2
$\begingroup$

It is well known that the set of density operators $\{\rho\}$ for a quantum theory form a convex set. As I have seen them defined, we simply say that a state corresponds to some linear operator $\rho$ which is self-adjoint, positive semidefinite, and of unit trace.

Now it is often taken in books as obvious that, for a situation (I'm struggling to make this precise which is part of the question I suppose) in which there is some "classical" uncertainty about the state of some quantum system which may be, with probability $p_n$, in some pure state $|\psi_n\rangle \langle \psi_n|$, we must represent the state of the system via a weighted sum of these pure states. More precisely, we take $$\rho = \sum_n p_n |\psi_n\rangle \langle \psi_n|, \sum p_n = 1, p_n \in [0,1] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (A)$$

Is there any reason other than "experiment confirms this choice" that we might take this? Obviously this also depends on exactly which physical contexts imply this choice for $\rho$. For instance, if I enter a laboratory and am unsure if a spin has been prepared up or down but I know it is in one or the other, is it correct for me to assign $\rho = \frac{1}{2}|+\rangle \langle +|+\frac{1}{2}|-\rangle \langle -|$? It seems absurd since this cannot reproduce the statistics of measurements of the actual pure state.

Thus to rephrase and try to be more precise, my questions are:

(1) When is (A) the correct choice for $\rho$?

(2) Can we prove that it is the correct choice without appealing to experiment? NB that I understand that this $\rho$ is a valid state since $\{\rho\}$ forms a convex set; thus my question here is not "why is it a valid state" but rather "why is it the valid state for this physical situation"?

$\endgroup$
6
  • 1
    $\begingroup$ "since this cannot reproduce the statistics of measurements of the actual pure state."-why do you think so /what do you mean? QM only yields probababilitistic predictions. So what you want to reproduce is the correct statistics of many measurements- $\endgroup$ Commented Jan 24, 2023 at 7:20
  • $\begingroup$ @TobiasFünke What I mean is that a "classical" mixture and a quantum superposition which look ostensibly the same will not in general produce the same statistics. Consider $\rho$ as I've given it versus the $\rho'$ formed from the pure state $\rho = (|+\rangle + |-\rangle)/\sqrt{2}$. While I understand that QM only yields probabilistic predictions, my point there is that, in that particular example, my classical mixture intuition yields different predictions than whatever will actually arise from the real pure state (supposing I ran this "experiment" many times). $\endgroup$
    – EE18
    Commented Jan 24, 2023 at 15:13
  • $\begingroup$ When you the "the valid state", etc., are you thinking of $\rho$, or of some specific decomposition of $\rho$? The former should be clear (it predicts all measurement outcomes), and the latter is unphysical (different decompositions cannot be distinguished by any experiement). $\endgroup$ Commented Jan 24, 2023 at 21:08
  • $\begingroup$ @NorbertSchuch Here the question is not about the mathematical framework. I am overloading the words "valid state" here, so my apologies for the lack of clarity. In my last sentence of the OP, the first "valid state" refers to "the mathematical object $\rho$ obeys the three requirements" whereas the second "valid state" refers to why this choice of $\rho$ is appropriate for the given physical system. Thus this question is about the correspondence rule between the physical world and the mathematical framework for these particular cases of "classical mixtures". $\endgroup$
    – EE18
    Commented Jan 24, 2023 at 21:12
  • $\begingroup$ @EE18 As far as I can tell there are two aspects to the question. (1) Why $\rho$ in such a situation? This question is answered in the accepted answer. (2) Why write $\rho$ as this sum? This is a matter of interpretation, and there is no "right" decomposition. The one you give is not better than any other, regardless of the situation. (Unless you would like to stick to some interpretation which gives meaning to unphysical things, of course.) $\endgroup$ Commented Jan 24, 2023 at 21:21

2 Answers 2

2
$\begingroup$

The heuristic I've seen before is to stipulate that there should be two kinds of averages,

  • a coherent (quantum) one in which we take expectation values as $\langle \psi | \hat{A} | \psi \rangle$, and

  • an incoherent (classical) one in which we take averages over a probability distribution in the usual way, i.e., $\bar{a} = \sum_n p_n a_n$.

