Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Classically, probability distributions are nonnegative real measures over the space of all possible outcomes which add up to 1. What they mean is open to debate between Bayesians, frequentists and ensemble interpretations. A degenerate distribution is the least random distribution with a probability of 1 for a given fixed event, and 0 for everything else.

What is the analog of a classical probability distribution in quantum mechanics? Is it a wave function augmented with the Born interpretation for probabilities, or is it the density matrix? Does a pure density matrix correspond to a degenerate distribution?

share|improve this question
1  
The point is perhaps that quantum theory states are not analogs of probability distributions -- at least not exactly. I suggest arxiv.org/abs/quant-ph/0101012v4 as an interesting attempt to squeeze the two as close together as they can be (it's 34 pages, but Lucien Hardy is relatively easy reading). I'm not as happy as I'd like with this response, which is why it's a comment, not an Answer. –  Peter Morgan May 27 '11 at 12:23
1  
One obligatory reference is that of the Wigner quasi-probability function: en.wikipedia.org/wiki/Wigner_quasi-probability_distribution . This object, which is the Wigner transform of the coordinate representation of the density matrix, is as close as you can get to the classical distribution function. You can even prove that it becomes the classical phase space distribution function in the classical limit. –  Olaf May 27 '11 at 13:15
    
Help my simple confusion. Isn't probability basically just the wave function squared? –  Mike Dunlavey Dec 6 '11 at 14:38
    
It's difficult to give an authoritative answer to this one. But one thing to note is that, formally, density matrices are exactly equivalent to probability distributions in the case where you never transform them to have non-zero elements off the diagonal, i.e. you always work in the same orthonormal basis. It's the ability to measure in different bases that gives density matrices their specifically "quantum" character. –  Nathaniel Mar 7 '12 at 18:34
add comment

8 Answers

I'll reproduce my comment from above here.

The point is perhaps that quantum theory states are not analogs of probability distributions -- at least not exactly. I suggest arxiv.org/abs/quant-ph/0101012v4 as an interesting attempt to squeeze the two as close together as they can be (it's 34 pages, but Lucien Hardy is relatively easy reading). I'm not as happy as I'd like with this response, which is why it's a comment, not an Answer.

I decided I was being lazy, and looked at Lucien's paper for what I might take to be its relevance to your Question. Lucien takes quantum pure states to be analogous enough to degenerate probabilities to use the idea. Where that becomes interesting is the way he can then characterize the difference between classical probabilistic states and quantum states, given this starting point. The distinguishing feature is his fifth axiom,

"Axiom 5 Continuity. There exists a continuous reversible transformation on a system between any two pure states of that system."

IMO, this is definitely a curious way to construct things. It has a distinct failing, that it's limited to finite-dimensional Hilbert spaces and probability distributions over a finite set of outcomes, and AFAIK no-one has extended Lucien's analysis to infinite dimensional Hilbert spaces and probability spaces, which somewhat diminishes its interest unless you in any case work only with finite dimensional Hilbert spaces (as you might if you work in quantum information).

The point I'd make about this is that this is an interesting partial analogy, although I do not know that any more directly related-to-experiment use has been made of it. It may well be worth thinking in terms of this partial analogy some more, but my personal assessment has been that this is not something worth hanging my hat on exclusively. On the other hand, it's only if one immerses oneself in a way of thinking in a committed way that new results come from anywhere, and it's because people have made different choices than I make that we get different results.

In any case, I think Lucien's paper is relevant for you. It has, to me, the same feel as the way you have asked your Question.

share|improve this answer
add comment

This question is rather broad, but I will try to address the main issue. A state $|\psi\rangle$ is alinear sum of eigenvectors $|n\rangle$ so that $$ |\psi\rangle~=~\sum_nc_n|n\rangle. $$ The probabilities maybe directly computed this way from $\langle\psi|\psi\rangle$ $=~1$ as $$ \langle\psi|\psi\rangle~=~\sum_{mn}c^*_mc_n\langle m|n\rangle $$ and since $\langle m|n\rangle~=~\delta_{mn}$ and the amplitudes define the probabilities as their modulus squared $ c^*_nc_n~=~P_n$ this expectation is then just a sum of the probabilities. The born rule then tells that for an observable ${\cal O}$ diagonal in this basis with ${\cal O}|n\rangle~=~o_n|n\rangle$ the expected value of the observable for $|\psi\rangle$ is then $$ \langle\psi|{\cal O}\psi\rangle~=~\sum_{mn}c^*_mc_no_n \langle m| n\rangle~=~\sum_no_nP_n. $$ This is just the sum of the eigenvalues of the observable times their probabilities for being observed.

