About nature of superposition of states

  • #106
HighPhy said:
Is this compatible with the fact that a particle (or the cat, in the paradox) can assume sort of two states at the same time?
Just about any random statement is "sort of" reasonable if you can get those words to do enough work - and you're asking "sort of" to do a lot of work here.
I know that a tossed coin will eventually land either heads or tails, so I can say that while it is spinning through the air it is sort of assuming two states at once - but it makes way more sense to say that while it is spinning through the air its state is neither "heads" nor "tails", but instead "spinning through the air".

Three pages into the thread and you are still treating natural-language statements couched in vague and imprecise language ("sort of" may be a new extreme of vagueness and imprecision) as if they are statements of fact. It won't work.
 
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  • #107
Nugatory said:
Just about any random statement is "sort of" reasonable if you can get those words to do enough work - and you're asking "sort of" to do a lot of work here.
I know that a tossed coin will eventually land either heads or tails, so I can say that while it is spinning through the air it is sort of assuming two states at once - but it makes way more sense to say that while it is spinning through the air its state is neither "heads" nor "tails", but instead "spinning through the air".

Three pages into the thread and you are still treating natural-language statements couched in vague and imprecise language ("sort of" may be a new extreme of vagueness and imprecision) as if they are statements of fact. It won't work.
Yes, you're right.

IMO, this happens because I still don't have clear in my mind the distinction between the states.

If a system has ##50\%## chance to be in state ##\left|\psi_1\right>## and ##50\%## to be in state ##\left|\psi_2\right>##, then this is a mixed state. Both states exist, and as we said, this is not the case with the cat in the box. Right?

Now, consider the state
$$\left|\Psi\right>=\frac{\left|\psi_1\right>+\left|\psi_2\right>}{\sqrt{2}},$$ which is a superposition of the states ##\left|\psi_1\right>## and ##\left|\psi_2\right>##.
It is a pure state. Meaning, there's not a 50% chance the system is in the state ##|\psi_1\rangle## and a 50% it is in the state ##|\psi_2\rangle##. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state ##|\Psi\rangle##. Right?

This is consistent with what PeterDonis says:
PeterDonis said:
(if you agree with Schrodinger) ... the entangled superposition wave function does not describe a cat that is dead or alive, we just don't know which. It describes a cat which is in some different state that, whatever it is, is not the state of a cat that is dead or a cat that is alive.
But in the course of the thread, it was mentioned that the term "superposition of states" is also incorrect. But is it incorrect only because the term "superposition" does not inherently provide for the notion of "entanglement"?
Or is there some more subtle reason?

Also, what I want to ask after pointing out this difference between pure state and mixed state is: is the problem in pop science (and also in my previous response) the confusion between mixed state and superposition of states?
 
  • #108
HighPhy said:
is the problem in pop science (and also in my previous response) the confusion between mixed state and superposition of states?
Not the problem but a problem, one of many. The solution to all of them is to stop wasting one’s time with pop-sci sources.
 
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  • #109
HighPhy said:
If a system has ##50\%## chance to be in state ##\left|\psi_1\right>## and ##50\%## to be in state ##\left|\psi_2\right>##, then this is a mixed state.
More precisely, this is one situation in which we model the system using a mixed state. (The other situation is modeling a system which is entangled with other systems and therefore does not have a well-defined state of its own; we can trace over the other systems and obtain a mixed state for the system we are interested in.)

HighPhy said:
Both states exist
No. A mixed state in the scenario you describe reflects our lack of knowledge of how the system was prepared. It does not reflect any quantum superposition with regard to the system itself.

HighPhy said:
this is not the case with the cat in the box. Right?
If you mean the cat in the box is not modeled using a mixed state, that is correct. We have full knowledge of how the system was prepared, and we are not interested in tracing out other subsystems besides the cat, so there is no need to use a mixed state; we can model everything using pure states.

