Well, zero of course. Because

$S = -\text{tr}(\rho \ln \rho)$ and $\rho$ for a pure state gives zero entropy.

But... all quantum states are really pure states right? A mixed state just describes our ignorance about a particular system. So how can properties like entropy and temperature come out of ignorance of information? That doesn't make sense to me.

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    $\begingroup$ where did you think entropy came from classically? $\endgroup$ Nov 9, 2014 at 22:05
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    $\begingroup$ entropy is not an easy concept! but the more you read and think about it, the more sense it makes. You can write an entire book discussing its meaning(s). $\endgroup$
    – user65081
    Nov 9, 2014 at 22:09
  • $\begingroup$ Classically? Basically the heat-engine way it was originally defined. $\endgroup$
    – Nick
    Nov 9, 2014 at 22:38
  • $\begingroup$ Your question stands at the border between quantum and classical mechanics. Nobody knows exactly where is this border. E.g., an object consisting in many particles, shall we describe it by quantum, or by classical mechanics? A classical object is in so rapid exchange of energy and particles with the environment, that we can't even say at every moment how many particles and what energy it possesses. So, your mixed state is a quantum superposition of states of a many-particle object and we don't know the phases, or we have to do with a classical many-body object? $\endgroup$
    – Sofia
    Nov 9, 2014 at 23:22
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    $\begingroup$ Isn't a pure state equivalent to a "microstate" in statistical mechanics? I think one can only talk about the entropy of the macrostate to which a particular microstate is a part of (and most macrostates consist of large sets of microstates, although in principle there can be macrostates consisting of a single microstate). Would "macrostates" in QM stat. mech. normally be mixed states, consisting of different statistical ensembles of pure states? I'm not sure but if so, the same idea could apply, in that you could only talk about the entropy of the mixed state the pure state was part of. $\endgroup$
    – Hypnosifl
    Nov 10, 2014 at 3:07

5 Answers 5


I think it is a mistake, in this case, to think of entropy as "a description of our ignorance." Rather, I would suggest that you think of entropy as a well-defined, objective property provided that you specify which degrees of freedom in the universe are inside and outside of your system. The content of this statement isn't really different, but it emphasizes that entropy is an objective property and not observer-dependent.

If your included list is "everything" (or at least everything that has ever interacted together in the history of your system), then what you said is true: if you started out with a pure state it will always remain so, and there isn't much thermodynamics to speak of.

The basic question of thermodynamics (and, more broadly, statistical mechanics) is what happens in any other case - most typically, the case in which the degrees of freedom you specify are continuously coupled to an open system in some way. Somewhat amazingly, there is a general answer to this question for many such arrangements.

More concretely, in classical thermodynamics one of the important things about entropy and temperature is that they tell you how much work you can extract of a system. So one way to reform your question is: "How can properties like maximum work extracted come out of ignorance of information?" But it is easy to think of situations when this is the case. As a toy model, imagine a sailor trying to navigate a sailboat in some storm, with the wind changing wildly and rapidly. If he somehow knows beforehand exactly when and how the wind will shift, he will have a much easier time moving in the direction he wants.

Ultimately, a similar game is being played on a microscopic level when one speaks, for example, of the maximum efficiency possible in a heat engine. The explicit connection is made by Landauer's Principle, which is the direct link you're looking for between the included degrees of freedom (or, if you insist, "knowledge") and work. This law was inspired by the famous thought experiment Maxwell's Demon, which is a microscopic equivalent to my weather-predicting sailor.

