Are superposition states observable?

[mike1000]

The way I am coming to understand it, the allowed states that an observable can be "observed/measured" in are defined by the eigenvectors (and associated eigenvalues) of the observable's operator. Since those eigenvectors form a basis and span the space of vectors defined by the operator, a linear combination of two or more eigenstates is also an allowed state of the observable i.e. superposition.

Does this mean that an observable can be observed/measured in a state which is a superposition of eigenstates?

Here is a quote from Dirac's book "The Principles of Quantum Mechanics". I do not have this book and I have not read this book, yet!) Someone in another thread mentioned it and that started me on a quest. In that quest I found the following quotation from Dirac's book. Here is that quote...

"...We are observing whether it is polarized parallel or perpendicular to the optic axis. The effect of making this observation is to force the photon entirely into the state of parallel or entirely into the state of perpendicular polarization. It has to make a sudden jump from being partly in each of those two states to being entirely in one or the other of them. Which of the two states it will jump into cannot be predicted, but is governed only by probability laws."

It seems to me that Dirac is saying, "No, we cannot observe/measure the particle in a superposition of states"

Or maybe he is saying that if we want to observe the photon in its superposition state we need a different way to measure it!

What if we did not want to observe whether the photon was polarized in one of only two states? Did we force it into one of those two states by the method we used to measure it?

 

[DrChinese]

...

What if we did not want to observe whether the photon was polarized in one of only two states? Did we force it into one of those two states by the method we used to measure it?

Usually, you expect a photon polarization observation to result in one of two values - on the selected basis. There are essentially an infinite number of those bases, for starters (rotating around 360 degrees). Also, if you have full knowledge of state on one basis, there is at least one basis which is now completely unknown. Further, there is no requirement that you have complete knowledge on any basis. You could theoretically have partial knowledge on one basis, and partial knowledge on another basis - that would not violate the HUP.

 

[Jilang]

So does the measurement force it to be in one of the available states for that basis?

 

[PeterDonis]

does the measurement force it to be in one of the available states for that basis?

It depends on whether you adopt a collapse or a no collapse interpretation of QM. On a collapse interpretation, yes, the measurement collapses the state onto one of the eigenstates of the measurement operator. On a no collapse interpretation, no, it doesn't; all of the branches of the superposition remain (but each branch gets entangled with the corresponding state of the measuring device, observers, etc.).

 

[Nugatory]

So does the measurement force it to be in one of the available states for that basis?

As PeterDonis says, that depends on your interpretation. You can avoid the interpretational swamp by saying that your subsequent measurements will behave AS IF the measurement had forced the system to be in one of the particular states, namely the one that you measured.

 

[mikeyork]

It seems to me that Dirac is saying, "No, we cannot observe/measure the particle in a superposition of states"

Or maybe he is saying that if we want to observe the photon in its superposition state we need a different way to measure it!

The latter. Any physical state is represented by an eigenvector in some observable and frame of reference -- but a superposition in others. In particular, any observable state will have been prepared (either by our apparatus or by nature) in an eigenstate of some basis even if a superposition in another. So that preparation basis is the basis in which a superposition in another basis could be "observed" as such.

As a fairly simple example, prepare an electron state with spin +1/2 along the z-axis. Measuring the spin projection on the z-axis is effectively the same as measuring a superposition in any other direction (except the opposite direction where it is an eigenstate with spin -1/2). Repeating such an experiment of measuring in another direction over and over will then give the appropriate statistical frequency that converges on that corresponding to the superposition.

Having said that, "observing a superposition state" is a very convoluted and strange way to describe this situation.

 

[mike1000]

The latter. Any physical state is represented by an eigenvector in some observable and frame of reference -- but a superposition in others. In particular, any observable state will have been prepared (either by our apparatus or by nature) in an eigenstate of some basis even if a superposition in another. So that preparation basis is the basis in which a superposition in another basis could be "observed" as such.

As a fairly simple example, prepare an electron state with spin +1/2 along the z-axis. Measuring the spin projection on the z-axis is effectively the same as measuring a superposition in any other direction (except the opposite direction where it is an eigenstate with spin -1/2). Repeating such an experiment of measuring in another direction over and over will then give the appropriate statistical frequency that converges on that corresponding to the superposition.

Having said that, "observing a superposition state" is a very convoluted and strange way to describe this situation.

