What does a superposition of states mean in quantum mechanics?

[vanhees71]

It's not simply oversimplified, it's plain wrong!

 

[Karolus]

No, it's in the state $\frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ (for example).

Sometimes people describe the state $\frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ as "simultaneously in both 0 and 1", but this is an just an oversimplified and misleading translation from the math into English. It's like saying that a diagonal line is "simultaneously both a horizontal line and a vertical line". That's a terrible description of what it means to be diagonal.

compare the superposition principle (which is not a mistake, but one of the principles of quantum mechanics) to a diagonal that is simultaneously horizontal and vertical, it is the most horrible thing I've ever heard in science. It 'so bad, that I had never come into my mind!

 

[vanhees71]

Indeed, it's as horrible as to claim the system was in the states $|0 \rangle$ and $|1 \rangle$ "at the same time", while in fact it is of course in a state given by the specific superposition of these two states.

 

[Ostrados]

Again and again and again, please don't use this nonsense talk about a system being in more than one state or an observable taking more than one value at the same time. That's NOT implied by anything in quantum theory, but it's claimed by bad popular-science texts, and there are more bad than good popular-science writings around. The problem is that to write a good popular-science article is among the most difficult tasks for a scientist; in some sense it's more difficult than to write a scientific article, because popular-science writing forbids one to use the only language that is really adequate to talk about quantum theory, i.e., math!

Quantum theory tells you that a system is always in one state, which can be proper or mixed. Even in a completely determined state, i.e., if the system is prepared such that a complete set of compatible observables takes determined values, other observables (that are not compatible with all observables of the complete set) do not necessarily take determined values, but than it's wrong to say that they take more than one value at once. In this case they simply don't take any determined value. The state, implies the probabilities to measure any possible value of that observable and nothing else. To explain this in a really understandable way, I'd have to use the mathematical formalism of quantum theory.

You are right about the how poorly QM is explained in pop science, but the problem is that this style is widespread and is even used by many "real" scientist, which not only resulted in spread of misconceptions but also led to confusion in the physics community itself. How many times did you read an article that says "but because the qubit is both 0 and 1" or "The qubit does not take 0 or 1 but it takes 0&1 simultaneously"?

 

[Strilanc]

compare the superposition principle (which is not a mistake, but one of the principles of quantum mechanics) to a diagonal that is simultaneously horizontal and vertical, it is the most horrible thing I've ever heard in science. It 'so bad, that I had never come into my mind!

The analogy is actually pretty direct. If the unit direction vector along a horizontal line is $|H\rangle$, and the unit direction vector along a vertical line is $|V\rangle$, then the unit direction vector for the diagonal line X=Y is $\frac{1}{\sqrt 2} |H\rangle + \frac{1}{\sqrt 2} |V\rangle$. Look familiar?

When you say that $\frac{1}{\sqrt 2} |0\rangle + \frac{1}{\sqrt 2} |1\rangle$ is simultaneously both 0 and 1, it's exactly like saying that the line along $\frac{1}{\sqrt 2} |H\rangle + \frac{1}{\sqrt 2} |V\rangle$ is simultaneously along H and V. It's a type error. You can decompose a diagonal line's direction into the HV basis, but that doesn't mean the diagonal line is secretly made up of horizontal and vertical line segments.

 

[mike1000]

Again and again and again, please don't use this nonsense talk about a system being in more than one state or an observable taking more than one value at the same time. That's NOT implied by anything in quantum theory, but it's claimed by bad popular-science texts, and there are more bad than good popular-science writings around. The problem is that to write a good popular-science article is among the most difficult tasks for a scientist; in some sense it's more difficult than to write a scientific article, because popular-science writing forbids one to use the only language that is really adequate to talk about quantum theory, i.e., math!

