Why the wave function collapse corresponds to a non-unitary quantum operation?

  • $\begingroup$ My question arose from this statement: "A general description of the evolution of quantum mechanical systems is possible by using density operators and quantum operations. In this formalism (which is closely related to the C*-algebraic formalism) the collapse of the wave function corresponds to a non-unitary quantum operation." en.wikipedia.org/wiki/Wave_function_collapse $\endgroup$
    – Andreas K.
    Oct 16 '11 at 18:52

Unitary operators are operators that satisfy some conditions. Among other things, they have to be linear:


The operation (or "an operation") that maps any $\psi(x)$ to $\delta(x-x_0)$ where $x_0$ is the random position resulting from a measurement can't be associated with any linear operator. It's easy to see why. Take two functions $\psi_1(x)$ and $\psi_2(x)$ that have different supports: for example, the first one is localized in the vicinity of Boston while the other sits near New York.

Linearity of the collapse operator $C$ requires $$ C (\psi_1 + \psi_2) = C(\psi_1) + C(\psi_2).$$ However, the first term of the right hand side is a delta-function localized somewhere near Boston while the second term of the right hand side is a delta-function localized near New York. Their sum therefore can't be a multiple of a single delta-function, so the left hand side can't be a "collapsed wave function", proving that an operator that maps anything to a single delta-function can't be linear.

There are of course other ways to prove that it can't be a unitary operator – which is a very strong condition.

Of course, the right resolution of this non-unitarity problem is that there's nothing such as the collapse of a wave function. The wave function is not a real wave: it's a set of complex amplitudes whose squared absolute values don't describe "the reality" but rather just the probabilities of "different realities". The probability distributions mean that you always get just one outcome and the values of the probability distribution just tell you what the probabilities of different outcomes are. Nothing has to "collapse" because the wave wasn't a "real observable wave" to start with.

The idea that the wave function has to "collapse" and one has to look for a "mechanism" how it collapses is an artifact of a misinterpretation of the wave function.

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    $\begingroup$ Hi Lubos, I don't believe in the validity of this interpretation of measurement as a merely notational artifact; intermediate measurements make probabilities behave different. Collapse, be it an effective result of complex phenomena or something fundamental, it is a real process. In any case, even if measurements are not linear and hence non-unitary by the book definition, they still conserve the norm of eigenstates $\endgroup$
    – lurscher
    Oct 16 '11 at 20:51
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    $\begingroup$ Dear @lurscher, the fact that the wave function is just a set of numbers to calculate probabilities from, and not a real observable, is an experimentally proven fact, not a matter for beliefs or disbeliefs. By the way, your second statement is also impossible. One can't define a prescription for such a collapse so that it would conserve the norm if we also require the physics to be local: one has to artificially "renormalize" the wave function after the collapse and it's clearly a non-local procedure because it depends on the magnitude of the wave function at other "places". $\endgroup$ Oct 17 '11 at 18:58
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    $\begingroup$ Dear @ANKU, I don't know how you proved that the inner product of $C\psi_i$ and $C\psi_j$ is the particular Kronecker-delta. I think that there's no operator that would satisfy your condition for every $\psi_i$, $\psi_j$. Moreover, I don't understand what's the difference between $\psi_j$ and $j$. What you wrote is just very confusing. $\endgroup$ Oct 17 '11 at 19:01
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    $\begingroup$ @Lurscher, let me also mention that if you want the post-collapse wave function to be proportional to a delta-function in the position representation, such an outcome would 1) have a very sick normalization because you need $\psi(x)=\sqrt{\delta(x-x_0)}$ for the squared wave function to have the right integral; the square root of a delta function isn't really an element of the Hilbert space; 2) if the wave function is strictly proportional to the delta-function, it carries an infinite average kinetic energy: the momentum is totally undetermined and the expectation value of $p^2$ diverges. $\endgroup$ Oct 17 '11 at 19:03
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    $\begingroup$ "measurements in the most basic formulation given in QM courses always assume that you are able to measure the state in a non-local way" - Nope, there is no contradiction between locality and the postulates of quantum mechanics, as demonstrated by the important example of local quantum field theories such as the Standard Model which are perfectly local. They also describe reality: all physical processes in this Universe are demonstrably local, which proves that all mechanisms that are nonlocal are unphysical. $\endgroup$ Oct 18 '11 at 5:29

I'm not sure what would be an answer to the why question. I can only comment on the known structure of the measurement as a operator.