Let me try saying it this way. I am really asking a very simple, and I think, very direct question, that to me, should only have one answer and the answer should be "Yes".

Reading a little of Dirac and beginning to learn the basics of Matrix Mechanics, it is the eigenstates of an particular operator that become the states in which the observable can be found (measured). Also, the eigenstates form an orthonormal basis for the observable. It has been pointed out many times that a state formed by a linear combination of eigenstates(superposition) is also a pure state for the observable. Doesn't this imply that we should be able to observe a particle in a superposed state?

But I gather from all of the responses, that the answer is no, we cannot observe a particle in a superposed state. This implies that a quantum particle in a superposed state is NOT an observable, doesn't it? And if a particle in a superposition of eigenstates is not an observable, then shouldn't the superposed state not be considered a "pure" state?

 

[mikeyork]

Let me try saying it this way. I am really asking a very simple, and I think, very direct question, that to me, should only have one answer and the answer should be "Yes".

Then the answer depends on what you mean by the question. If by "observing a superposition" you mean what I just described, then the answer is yes. If you mean something else like, for instance "measuring several different states at the same time" then the answer is no because it is obviously self-contradictory nonsense.

But I gather from all of the responses, that the answer is no, we cannot observe a particle in a superposed state. This implies that a quantum particle in a superposed state is NOT an observable, doesn't it?

No. It is an observable that might be but has not yet been observed.

It seems to me you are getting tied up in the limitations of English. Study the math and be content with that.

 

[mike1000]

Then the answer depends on what you mean by the question. If by "observing a superposition" you mean what I just described, then the answer is yes. If you mean something else like, for instance "measuring several different states at the same time" then the answer is no because it is obviously self-contradictory nonsense.No. It is an observable that might be but has not yet been observed.

It seems to me you are getting tied up in the limitations of English. Study the math and be content with that.

No, I do not think I am getting tied up in the limitations of English. I find your response convoluted. It is a very simple question, can we observe a particle in a state which is a superposition of eigenstates?

The reason I said before that the answer should be "Yes" is because of the definition of observable and the foundation of Matrix Mechanics ( as I understand it as derived for observable quantities only) All pure states should be observable. That is why I said before the answer should be "yes".

 

[mikeyork]

No, I do not think I am getting tied up in the limitations of English. I find your response convoluted. It is a very simple question, can we observe a particle in a state which is a superposition of eigenstates?

And I gave a simple answer. If you mean can you measure the relative frequencies of different eigenstates over multiple repetitions of the same experiment, or if you mean can a superposition be an eigenstate in another observable, then the answer is yes. If you mean something else then you had better be more specific about what you mean.

If you repeat the same question again, I shan't bother to repeat my answer.

 

[mike1000]

And I gave a simple answer. If you mean can you measure the relative frequencies of different eigenstates over multiple repetitions of the same experiment, or if you mean can a superposition be an eigenstate in another observable, then the answer is yes. If you mean something else then you had better be more specific about what you mean.

If you repeat the same question again, I shan't bother to repeat my answer.

It appears to me that you are changing the definition. I am not talking about the outcomes of many trials. I am talking about the outcome of a single trial. I think that is what is implied by the definition of eigenstates and superposition for an operator. Please look at the Dirac quote above, I will repeat it here

"...We are observing whether it is polarized parallel or perpendicular to the optic axis. The effect of making this observation is to force the photon entirely into the state of parallel or entirely into the state of perpendicular polarization. It has to make a sudden jump from being partly in each of those two states to being entirely in one or the other of them. Which of the two states it will jump into cannot be predicted, but is governed only by probability laws."

 

[mikeyork]

It appears to me that you are changing the definition. I am not talking about the outcomes of many trials. I am talking about the outcome of a single trial.

Then as I said 3 posts ago, and repeated in my last post, the answer is yes if you can find the right observable for which the state is an eigenstate*. Otherwise the answer is no.

*Or, if we understand that as far as the logic of QM is concerned, observation can mean either detection or preparation, then the appropriate observation of the superposition was already made when the state was prepared.

 

[mike1000]

[

Then as I said 3 posts ago, and repeated in my last post, the answer is yes if you can find the right observable for which the state is an eigenstate. Otherwise the answer is no.

I am sorry but I do not understand that answer. It should be either yes or no, there should be no "if" in it.

I think what you are saying is that for every superposition in one basis there must be a different basis, in which, the superposition in the first basis becomes an eigenvector in the new basis.