The density matrix is really a form generalization of this. We consider the outer product of the states as the density operator $$ {\hat\rho}~=~|\psi\rangle\langle\psi|~=~\sum_{mn}c^*_mc_n|n\rangle\langle m| $$ which is a matrix representation of this operator. The trace of the density matrix, $Tr{\hat\rho}~=~\sum_n\langle n|{\hat\rho}|n\rangle$ then gives the sum of probabilities. I leave it as an exercise to evaluate an observable in the trace $Tr{\cal O}{\hat\rho}$ to derive the Born rule.

The modulus squares of the amplitudes define the probabilities, which are real valued. So the connection between the states and the probabilities is not one to one. One reason people like the density matrix is that there is more of a connection between a linear quantum operator and the probabilities in the trace operation.

share|improve this answer
add comment

You can make a very strong formal analogy between the density matrix and classical probability distributions. An alternate way to think of random variables from a particular distribution is instead to focus on an algebra of observables. Random variables are closed under products, sums, complex conjugation, and so forth, and expectation (with respect to a given distribution) must be a linear functional. This can be modeled perfectly well with the standard $\langle X \rangle = \operatorname{Tr} (\rho X)$ of quantum mechanics.

Seen this way, classical probability distributions are the special case where all of the operators commute. Since they all commute, they can be simultaneously diagonalized, and thus evrything observable about the distribution is entirely captured by the diagonal elements in this basis. The possible non-commutativity of operators in the quantum case allows for behaviors different than the classical case, of course. (If it didn't, it would be the classical case, rather than analogous to the classical case.) In this picture, wavefunctions are more analogous to classical ontic states, but with the curious feature that these pure state have questions with no definite answers.

share|improve this answer
add comment

The density matrix is the quantum analogue of a classical probability density.

If you write quantum mechanics from the start in terms of an algebra of observables and density matrices (rather than wave functions and operators), it looks very much like a direct generalization of classical mechanics, in the sense that almost everything is the same except that commutativity of multiplication is lost. (Indeed, in mathematics, the subject studying the probabilistic aspects of the quantum mechanical formalism is usually called ''noncommutative probability''.)

You can convince yourself of this by looking at my book ''Classical and Quantum Mechanics via Lie algebras'' (http://lanl.arxiv.org/abs/0810.1019), where this approach is followed systematically.

For example (directly relating to your question), in classical probability theory a pure state (with a degenerate distribution in your terminology) is characterized by zero entropy, and in quantum mechanics, a pure state (i.e., one in which the density matrix can be written as $\psi\psi^*$ with a wave function $\psi$ is also characterized by zero entropy. The formula for the entropy is also the same; $S=\langle -\log\rho\rangle$, with expectation taken in the state $\rho$.


Expressed directly in terms of the density matrix, quantum mechanics is governed by the following six axioms and their explanation. (This is taken from the Section ''Postulates for the formal core of quantum mechanics'' of Chapter A1: Fundamental concepts in quantum mechanics of my theoretical physics FAQ.)

Note that the only difference between classical and quantum mechanics in this axiomatic setting is that

  • the classical case only works with diagonal operators, where all operations happen pointwise on the diagonal elements. Thus multiplication is commutative, and one can identify operators and functions. In particular, the density mattrix degenerates into a probability density.
  • the quantum case allows for noncommutative operators, hence both observable quantities and the density are (usually infinite-dimensional) matrices.

    A1. A generic system (e.g., a 'hydrogen molecule') is defined by specifying a Hilbert space $K$ and a (densely defined, self-adjoint) Hermitian linear operator $H$ called the Hamiltonian or the energy.

    A2. A particular system (e.g., 'the ion in the ion trap on this particular desk') is characterized by its state $\rho(t)$ at every time $t \in R$ (the set of real numbers). Here $\rho(t)$ is a Hermitian, positive semidefinite, linear trace class operator on $K$ satisfying at all times the condition

    $Tr\ \rho(t) = 1$, (normalization)

    where $Tr$ denotes the trace.

    A3. A system is called closed in a time interval $[t_1,t_2]$ if it satisfies the evolution equation

    $d/dt\ \rho(t) = i/\hbar [\rho(t),H] \mbox{ for } t \in [t_1,t_2]$,

    and open otherwise. ($\hbar$ is Planck's constant, and is often set to 1.) If nothing else is apparent from the context, a system is assumed to be closed.

    A4. Besides the energy $H$, certain other (densely defined, self-adjoint) Hermitian operators (or vectors of such operators) are distinguished as observables. (E.g., the observables for a system of $N$ distinguishable particles conventionally include for each particle several 3-dimensional vectors: the position $x^a$, momentum $p^a$, _orbital_angular_momentum_ $L^a$ and the _spin_vector_ (or Bloch vector) $\sigma^a$ of the particle with label $a$. If $u$ is a 3-vector of unit length then $u \cdot p^a$, $u \cdot L^a$ and $u \cdot \sigma^a$ define the momentum, orbital angular momentum, and spin of particle $a$ in direction $u$.)