Quantum theory tells you that a system is always in one state, which can be proper or mixed. Even in a completely determined state, i.e., if the system is prepared such that a complete set of compatible observables takes determined values, other observables (that are not compatible with all observables of the complete set) do not necessarily take determined values, but than it's wrong to say that they take more than one value at once. In this case they simply don't take any determined value. The state, implies the probabilities to measure any possible value of that observable and nothing else. To explain this in a really understandable way, I'd have to use the mathematical formalism of quantum theory.

First to explain my comment "Does this mean the electron spin can be both in the up and down state"? That was a rhetorical comment I made to the other poster in response to something he said. Even I know now that the electron spin can only be either up or down, not both at the same time. I had made a previous comment that the electron spin could only be measured in one of its eigenstates. He responded that, no, the allowed states were the states of the system which could be any linear combination of its eigenstates. That answer prompted me to respond with the question I stated above.

If you go back to my original post in this thread, I think you will find that I was trying to describe the situation that you describe in your second paragraph. Here is the way I am beginning to see the situation...The eigenstates of the operator form a basis for the Hilbert Space. In the Hilbert Space any linear combination of the eigenstates is also a member of the Hilbert Space. The wave function is given by the following equation, $$\psi(x) =\langle \hat{\vec{x}}|\psi\rangle,$$ where $\hat{\vec{x}}$ are the position eigenstates of the position operator and $\psi$ is probability amplitudes associated with a particular eigenstate. The function $\psi(x)$ is the corresponding apriori wave function and may be a number which respresents a linear combination of many different eigenstates. It is used to calculate the probability only. It says nothing about the system being in two or more states simultaneously. A linear combination of eigenstates is the superposition. However, we know that when we make a measurement the measured property is only going to be found in one and only one state. I do not think there is any decoherence going on. The wave function is not collapsing. The apriori wave function is used to calculate the probabilities only and is a statement of the apriori state of the system. Kind of like shuffling a deck of cards if you want to place the system into a mixed state prior to experiment.

Regarding mixed states. It seems to me, as I think about it, that mixed states which involve linear combinations of eigenstates(superposition) should refer to populations of particles and not a single particle. In a population of particles, the mixed state wave function would indicate how the population is partitioned among eigenstates. Each individual particle is only in one eigenstate but not all particles are in the same eigenstate. The mixed wave function tells us how the particles are partitioned among the eigenstates, before any measurement. When you measure a property of one of the electrons, you find it is in a particular eigenstate, not in a mixed eigenstate.

However, I realize that I may not understand the role of mixed states so all of the above can change for me as I am still learning!

 

[Karolus]

Again and again and again, please don't use this nonsense talk about a system being in more than one state or an observable taking more than one value at the same time. That's NOT implied by anything in quantum theory, but it's claimed by bad popular-science texts, and there are more bad than good popular-science writings around. The problem is that to write a good popular-science article is among the most difficult tasks for a scientist; in some sense it's more difficult than to write a scientific article, because popular-science writing forbids one to use the only language that is really adequate to talk about quantum theory, i.e., math!
Quantum theory tells you that a system is always in one state, which can be proper or mixed. Even in a completely determined state, i.e., if the system is prepared such that a complete set of compatible observables takes determined values, other observables (that are not compatible with all observables of the complete set) do not necessarily take determined values, but than it's wrong to say that they take more than one value at once. In this case they simply don't take any determined value. The state, implies the probabilities to measure any possible value of that observable and nothing else. To explain this in a really understandable way, I'd have to use the mathematical formalism of quantum theory.

Your answer reveals itself in the confusion you have about quantum mechanics.
I shall focus on two points.
1.

Again and again and again, please don't use this nonsense talk about a system being in more than one state or an observable taking more than one value at the same time

I never claimed that an observable takes different values or assumes different values simultaneously! Quite the opposite, when I make an observation I can only measure a specific value at a time, or rather on the same observable sometimes I will get | 0> and sometimes | 1>, but never | 0> and | 1> at the same time!

2.

Quantum theory tells you that a system is always in one state, which can be proper or mixed.