 

[mikeyork]

I think what you are saying is that for every superposition in one basis there must be a different basis, in which, the superposition in the first basis becomes an eigenvector in the new basis.

That is correct.

 

[mike1000]

That is correct.

Well, it is not quite correct, because I forgot to specify that there must be an observable that has the new eigenvector as one of its states.

Lets take the spin direction for an electron. It can be up or down. What you are saying implies that there must be an observable for which the electron can be measured both up and down simultaneously.

 

[mikeyork]

Well, it is not quite correct, because I forgot to specify that there must be an observable that has the new eigenvector as one of its states.

Lets take the spin direction for an electron. It can be up or down. What you are saying implies that there must be an observable for which the electron can be measured both up and down simultaneously.

No. The -z direction is not the same as the +z direction. But detecting one is equivalent to detecting the other.

 

[mike1000]

No. The -z direction is not the same as the +z direction. But detecting one is equivalent to detecting the other.

That is not the same thing. You can try to use it in this case because its a binary system, ie there are only two , mutually exclusive, states. For any observable where there are more than two eigenstates your answer would not be correct.

 

[mikeyork]

That is not the same thing. You can try to use it in this case because its a binary system, ie there are only two , mutually exclusive, states. For any observable where there are more than two eigenstates your answer would not be correct.

Ok. If you meant two different eigenstates, then each has a different superposition in any other direction (obtained by a rotation) so a unique superposition in the original basis uniquely selects an orientation in which s = +1/2 and not -1/2. I do not understand why you think the number of eigenstates makes any difference. Even a position eigenstate generates a unique superposition in a momentum basis.

 

[PeterDonis]

It is a very simple question, can we observe a particle in a state which is a superposition of eigenstates?

No, it isn't a simple question, because the question is ambiguous. See below.

I forgot to specify that there must be an observable that has the new eigenvector as one of its states.

You don't have to specify it. It's already a theorem in linear algebra.

What you are saying implies that there must be an observable for which the electron can be measured both up and down simultaneously.

No, it doesn't. It just implies that, if "up" and "down" are orthogonal eigenstates of some observable, then there will be some other observable that has the linear combination "up plus down" (with appropriate normalization) as an eigenstate. In the case of a spin-1/2 particle, that observable is spin-x (assuming that "up" and "down" refer to eigenstates of the spin-z observable). But spin-x "+" (or "left", or whatever you want to call it) is not the same as "spin-z up and spin-z down simultaneously".

So the question I quoted from you at the top of this post, as I said there, is ambiguous; it can be interpreted two ways, one of which leads to the answer "yes" and the other of which leads to the answer "no":

(1) Given an observable O, can we make a measurement of that observable which gives a result that is a superposition of eigenstates of that observable? The answer to that is "no".

(2) Given an observable O and a state which is a superposition of eigenstates of that observable, can we make a measurement of some other observable which will give that state as a possible result (i.e., the state is an eigenstate of the other observable)? The answer to that is "yes".

 

[mike1000]

The answer is "yes", but that "yes" doesn't mean what you think it means. See below.

I am not talking about changing the basis for a given observable and I do not think that is what Dirac is talking about in this quote.

"...We are observing whether it is polarized parallel or perpendicular to the optic axis. The effect of making this observation is to force the photon entirely into the state of parallel or entirely into the state of perpendicular polarization. It has to make a sudden jump from being partly in each of those two states to being entirely in one or the other of them. Which of the two states it will jump into cannot be predicted, but is governed only by probability laws."

I think the answer is "no". States which are superpositions of eigenstates are not observable, meaning that the observable in question cannot be observed in that superpositioned state.

The best analogy I have is, again, flipping a coin. When the coin is flipping you can say that it is in the superposition of the head and tail state, because that is the basis, or the only two states the coin can actually be in. But when it is flipping it is in an underdermined state which is represented as 50% heads and 50% tails. When we measure it, by stopping the flipping, we find one of the eigenstates. The superposition allows us to calculate the probability but it is not an observable state that the coin can be found in.

 

[PeterDonis]

I think the answer is "no".

See the edit I just made to post #19. I had hit "post" too early and had to revise it.

I am not talking about changing the basis for a give observable

In that case, yes, the answer is "no" since you are adopting the first of the two possible interpretations I gave in post #19. But in post #17, you said:

I am really asking a very simple, and I think, very direct question, that to me, should only have one answer and the answer should be "Yes".