Then, if you have a system that could be in multiple quantum states $|\psi_n\rangle$---and you don't know which one but you can assign a probability distribution $p_n$ to them---the average is given by the "double" average $$ \overline{\langle \hat{A} \rangle} = \sum_n p_n \langle\psi_n|\hat{A}|\psi_n\rangle\,. $$ From here, it's a matter of rearranging things. First, choose any orthonormal basis $|m\rangle$, and resolve the identity in this expression as $$ \overline{\langle \hat{A} \rangle} = \sum_n p_n \langle\psi_n|\hat{A}|\psi_n\rangle = \sum_n p_n \langle\psi_n|\hat{A}\left( \sum_m |m\rangle\langle m | \right)|\psi_n\rangle\,. $$ Rearranging, we get $$ \overline{\langle \hat{A} \rangle} = \sum_m\sum_n p_n \langle\psi_n|\hat{A} |m\rangle \langle m |\psi_n\rangle = \sum_m\sum_n p_n \langle m |\psi_n\rangle\langle\psi_n|\hat{A} |m\rangle \,. $$ Finally, push the sum over $n$ through to get $$ \overline{\langle \hat{A} \rangle} = \sum_m \langle m |\left( \sum_n p_n|\psi_n\rangle\langle\psi_n|\right)\hat{A} |m\rangle =\operatorname{Tr}\left(\rho\hat{A}\right)\,, $$ where $$ \rho = \sum_n p_n|\psi_n\rangle\langle\psi_n|\,. $$

$\endgroup$
8
  • $\begingroup$ If I am following your logic correctly, your argument is that we can obtain the density operator from the following heuristic: (1) Suppose we don't know the structure of the density operator a priori, but we do know that it will be in one among some set of pure states with various probabilities, then (2) We can use probability theory (in particular, the Law of total expectation -- en.wikipedia.org/wiki/Law_of_total_expectation) to develop an expression for the expectation value in a measurement of $A$ (3) Perform the manipulations you've done, thus... $\endgroup$
    – EE18
    Commented Jan 24, 2023 at 17:08
  • $\begingroup$ ...arriving at a possible form for $\rho$. But of course, all we wanted was motivation for the structure of $\rho$ for this particular case. Does this sound right? The law of total expectation comes from using "quantum expectation given that it's in the given pure state". $\endgroup$
    – EE18
    Commented Jan 24, 2023 at 17:08
  • $\begingroup$ @EE18. This isn't right: "Suppose we don't know the structure of the density operator a priori, but we do know that it will be in one among some set of pure states with various probabilities", but maybe you just phrased it imprecisely, in the sense that "it" should refer to the "system" not the "density matrix". And yes, I feel like all this does is motivate the definition of the density matrix. (But yes, I feel like that's all we need, and then we compare to experiments.) $\endgroup$
    – march
    Commented Jan 24, 2023 at 17:12
  • $\begingroup$ @EE18. You might also want to check out Gleason's theorem $\endgroup$
    – march
    Commented Jan 24, 2023 at 17:12
  • $\begingroup$ You are very right, I should have said "system" and not "density operator". We of course end up deducing/motivating the structure of the density operator at the end. $\endgroup$
    – EE18
    Commented Jan 24, 2023 at 17:17
0
$\begingroup$

Short answers : yes.

I guess that your confusion comes from the different natures of randomness.

  1. Quantum randomness : as you raised it, quantum systems possess an intrinsic randomness, absent from classiscal systems. Indeed, even when there is no uncertainty about the state of the considered system, that is when we know its state is $|\psi\rangle$, hence a pure state density matrix $\hat{\rho} = |\psi\rangle\langle\psi|$, it is still described by a propability distribution carried by $\psi$.

  2. Classical randomness : this type of randomness corresponds to a mere statistical mixture (of pure states), hence the form $\hat{\rho} = \sum_np_n|\psi_n\rangle\langle\psi_n|$. In other words, when doing measurements, you pick a state $|\psi_n\rangle$ in the same manner you would pick a given ball from an urn. Nevertheless, another specificity of quantum statistical mixtures over classical ones in that case is the superposition principle; indeed, the quantum system is in every state of the mixture at the same time, hence the linear combination structure, such that the individual (entangled) states cannot be separated as the balls in the urn could be $-$ even by measurements, since the wavefunction collapse "destroys" the other states of the mixture.

  3. "Bayesian randomness" : the last type of randomness, the one modelled by the $\frac{1}{2}$-weights in $\hat{\rho} = \frac{1}{2}|+\rangle\langle+| + \frac{1}{2}|-\rangle\langle-|$, doesn't describe the system as such but your guess about it, because of your lack of knowledge when entering the room, in the same way as the a priori distribution of bayesian theory, which you will update afterwards, thanks the additional informations due to measurements.

$\endgroup$
1
  • $\begingroup$ What about "randomness because you only have access to part of the system"? -- Independent of that, the issue I have with this answer is that it it makes the impression these "different natures of randomness" can be distinguished in some way. $\endgroup$ Commented Jan 24, 2023 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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