You use the word "mixed" and then you say that you only need to use the language of mathematics! That "mixed" mean? Which is a bit | 0> and a little | 1>? As in a recipe where we do a mixed salad ? Your language, rather than a scientist, it seems that of a cook ...

 

[Strilanc]

You use the word "mixed" and then you say that you only need to use the language of mathematics! That "mixed" mean? Which is a bit | 0> and a little | 1>? As in a recipe where we do a mixed salad ? Your language, rather than a scientist, it seems that of a cook ...

Are you not familiar with mixed states? It's standard nomenclature.

 

[PeterDonis]

Even I know now that the electron spin can only be either up or down, not both at the same time.

I'm sorry, but I'm going to comment on this statement once again. What you should be saying is: if you measure the spin of the electron about a given axis, you will always get one of two results: up or down. But that does not mean that the electron can only be in one of two states: up or down about that particular axis. There are an infinite number of possible states the electron can be in, but if it is in any of those states, and you measure its spin about a given axis, you will always get one of two results: up or down. Or, more briefly: the electron spin can only be measured to be either up or down, not both at the same time.

The eigenstates of the operator form a basis for the Hilbert Space. In the Hilbert Space any linear combination of the eigenstates is also a member of the Hilbert Space.

This is correct.

The wave function is given by the following equation

This is only correct if we are talking about the position degree of freedom for a single particle. But we have been talking about spin. Spin does not have a position representation, and the equation you give does not apply to spin.

I do not think there is any is decoherence going on.

This is a serious error. Decoherence is a well-established phenomenon.

The wave function is not collapsing.

This depends on which interpretation of QM you adopt. There are interpretations that have collapse and interpretations that don't. Both kinds of interpretation make all of the same predictions, so we have no way at present of testing which one is right.

mixed states which involve linear combinations of eigenstates(superposition)

Mixed states are not the same as superpositions. I strongly suggest taking some time to learn the difference: the key term to understand is "density matrix" vs. "wave function". A superposition is a pure state, which is a state that can be represented by a wave function. A mixed state cannot be represented by a wave function: it can only be represented by a density matrix. Most modern QM textbooks discuss this.

 

[mike1000]

I'm sorry, but I'm going to comment on this statement once again. What you should be saying is: if you measure the spin of the electron about a given axis, you will always get one of two results: up or down. But that does not mean that the electron can only be in one of two states: up or down about that particular axis. There are an infinite number of possible states the electron can be in, but if it is in any of those states, and you measure its spin about a given axis, you will always get one of two results: up or down. Or, more briefly: the electron spin can only be measured to be either up or down, not both at the same time.
This is correct.
This is only correct if we are talking about the position degree of freedom for a single particle. But we have been talking about spin. Spin does not have a position representation, and the equation you give does not apply to spin.
This is a serious error. Decoherence is a well-established phenomenon.
This depends on which interpretation of QM you adopt. There are interpretations that have collapse and interpretations that don't. Both kinds of interpretation make all of the same predictions, so we have no way at present of testing which one is right.
Mixed states are not the same as superpositions. I strongly suggest taking some time to learn the difference: the key term to understand is "density matrix" vs. "wave function". A superposition is a pure state, which is a state that can be represented by a wave function. A mixed state cannot be represented by a wave function: it can only be represented by a density matrix. Most modern QM textbooks discuss this.

Your method of piecemealing my statements is not helping me. I am quite sure that you can piecemeal even the statements of Dirac if you chose to do so. I am not saying I am Dirac. I am saying that your method really does not help me very much. I think you have an academic understanding but not a practical understanding. This is not a debate. I am trying to learn and you are not helping much. You always say what I get wrong and not what I get right.

Ok, so mixed states are not the same as superpositions. Why did you stop there? If you are going to be critical then explain what you mean.

Are mixed states what we have when we have a population of particles?

 

[PeterDonis]

You always say what I get wrong and not what I get right.

I have pointed out things you got right as well.

so mixed states are not the same as superpositions. Why did you stop there?