The answer is "yes" if you use the second interpretation that I gave in post #19. But that requires you to change the basis, which you have now said you didn't intend to do.

 

[mike1000]

See the edit I just made to post #19. I had hit "post" too early and had to revise it.
In that case, yes, the answer is "no" since you are adopting the first of the two possible interpretations I gave in post #19. But in post #17, you said:
The answer is "yes" if you use the second interpretation that I gave in post #19. But that requires you to change the basis, which you have now said you didn't intend to do.

The best analogy I have is, again, flipping a coin. When the coin is flipping you can say that it is in the superposition of the head and tail state, because that is the basis, or the only two states the coin can actually be in. But when it is flipping it is in an underdermined state which is represented as 50% heads and 50% tails. When we measure it, by stopping the flipping, we find one of the eigenstates. The superposition allows us to calculate the probability but it is not an observable state that the coin can be found in.

 

[PeterDonis]

When the coin is flipping you can say that it is in the superposition of the head and tail state, because that is the basis, or the only two states the coin can actually be in.

Not for a real coin. You can say that "heads" and "tails" are the only states of the coin you are interested in, but that doesn't make those two states a basis of the coin's complete state space. There are many coin states other than "heads" and "tails" and which are not expressible as linear combinations of "heads" and "tails"; that contradicts the hypothesis that "heads" and "tails" are a basis of the coin's state space.

If you want to consider a hypothetical "quantum coin" that acts like an actual spin-1/2 particle, you can as a heuristic analogy, but you should be aware of the limitations of such an analogy.

when it is flipping it is in an underdermined state which is represented as 50% heads and 50% tails.

Even in the limited "quantum coin" analogy, this statement is not quite correct. Assuming that the coin starts out in a superposition, if the probabilities of observing heads and tails are equal, the superposition we have been discussing up to now (but not the only possible one--see below) is $1 / \sqrt{2}$ times heads plus $1 / \sqrt{2}$ times tails, because the squares of the coefficients in the superposition are what give the probabilities and have to add up to 1 (or 100%). But the "representation" of the state is the superposition itself, with the $1 / \sqrt{2}$ coefficients.

This matters because there is not just one state that gives 50% probabilities for heads or tails. The most general state that does so would be expressed as

$$
\vert \Psi \rangle = \frac{1}{\sqrt{2}} e^{i \theta} \vert \text{heads} \rangle + \frac{1}{\sqrt{2}} e^{i \phi} \vert \text{tails} \rangle
$$

where $\theta$ and $\phi$ are arbitrary phase angles. So there are an infinite number of possible states that can all be described as "50% heads and 50% tails". But all of those states behave differently when we perform other operations besides the heads/tails measurement. (A good recent text on quantum computing will go into this in great detail, since I am just describing the possible qubit states and operations.)

(Also, the two basis states for a spin-1/2 particle are not "the only two states the coin can actually be in". They are just "the only two states that can result from a measurement in that basis", i.e., a measurement of the observable that has those two states as eigenstates. But there are an infinite number of possible observables and correspondingly an infinite number of possible pairs of basis states you could choose, and each choice corresponds to a distinct observable.)

 

[bhobba]

I am not talking about changing the basis for a given observable and I do not think that is what Dirac is talking about in this quote.

That looks like a quote from Dirac - Principles of QM.

A classic it is, and the best presentation of the early pioneers (with the possible exception of Von-Neumann if you want mathematical rigor), but it has errors that have been discussed here in the past. Care is required since things have moved on a lot since then. At that level Ballentine is a far better choice - then read Dirac. I know from personal experience - I did the reverse and got into all sorts of trouble. It sent me on a long sojourn into exotica like Rigged Hilbert Spaces etc, and I came out the other end with a very good grasp of certain mathematical technicalities, but it was NOT the best approach.

Dirac states the principle of superposition very clearly - there is no ambiguity - and if you know linear algebra it simply says the states, in principle, form a vector space. He also states the space he uses is more general than a Hilbert space, but doesn't say more than that and that's exactly where certain technical mathematical issues arise that RHS's are required to resolve and what I spent a long time sorting out. Regardless they are both vector spaces which is all you need to know for the principle of superposition.

Ballentine states the 2 axioms of QM. Get a copy and read the first 2 chapters.

There is no collapse. The second axiom is partly dependent on the first as I explain here:
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

See post 137.