HighPhy said:
Now, consider the state
$$\left|\Psi\right>=\frac{\left|\psi_1\right>+\left|\psi_2\right>}{\sqrt{2}},$$ which is a superposition of the states ##\left|\psi_1\right>## and ##\left|\psi_2\right>##.
It is a pure state. Meaning, there's not a 50% chance the system is in the state ##|\psi_1\rangle## and a 50% it is in the state ##|\psi_2\rangle##. There is a 0% chance that the system is in either of those states, and a 100% chance the system is in the state ##|\Psi\rangle##. Right?
No.

First, this state is not entangled, so it is not the same as the entangled superposition cat states we have been looking at.

Second, because the state is not entangled, calling it a "superposition" is basis dependent. If you are representing the state as a superposition of ##\psi_1## and ##\psi_2##, that implies that you intend to measure the system and that ##\psi_1## and ##\psi_2## are eigenstates representing possible measurement results. And if you are doing that, then the relative amplitudes for ##\psi_1## and ##\psi_2## give you the probability of getting the measurement results corresponding to ##\psi_1## and ##\psi_2##. In the state you wrote down, these probabilities are 50% each. But you could choose to make some different measurement which would lead you to write down the state in a different basis and give different probabilities. You could even choose a measurement for which ##\Psi## itself is an eigenstate, in which case the probability would be 100% that you would get the corresponding measurement result.

Third, what the state represents apart from probabilities for possible measurement results in whatever basis you have chosen is interpretation dependent. Some interpretations treat the state as representing the physically real state of an individual quantum system; in such an interpretation, the physically real state before any measurement is made would indeed be ##\Psi##, not ##\psi_1## or ##\psi_2##. Our basis for saying that would be that we prepared the system in such a way that we know the state ##\Psi## is what came out of our preparation process.

But other interpretations do not treat the state as representing the physically real state of an individual quantum system. They might treat it as representing an abstract ensemble of systems all prepared by the same process, or as representing the preparation process itself. In such an interpretation, you can't say anything about the state of an individual quantum system.

HighPhy said:
This is consistent with what PeterDonis says:
No, it has nothing to do with what I said in what you quote, since, as noted above, the state you wrote down is not an entangled superposition.

HighPhy said:
it was mentioned that the term "superposition of states" is also incorrect. But is it incorrect only because the term "superposition" does not inherently provide for the notion of "entanglement"?
Yes. See above.

HighPhy said:
Also, what I want to ask after pointing out this difference between pure state and mixed state is: is the problem in pop science (and also in my previous response) the confusion between mixed state and superposition of states?
The problem with pop science is that it is pop science and is not a reliable source if you want to learn actual science.

Apart from that, you appear to be placing way, way too much weight on ordinary language descriptions. Physics is not done in ordinary language. It is done in math. You keep trying to make hairsplitting distinctions about ordinary language that is already known to be inadequate to describe the physics anyway. You would be far better served by forgetting all about ordinary language and learning how to describe the physics with math.
 
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  • #110
PeterDonis said:
Second, because the state is not entangled, calling it a "superposition" is basis dependent. If you are representing the state as a superposition of ##\psi_1## and ##\psi_2##, that implies that you intend to measure the system and that ##\psi_1## and ##\psi_2## are eigenstates representing possible measurement results. And if you are doing that, then the relative amplitudes for ##\psi_1## and ##\psi_2## give you the probability of getting the measurement results corresponding to ##\psi_1## and ##\psi_2##. In the state you wrote down, these probabilities are 50% each. But you could choose to make some different measurement which would lead you to write down the state in a different basis and give different probabilities. You could even choose a measurement for which ##\Psi## itself is an eigenstate, in which case the probability would be 100% that you would get the corresponding measurement result.
Interpretations aside (which, by the way, are not to be neglected), I omitted much of the context in which I wanted to present my response. My fault. I'll try to rectify that.
I would like to know if my claims can be fixed by the following reasoning. Or alternatively, if there is still something wrong.

The point I omitted is that these statements are all made before I make any measurements. Yes?

It is true that if I measure the observable corresponding to ##\psi##, then there is a ##50 \%## chance after collapse the system will end up in the state ##|\psi_1\rangle##.