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    $\begingroup$ "Rather, I would suggest that you think of entropy as a well-defined, objective property provided that you specify which degrees of freedom in the universe are inside and outside of your system." Bingo. $\endgroup$
    – DanielSank
    Nov 12, 2014 at 5:45
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    $\begingroup$ @Nick, regarding your frying pan question (sorry, I don't have enough reputation to comment under that question), you may be interested in a recent proposal to modify the standard definition of entropy to apply to closed systems. However, see also this rebuttal, which argues the conventional view: that pure states have an entropy of 0 and a temperature that is ill-defined. $\endgroup$
    – Rococo
    Nov 12, 2014 at 6:14
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    $\begingroup$ I like the idea behind your first sentence, but it does not really follow that entropy is not observer-dependent (it is observer-dependent!). Rather, you are defining a "standard observer" who has access to a given set of observables (or technically, a thermodynamic state space). Then, for all standard observers, the entropy is uniquely defined. However, you can still imagine observers with access to more information (e.g. Maxwell demons) for whom the observed entropy is different. $\endgroup$ Nov 21, 2014 at 12:14
  • $\begingroup$ Hi @MarkMitchison- I'm not sure we disagree in any deep way, but there were two things I wanted to emphasize: $\endgroup$
    – Rococo
    Dec 1, 2014 at 22:13
  • $\begingroup$ 1. The definition of entropy does not in any way require the notion of an observer, but does require a specification of the subspace considered, to get the density matrix. 2. An observer may measure different entropy depending on which aspects of the system he/she considers. Concretely, for a system of two entangled particles one will measure a different entropy for each of the particles independently than the full entangled state. $\endgroup$
    – Rococo
    Dec 1, 2014 at 22:25

There is a difference between the 'fine grained' and the 'coarse grained' entropy. If we start with a pure state (zero entropy), and we time-evolve it, the entropy indeed stays zero by unitary of time evolution. The fine grained entropy did not change.

The coarse grained entropy is what we usually call the thermal entropy, and is the thing that always increases (or stays equal) with time. Consider a system with more than one subsystem. The thermal entropy is defined as the sum of all the entropies of the subsystems. Say, for example, we start in a pure state of the system as a whole, and all the subsystems are in their own pure states as well (not necessarily so, but we choose it to be so in this example). At a later time, the subsystems will develop correlations because of interactions with other subsystems. Since the subsystems now have an (entanglement) entropy, the overall thermal entropy will not be zero.

  • $\begingroup$ I basically know this from the book "An Introduction To Black Holes, Information And The String Theory Revolution: The Holographic Universe" by L. Susskind and J. Lindesay, chapter "the laws of nature." (You don't need to know anything about the things in the title to understand the discussion about entropy) There, you can also find a nice explanation that involves the evolution of a certain volume in phase space. $\endgroup$
    – Georg
    Nov 9, 2014 at 23:25
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    $\begingroup$ Sorry, the answer is not satisfactory. If you assign to an ideal gas a quantum state, then if the many-particle object doesn't interact with the environment, it evolves unitarily, and a quantum state remains a single quantum state. $\endgroup$
    – Sofia
    Nov 9, 2014 at 23:40
  • $\begingroup$ @Sofia True, but what is the problem with that? In my example it also stays a pure quantum state (although the thermal entropy increases). $\endgroup$
    – Georg
    Nov 9, 2014 at 23:58
  • $\begingroup$ "Consider a system with more than one subsystem." What is the rigorous definition of a subsystem? How is a multidimensional wavefunction broken down, mathematically, into subsystems? $\endgroup$
    – Nick
    Nov 10, 2014 at 7:33
  • $\begingroup$ Explaining everything precisely would be too elaborate here. For the details you'll have to read a book about it. (Like the one I mentioned earlier). In short: when you have a Hilbert space that can be written as a tensor product of 'smaller' Hilbert spaces, each of these smaller Hilbert spaces is a subsystem. $\endgroup$
    – Georg
    Nov 10, 2014 at 10:40

The issue is that when a wavefunction collapses it has an inherent randomness to it. Since entropy is fundamentally related to information, I'll start with information to explain why this is significant.

Information, Randomness, and Entropy

If you think of the information of a state as what is needed to completely define a system something interesting happens. This is best illustrated by taking a picture and trying to compress it. Let's take this image:

enter image description here

You can think about the information necessary for us to completely define a system as being the total amount of details we need to describe the system. For this image, you CAN represent it as a collection of 2000 x 2000 pixels which each require 32 bits of information (RGB+Alpha), or $\ 128$ $10^8$ bits of data.