The key single axiom that implies the two of Ballentine is:
An observation/measurement with possible outcomes i = 1, 2, 3 ... is described by a POVM Ei such that the probability of outcome i is determined by Ei, and only by Ei, in particular it does not depend on what POVM it is part of.

Note nothing is mentioned about collapse. All that is mentioned is you can find this mathematical thing called a POVM and you can associate the outcomes of an observation with it. That is you take whatever the outcomes are and you can find a POVM such that each outcome can be associated with an element of the POVM. To be 100% clear I will restate it another way. Suppose you have 1, 2, 3 ...n outcomes. Then you can find a POVM with n elements E1, E2, E3, ... En that you associate with the outcomes - E1 is associated with outcome 1, E2 outcome 2 etc. That's all there is to it formalism wise. The other implication is all you can do is predict probabilities of outcomes - but that is just a general case of determinism which is simply probabilities of 0 or 1 so whether its an actual assumption is debatable.

Now what it means is another matter - that's where interpretations come in. In some of them you have this thing called collapse - but its not part of the formalism.

Thanks
Bill

 

[Nugatory]

The best analogy I have is, again, flipping a coin. When the coin is flipping you can say that it is in the superposition of the head and tail state, because that is the basis, or the only two states the coin can actually be in. But when it is flipping it is in an underdermined state which is represented as 50% heads and 50% tails. When we measure it, by stopping the flipping, we find one of the eigenstates. The superposition allows us to calculate the probability but it is not an observable state that the coin can be found in.

There are certainly worse analogies, but like all analogies this one is still quite misleading. Consider that we stop the coin from spinning by clapping it between our hands. The coin will end up with one side against our left palm and the other side against our right palm; we'll call the measurement result heads or tails according to which side is against our left palm. But we can hold our hands either vertically or horizontally when we clap them together, so we actually have two observables: heads/tails horizontally and heads/tails vertically. With the spinning coin both are random, so much so that you might wonder why I bother making the distinction between the two observables - when I'm tossing a coin to settle an argument no one cares how I hold my hands when I grab it. But here the analogy has failed to capture an essential characteristic of superposition: if the coin behaved like a superposition, I could start it spinning in such a way that it comes up vertical-heads every single time even though the horizontal measurement is 50/50 random between heads and tails.

 

[vanhees71]

I've not read the entire thread, but I think, there's a lot of confusion already in the OP. So here's what I think about that question.

One has to clearly distinguish between states and observables in quantum theory. I think, that's the most complicated issue to get used to when starting to learn QT.

In classical mechanics the state of a system is simply a point in phase space. For a set of point particles it's represented, e.g., by giving all positions and momenta of the particles, and knowing these values you know the values of all observables, because observables are functions of the positions and momenta. All observables thus have determined values at any time.

The most radical difference between QT and classical physics is that this is found not to be true! It took (amazingly only!) around 25 years from the first discovery of quantum effects by Planck till Dirac in 1925 (and John von Neumann to make it mathematically rigorous) to clearly formulate modern QT. The reason, why I refer to Dirac is that he was the one who formulated it in the most clear way without any a priori preferred representation (wave mechanics="position representation", matrix mechanics="harmonic-oscillator representation" in a somewhat superficial sense). So we are in the fortunate position that we can clearly formulate the distinction between the description of states and observables, which encodes all the mental troubles we have when learning quantum theory. It's also important to know that there are no physics problems with this notion, although some people like to bring up philosophical issues, which are nice entertainment but irrelevant to physics!

So here's the physics in a nutshell:

(1) A quantum system is described by a (rigged) Hilbert space $\mathcal{H}$

(2) States are described by a self-adjoint positive semidefinite operator $\hat{\rho}$ (the statistical operator) with $\mathrm{Tr} \hat{\rho}=1$.

(3) Observables are described by self-adjoint operators. When measuring an observable $A$, represented by the self-adjoint operator $\hat{A}$, the outcome is an eigenvalue of $\hat{A}$.