    A5. For any particular system, and for every vector $X$ of observables with commuting components, one associates a time-dependent monotone linear functional $\langle \cdot\rangle_t$ defining the expectation

    $\langle f(X)\rangle_t:=Tr\ \rho(t) f(X)$

    of bounded continuous functions $f(X)$ at time $t$. (This is equivalent to a multivariate probability measure $d\mu_t(X)$ on a suitable sigma algebra over the spectrum $spec(X)$ of $X$) defined by

    $\int d\mu_t(X) f(X) := Tr\ \rho(t) f(X) =\langle f(X)\rangle _t$.

    This sigma algebra is uniquely determined.)

    A6. Quantum mechanical predictions consist of predicting properties (typically expectations or conditional probabilities) of the measures defined in Axiom A5, given reasonable assumptions about the states (e.g., ground state, equilibrium state, etc.)

    Axiom A6 specifies that the formal content of quantum mechanics is covered exactly by what can be deduced from Axioms A1-A5 without anything else added (except for restrictions defining the specific nature of the states and observables), and hence says that Axioms A1-A5 are complete.

    The description of a particular closed system is therefore given by the specification of a particular Hilbert space in A1, the specification of the observable quantities in A4, and the specification of conditions singling out a particular class of states (in A6). Given this, everything else is determined by the theory, and hence is (in principle) predicted by the theory.

    The description of an open system involves, in addition, the specification of the details of the dynamical law.

  • share|improve this answer
    add comment

    It is wrong to think of a mixed density state as a probability distribution over "ontic" wavefunctions. See the question Are these two quantum systems distinguishable? . Two different ways of preparing a quantum state are given and the "ontic" wavefunctions don't agree, but you know what? The density matrices still agree and we can never tell the difference.

    share|improve this answer
    add comment

    In the orthodox view, the wave function is the quantum analogue to a physical state in classical mechanics, and hence a degenerate probability distribution in classical probability. A mixed state is the quantum analogue to a probability distribution on phase space. A density matrix is the usual way to represent a mixed state, nowadays, it was not part of the original formulation of QM but was invented by von Neumann. It differs from a classical proability distribution, just like all quantum notions differ from classical notions, and leads to the formulation of a new, non-classical, probability, called either Quantum Probability or Non-Commutative Probability.

    Not everyone will agree with me. Within QIT, and especially within one of Lucien Hardy's early 're-constructions' of QM, see

    Quantum Theory from Five Reasonable Axioms. http://arxiv.org/abs/quant-ph/0101012

    he takes mixed states or density matrices as the fundamental 'physical states', perhaps without caring whether this is an alteration of the orthodox QM or not. Admittedly, this revision of QM is very commonly accepted by physicists now.

    Elsewhere, (just google on Axiomatisation of Physics) I have argued
    that he would need to seriously revise even classical probability to make his ideas work, and I doubt it is possible, hence I doubt that his axioms are physically true. This at least shows, whether they are true or not, that they are different from normal QM, even for finite dimensional Hilbert spaces. I do not think very many QIT people understand even classical probability, even for discrete spaces, but then, it is awfully tricky. For the foundations of classical probability, see my own

    "The Logical Structure of Physical Probability Assertions", http://arxiv.org/abs/quant-ph/0508059

    share|improve this answer
    add comment

    The wavefunction does not have the same ontic status as a classical probability distribution of a Markovian process. Ignore classical probability distributions which can take on any form subject to the axioms of probability. If one just focuses upon Markovian processes, there is no need to specify the entire probability function for every single time step. All one needs to postulate are the existence of genuine randomness generators of the simple sort. All one needs to keep track of at each time step is the actual state of the system. The following state is determined from the Markov transition matrix and the random generator according to the Monte Carlo method. Unfortunately, this can't work in quantum mechanics because of interference. Pusey et al. proved one needs the entire wavefunction. The wavefunction is not analogous to the classical probability distribution of a Markovian process.

    share|improve this answer
        
    this does not answer the question –  Arnold Neumaier Nov 15 '12 at 14:54
    add comment

    The density matrix has a property not shared by classical probability distributions. If you take the marginal trace of a classical degenerate distribution, you end up with another degenerate distribution, but if you take the partial trace of a pure state, you can easily end up with a mixed state. Such is the CRAZY nature of entanglement. This is the property used by decoherence guys to explain how probabilities emerge from a wavefunction. This does not explain away macroscopic superpositions by any means. A macroscopic superposition with suppressed interference terms is still a superposition. A decohered superposition, but a superposition nonetheless.

    share|improve this answer
        
    doesn't answer the question –  Arnold Neumaier Nov 15 '12 at 15:07
    add comment

    Your Answer

     
    discard

    By posting your answer, you agree to the privacy policy and terms of service.

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