However, let's say we start compressing it (losslessly), such that we can fully reconstruct it from whatever information we have remaining. We can easily see for all pixels but the shadow we either have full transparency or fully opaque, so we can remove 7 bits of information from all other points and represent the alpha as either 1 or 0. We also see that anywhere with full transparency doesnt require RGB values, so we can remove 24 bits of information from each spot with full transparency.

Since this image was actually originally an svg, it turns out if we keep on getting smarter and smarter how we compress this we could actually reduce the image down to basically 2 shapes (white outline and the petal shape), 1 radius and 4 angles about the center to position each shape, 4 base colors (5th for the white), an equation for the gradient color change, and an equation for the shadow effect.

It turns out this system actually only has very little information. The amazing part is that this is the case only because the image we are looking at isn't very random. It can be represented with few bits of information due to it's uniformity. We were able to fully reduce all the redundant data because it was essentially not random.

It turns out that for an image like this:

enter image description here

It would take far more information to fully describe since it is essentially random all throughout. We are not able to really do any compression because we almost all the information we have is not redundant. The information of a system that has $n_i$ many states that require $\sigma_i$ bits of information (e.g: colors, equation of shape) to be described boils down to something approximately like

$\sum\limits_{i=1}^N n_i\sigma_i$ + (random information)

Quantum Mechanics/Summary

What in the world does this have to do with quantum mechanics and the entropy of a quantum system?

QM is a probabilistic theory at it's core. When a wave function collapses it does so randomly, so completely describing the system after a collapse of a wavefunction takes more and more exact information if you want an exact model of what's happening. This increasing randomness turns out to be the Law of Entropy in action. As systems interact random quantum effects increasingly perturb the system. If you want you can say that it comes out of "hidden variables" but many physicists have worked to disprove the orthodox viewpoint of "hidden variables" (to great success might I add). A collapsing wave function is inherently random, so it introduces randomness as the clock keeps ticking on.

Related Media: What is NOT Random


But... all quantum states are really pure states right?

There exists the density matrix formulation for the many body system quantum mechanically. What happens in a many body problem, order of 10^23 molecules per mole that are the appropriate numbers for a thermodynamic formulation, is that the off diagonal elements are so small that they are effectively zero and the many body system has lost the quantum mechanical correlations. The ensemble is not in a pure state as far as our measurement accuracies go.

A mixed state just describes our ignorance about a particular system.

Also our inability to measure the quantum mechanical correlations due to their infinitesimal size. Note that in certain conditions, superconductivity for example, the correlations exist strongly and pure states persists even in the large number many body state. In the many body state where the classical thermodynamic quantities have been theoretically defined and identified with measured variables, the quantum mechanical correlations are lost.

So how can properties like entropy and temperature come out of ignorance of information?

Temperature, pressure etc thermodynamic quantities are defined by measurements and experiments. The thermodynamic model/theory was developed to describe the experimental measurements and predict new situations. It works beautifully in the framework it was defined, except for situations where the quantized nature of the underlying framework created anomalies, as in the black body radiation. This last disagreement with the elegant theory of thermodynamics was one of the reasons that the quantized nature of radiation was postulated and added to the necessity of the theory of quantum mechanics.

When studying physics at different dimensions and energies, one has to keep in mind that all the theories are really models fitted on observations and data. The theories are appropriate for the framework they were developed and a hierarchy exists going from the quantized microcosm of elementary particles, atoms and molecules to the many body problem of macroscopic dimensions. Classical thermodynamics is an emergent theory from the many body state of the particulate nature of matter.


A mixed state just describes our ignorance about a particular system

I don't think you can call our inability to access a pure state of any system an ignorance about a particular system. Because I think of pure state as a mathematical abstruction that can only be related to reality by application of Born rule - wich either reflects our fundamental ignorance (akin to Kant's numenal) or indeed a lack of adequate theory (hidden variables), but in any case not an ignorance about a particular system. Termodinamic properties arise not from ignorance about pure state, but from discarding or ignoring the accessible information, wich translates into entropy.


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