I didn't. Did you read the rest of my post? Have you tried Googling "density matrix" to see what you find? Or looking at a textbook?

I have already suggested, several times now IIRC, that you take some time to work through a good QM textbook (I think I suggested Ballentine). What you are basically asking for at this point amounts to a course in QM 101. That is way beyond the scope of a PF thread. If you are really trying to learn, you need to put in the time on your own; it is simply not possible to teach you everything you need to know here at PF.

 

[Nugatory]

Ok, so mixed states are not the same as superpositions. Why did you stop there? If you are going to be critical then explain what you mean.

There is an explanation of the difference, with examples, in post #30 of this thread.

 

[mike1000]

I have pointed out things you got right as well.
I didn't. Did you read the rest of my post? Have you tried Googling "density matrix" to see what you find? Or looking at a textbook?

I have already suggested, several times now IIRC, that you take some time to work through a good QM textbook (I think I suggested Ballentine). What you are basically asking for at this point amounts to a course in QM 101. That is way beyond the scope of a PF thread. If you are really trying to learn, you need to put in the time on your own; it is simply not possible to teach you everything you need to know here at PF.

I am not asking for a course in QM 101.

BTW, I was wrong about quantum decoherence. Somehow I thought that wave collapse and decoherence were associated. I realize now they are two different concepts. It is wave collapse that I do not think happens. If the probabilities are apriori probabilities there is no need for a wave function collapse.

The main thrust of all my posts has been the idea that the wave function is really giving us the apriori probabilities and says nothing about particles being in two or more states at the same time prior to the measurement or observation.

 

[PeterDonis]

It is wave collapse that I do not think happens.

This is a permissible interpretation, yes. But there are also permissible interpretations in which collapse does happen. As I said before, we don't currently have any way of testing this experimentally, since both kinds of interpretations make the same predictions. Even if you favor a no collapse interpretation, it's still good to be aware that collapse interpretations exist and can't currently be ruled out by experiment.

the wave function is really giving us the apriori probabilities

If you favor a no collapse interpretation, you might run into problems trying to interpret the wave function this way. The most common no collapse interpretation, the many worlds interpretation, views the wave function as the actual physical state of the system. When a measurement happens, the state of the measuring device becomes entangled with the state of the measured system, so that there are multiple branches of the overall wave function, one branch for each possible result. For example, in the case of a spin-z measurement on an electron that is in a superposition of spin-z eigenstates, the evolution of the state through the measurement would look like this:

$$
\frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle + \vert \downarrow \rangle \right) \vert R \rangle \rightarrow \frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle \vert U \rangle + \vert \downarrow \rangle \vert D \rangle \right)
$$

where $\uparrow$ is the spin-z up eigenstate of the electron, $\downarrow$ is the spin-z down eigenstate of the electron, $R$ is the "ready to measure" state of the measuring device, $U$ is the "measured spin-z up" state of the measuring device, and $D$ is the "measured spin-z down" state of the measuring device. So the state after the measurement has two branches, one in which the electron is spin-z up and the measuring device has measured spin-z up, and the other in which the electron is spin-z down and the measuring device has measured spin-z down.

It is important to understand that the above expression, taken by itself, is not actually interpretation dependent; it is just the straight math of QM, unitary evolution, applied to the initial state, under the assumption that the measuring device is a valid measuring device for spin-z. (That assumption is what ensures that the $U$ state gets entangled with the $\uparrow$ state, and the $D$ state with the $\downarrow$ state.) The interpretation comes in when you ask, what happens next? The two possibilities, broadly speaking, are:

(1) Collapse: The state of the entire system collapses to one or the other of the above branches. This is probably most lay people's intuitive interpretation of what happens, since intuitively we observe a single measurement result, not a superposition. (For how the MWI explains that, see below.) The problem with it is the collapse: how does it happen? It would have to be a non-unitary process, since unitary evolution can't make either branch just disappear. It would also be very difficult to see how it could be relativistically covariant, since the two branches could be spatially separated. There is a huge literature about this, but no resolution, it is an open question.

