(W Pa) = Tr(W Σa Pa) = Tr(W) = 1.

The other way to regard a mixed state as being a thing in its own right, so that the transition has two steps: the first producing the mixed state itself, and the second producing one of its pure state components with the associated probability. Personally, I think that way of looking at it introduces unnecessary redundancy, so I will keep to the first interpretation.

If a subsequent measurement is made for a second quantity C whose quantized form Ĉ decomposes into Ĉ = Σc P'c, then the result will be a reduction to the state WA → WAC = Σc P'c WA P'c = Σa,c P'c Pa W Pa P'c.

This can be wrapped up by introducing the time-ordering pseudo-operator T[] and its dual T'[], defined with the properties T[UV] = UV if U occurs before V, VU if V occurs after U; and T'[UV] = UV if U occurs after V, T'[UV] = VU if U occurs before V. Then, you can write the reduction as W → WAC = Σa,c T'[Pa P'c] W T[Pa P'c].

The generalization to more than 2 quantum events should be fairly obvious, by now.

When this is generalized to field theory, each operator is no longer associated with a specific time, but with a specific space-time point. The Born rule then applies to a finite set of quantum events situated over a point cloud in a compact region of space-time. If the events correspond to the measurements of A = A1, A2, ..., An, then the transition is W → WA = Σa T'[Pa] W T[Pa], where I'm writing the projections more concisely as Pa = Pa1 ... Pan.

In effect, the Born rule then introduces a space-time evolution of the state, even though time is removed out of the picture by making operators (instead of wave vectors) dynamic. If you treat all of space-time as being populated by a (possibly infinite) point cloud, each point associated with a quantum event, then each state W will be associated with a partition of this point cloud into a "before set" and "after set", with the property that none of the points in the after set can have a future time-like or null curve that leads to any of the points in the "before set". The ones in the before set are the ones that state may be considered to have already undergone a Born reduction for, while the ones in the after set are those which it has not undergone such a reduction.

Then an effective Born evolution can be defined that produces a transition from any two states whose partitions agree on all but a finite number of points in the point cloud, whenever the before set of the one state (the "later" state) contains the before set of the other state (the "earlier" state). Then the Born rule is applied by taking the quantum events (and their associated operators) that the two states disagree on being before or after. The transition then goes from the earlier state to the later state by applying the reduction that's just been described.

So, in total: this is the closest you get to a straightforward translation of the Born Rule from the Schrödinger Picture to the Heisenberg Picture. The formalism this most closely resembles is what's known as Consistent Histories (https://en.wikipedia.org/wiki/Consistent_histories); whose math is similar. So, you might be able to cannibalize some of their formulae and apply them here, after stripping out all the extra small print they attach to them.

An interesting related reference that I found a few days before this https://arxiv.org/pdf/1308.5290.pdf treats the Born rule in a similar fashion, but also brings POVM's into the picture. Sections 3 and 6 are where the issues are dealt with.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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 :

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.