(4) Born's Rule: Let $|a,\beta \rangle$ denote all (generalized) eigenvectors of $\hat{A}$. The values of $a$ can be a continuous and/or discrete subset of real numbers, and $\beta$ is a label consisting of one or more also continuous and/or discrete subset of real numbers. In the following I assume that all these are discrete, and I write sums. The math tells us that these eigenvectors can be choosen as a complete orthonormalized set,
$$\langle a,\beta|a',\beta' \rangle=\delta_{a,a'} \delta_{\beta,\beta'}, \quad \sum_{a,\beta} |a,\beta \rangle \langle a,\beta|=\hat{1}.$$
Then, if the system is prepared such that its state is described by the statistical operator $\hat{\rho}$, then the probability to find the eigenvalue $a$ of $\hat{A}$, when measuring the observable $A$ is given by
$$P(a)=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle.$$
That's it. There's no more meaning to states and observables of relevance for physics in the formalism of QT than this probabilistic/statistical meaning (minimal statistical interpretation). There's much ado about Born's Rule, and a lot is written about the meaning of probabilities and a lot of more or less clever ideas are developed around it. The good thing of this is that one can fairly say that QT is the best tested and most thought-about theory of physics ever. At the same time it's also the most successful. There's no single observation contradicting it, and that's why from a point of view of physics it's considered to be the valid description of (almost) all of physics known today. The only real physics problem with QT is that there is no satisfactory description of gravitation within its realm (and only very recently one has hope to find some observable consequences of possible quantum effects regarding gravity).

The mind-boggling consequence of the above (quite complicated sounding but in fact not so difficult) mathematical foundation is that even when one has complete knowledge about a system, which is the case, if $\hat{\rho} =|\psi \rangle \langle \psi|$ with some normalized Hilbert-space vector $|\psi \rangle$. That's called a pure state, and it means that the system is in an eigenstate of an observable described by $\hat{A}=|\psi \rangle \langle\psi |$.

The Hilbert-space structure of the entire construct also implies that in principle any (generalized) superposition of vectors leading to such a normalized ket can also define such a pure state. If you have a superposition of, e.g., the form $|psi \rangle = \lambda_1 |a_1,\beta_1 \rangle + \lambda_2 |a_2,\beta_2 \rangle$ (and this is a normalizable true Hilbert-space vector), where $a_1$ and $a_2$ are two unequal eigenvectors of $\hat{A}$, then the observable has no determined value, and the probabilities to measure $a_1$ or $a_2$ are $P(a_1)=|\lambda_1|^2$ and $P(a_2)=|\lambda_2|^2$. That's all there is to know about superpositions of eigenvectors. Often one reads: "The system is in a state such that the observable $A$ takes the values $a_1$ and $a_2$ at the same time." From the formalism it's clear that this is simply utterly wrong! Rather, the value of $A$ is not determined due to the preparation of the system in this superposition. All there is are probabilities to find the one or the other value.

As I said, that's difficult to accept for a physicists trained in classical mechanics, and we all are as kids in a sense trained as "classical physicists" due to our daily experience with the world, which is macroscopic, and macrocopic systems are almost always completely well described by classical physics. To understand, why this is so, is an interesting problem of quantum many-body theory, and the answer is "coarse graining", but that's another story.

 

[zonde]

@vanhees71, I do not quite understand how looking at statistical operator makes it easier to understand superpositions. In superpositions phase relationship between different components of complete state are important. How would you track them across experiment using statistical operator?
For example look at this experiment: Can Two-Photon Interference be Considered the Interference of Two Photons?
There are H and V photons before beamsplitter and after beamsplitter you have two components in each output. And analyzers before detectors are mixing these components.
I don't understand how using statistical operator as description of state makes it possible to come up with correct prediction for this experiment. On the other hand it seems quite clear how to get prediction using state vectors for different components.

 

[vanhees71]

The statistical operator is simply the most general way to describe states in a unique way. For pure states they take the form $|\psi \rangle \langle \psi |$ for any normalized $|\psi \rangle$ in the ray, representing pure states. I just gave the most general description of the structure of QT to avoid discussing rays and all these complications. If you deal with pure states, you can always write out the statistical operator as a projection operator to get the probabilities in terms of "wave functions squared".

 

[zonde]

The statistical operator is simply the most general way to describe states in a unique way.

That I understand.
My question is more about how one can use this description to gain insight about things happening in experiments.

 

[vanhees71]

I don't understand this question. We always calculate probabilities, expectation values of operators (in quantum optics everything is in correlation functions of the photon-field operator etc. etc.) and compare to experiment. What has this to do whether I use the general description with the statistical operator rather than the special case of pure states, where you can work with the state vectors? All results using the formalism with the statistical operator and using state kets in the case of pure states are identical, of course!