Operators of a Hillbert space take vectors into vectors; they are an endomorphism of the Hillbert space. When you take a single measurement of a observable on a arbitrary vector, you obtain a random eigenvector of the observable. This is, conventionally speaking, a mapping between a vector and a vector. However the mapping is not deterministic, so there is not a single operator. In fact, there is one such operator for each eigenstate. They are better known as projection operators:

$$ M_{i} = | \psi_{i} \rangle \langle \psi_{i} |$$

When a measurement "happens" we can a posteriori state what specific projection operator took place on the system. We cannot use the operator formalism to describe the overall process in general.

Well, not quite. We have a formalism to describe classical probabilities distributions; its called mixed states. Mixed states are anything that is not a pure state, which can always be described with a density matrix whose entries are of the following form factor:

$$ \rho_{ij} = \psi_{i} \psi^{*}_{j} $$

as a density matrix, a classical probability distribution is seen as a purely diagonal matrix, whose entries are the probabilities of each eigenstate

so a measurement can be seen as the following map:

$$ \psi_{i} \psi_{j}^{*} \longrightarrow \psi_{i} \psi_{j}^{*} \delta_{ij} $$

So what the measurement does in general is kill all off-diagonal components of the density matrix, and only leave the diagonal, that represent classical, actual probabilities.

Even if this map is not linear (as clearly stated by @Lubos), it preserves the trace of the density matrix. In other words, even if its not unitary, it is still a isometric transformation.

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    $\begingroup$ "So what the measurement does in general is kill all off-diagonal components of the density matrix": you wanted to say "So what decoherence does...", right? Your description of what measurement does isn't valid. The measurement, as understood in the incorrect interpretation of QM that you're promoting here, is not only bringing the density matrix to a diagonal form: it also sets to zero all the diagonal entries except for the chosen one. $\endgroup$ Oct 17 '11 at 19:09
  • $\begingroup$ Because the text of your answer makes it clear that you actually don't want to pick the "measured outcome" or explain how it is done - you're really trying to explain decoherence and not collapse (which is why your answer has no relevance to the original question which was about the collapse, but let's discuss your answer anyway) - you're talking about decoherence. But decoherence produces a map on the space of density matrices, not on the Hilbert space only. $\endgroup$ Oct 17 '11 at 19:12
  • $\begingroup$ So it makes no sense to ask whether this operation (elimination of off-diag. entries) is an isometry on the Hilbert space itself: it's not a map on the Hilbert space at all, it's a map on the space of density matrices (roughly speaking the tensor product of the Hilbert space and its conjugate copy) only. And on this space, the elimination of the off-diagonal elements is clearly not an isometry, either. So whatever way you look at your statements about the collapse's being an "isometry", they're invalid. $\endgroup$ Oct 17 '11 at 19:14
  • $\begingroup$ To show that the "decoherence map" (elimination of off-diagonal entries) isn't an isometry on the space of matrices, just consider what this map does with matrices $((a,b),(b,c))$ for different values of $b$. These matrices are clearly very far from each other in the natural metric on the space of matrices, especially if you pick a large $b$. But all these matrices get mapped to $((a,0),(0,b))$ so the distance of the values of the map is zero. ;-) $\endgroup$ Oct 17 '11 at 19:17
  • $\begingroup$ the question is about measurements on the global state vector of a system, the outcome is by definition isometric (the final state has the same norm as the original state vector) $\endgroup$
    – lurscher
    Oct 17 '11 at 19:33

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