(2) No Collapse: Both branches continue to exist. This is the straightforward interpretation of the actual math, based on unitary evolution always being correct. At first it was thought that there was an obvious problem of us not observing a superposition but a single measurement result; but Hugh Everett resolved that when he published his Ph.D. thesis in 1957. He pointed out that, to be completely consistent, we would have to treat ourselves as quantum systems, so that our own quantum states, when we observed the results of a measuring device, would become entangled with the state of the device. So really, in the above evolution, the $U$ and $D$ states should mean, not just "measuring device measured spin-z up (or down)", but "all observers observe and agree that the measuring device measured spin-z up (or down)". In other words, you, I, and all other observers branch just like the measuring device branches: there are multiple copies of all of us, and each copy observes a single measurement result, and our own conscious experience is the conscious experience of a single copy.

The open problem with the MWI is actually how it can explain probabilities: how do you get the Born rule (that the probabilities of the possible results are proportional to the squares of their amplitudes in the wave function) out of the straight math of QM unitary evolution? Nobody has come up with a resolution to this, although there is a lot of literature on it. (Note that in collapse interpretations, the Born rule is not derived: it's just added as an extra postulate.) This is why I said you need to be careful about interpreting the wave function as giving probabilities, if you prefer a no collapse interpretation.

 

[weirdoguy]

That "mixed" mean? Which is a bit | 0> and a little | 1>? As in a recipe where we do a mixed salad ? Your language, rather than a scientist, it seems that of a cook ...

Are you even serious? You still have that "I know better than physicists" attitude and you still don't have enough konwledge to have it and you won't have it (the knowledge) unless you really read a proper textbook on the subject. Read this: https://en.wikipedia.org/wiki/Density_matrix and you'll know what "mixed state" means. You can't avoid the math, whether you like it or not.

 

[Karolus]

ok , "mixed" states ( density matrix etc) are different from superposition state, not overcomplicate with english..

 

[Karolus]

The analogy is actually pretty direct. If the unit direction vector along a horizontal line is $|H\rangle$, and the unit direction vector along a vertical line is $|V\rangle$, then the unit direction vector for the diagonal line X=Y is $\frac{1}{\sqrt 2} |H\rangle + \frac{1}{\sqrt 2} |V\rangle$. Look familiar?

When you say that $\frac{1}{\sqrt 2} |0\rangle + \frac{1}{\sqrt 2} |1\rangle$ is simultaneously both 0 and 1, it's exactly like saying that the line along $\frac{1}{\sqrt 2} |H\rangle + \frac{1}{\sqrt 2} |V\rangle$ is simultaneously along H and V. It's a type error. You can decompose a diagonal line's direction into the HV basis, but that doesn't mean the diagonal line is secretly made up of horizontal and vertical line segments.

Here there is some error, some approximation and some terminological confusion ... "simultaneously"! To say that a diagonal is the superposition of horizontal and vertical, mmm ... seems Hegel ... But if we speak of a vector, and its decomposition into two "basic" vectors, horizontal and vertical, or if you prefer, i and j with factors appropriate, then we are talking about linear algebra, and we found the hot water ... so, the principle of superposition has a clear formulation as a linear combination of vectors: a vector (state) can be expressed as a superposition (linear combination) of other states (vectors). That's all? It seems to you, according to your conception of the "superposition principle" is nothing more than an elementary math simple rules ... so it seems ...
I think indeed that the general superposition principle of quantum mechanics applies to the states of any dynamic system. It consists in the hypothesis that relations between these states exist characteristics such that, whenever said system is in a defined state, it can be considered as belonging simultaneously to two or more others.

 

[mike1000]

I deleted this post and I used it to create a new thread. Here is a link to that thread.

https://www.physicsforums.com/threads/more-on-superposition.909502/