However, let's say I choose to measure a different observable. Let's say the observable is called ##\phi##, and let's say that ##\phi## and ##\psi## are incompatible observables in the sense that as operators ##[\hat{\psi},\hat{\phi}]\neq0##. (I realize I'm using ##\psi## in a sense you would never use it). The incompatibility means that ##|\psi_1 \rangle## is not just proportional to ##|\phi_1\rangle##, it is a superposition of ##|\phi_1\rangle## and ##|\phi_2\rangle## (the two operators are not simultaneously diagonalized).

Then we want to re-express ##|\Psi\rangle## in the ##\phi## basis. Let's say that we find
$$
|\Psi\rangle = |\phi_1\rangle
$$

For example, this would happen if
$$
|\psi_1\rangle = \frac{1}{\sqrt{2}}(|\phi_1\rangle+|\phi_2\rangle)
$$
$$
|\psi_2\rangle = \frac{1}{\sqrt{2}}(|\phi_1\rangle-|\phi_2\rangle)
$$
Then I can ask for the probability of measuring ##\phi## and having the system collapse to the state ##|\phi_1\rangle##, given that the state is ##|\Psi\rangle##, it's ##100\%##.
So I have predictions for the two experiments, one measuring ##\psi## and the other ##\phi##, given knowledge that the state is ##\Psi##.

But now let's say that there's a ##50\%## chance that the system is in the pure state ##|\psi_1\rangle##, and a ##50\%## chance the system is in the pure state ##|\psi_2\rangle##. Not a superposition, a genuine uncertainty as to what the state of the system is. If the state is ##|\psi_1 \rangle##, then there is a ##50\%## chance that measuring ##\phi## will collapse the system into the state ##|\phi_1\rangle##. Meanwhile, if the state is ##|\psi_2\rangle##, I get a ##50\%## chance of finding the system in ##|\phi_1\rangle## after measuring. So the probability of measuring the system in the state ##|\phi_1\rangle## after measuring ##\phi##, is
$$(50\% \ \mathrm{being \ in} \ \psi_1)(50\% \ \mathrm{measuring} \ \phi_1) + (50\% \ \mathrm{being \ in} \ \psi_2)(50\% \ \mathrm{measuring} \ \phi_1)=50\%$$
This is different than the pure state case.
 
  • #111
HighPhy said:
The point I omitted is that these statements are all made before I make any measurements. Yes?
If that is what you want the context of your statements to be, then my remarks about what things are interpretation dependent applies. And any discussion of things that are interpretation dependent belongs in a separate thread in the interpretations subforum, not in this thread.

HighPhy said:
I have predictions for the two experiments, one measuring ##\psi## and the other ##\phi##, given knowledge that the state is ##\Psi##.
Yes, all that looks fine.

HighPhy said:
But now let's say that there's a ##50\%## chance that the system is in the pure state ##|\psi_1\rangle##, and a ##50\%## chance the system is in the pure state ##|\psi_2\rangle##.
But let's say it with math: you have a mixed state which is 50% ##\psi_1## and 50% ##\psi_2##. That means you have the following density matrix, written in the ##\psi_1##, ##\psi_2## basis:

$$
\rho = \begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
$$

HighPhy said:
So the probability of measuring the system in the state ##|\phi_1\rangle## after measuring ##\phi##, is
$$(50\% \ \mathrm{being \ in} \ \psi_1)(50\% \ \mathrm{measuring} \ \phi_1) + (50\% \ \mathrm{being \ in} \ \psi_2)(50\% \ \mathrm{measuring} \ \phi_1)=50\%$$
And another way of seeing that is to transform the above density matrix to the ##\phi_1##, ##\phi_2## basis, and find that it looks the same.

HighPhy said:
This is different than the pure state case.
Yes, because you have a mixed state: as I said before, you don't have full knowledge of how the system is prepared, so you can't make the same predictions you would make if you did know that.
 
  • #112
sillyputty said:
How can the existence of an interpretation enable that you're correct in speaking that way, unless it was true, that what makes you incorrect in speaking that way, is some other kind of interpretation?
I'm not sure what you're asking here, but discussion of interpretations is off topic in this thread; it belongs in a separate thread in the interpretations subforum.
 
  • #114
PeterDonis said:
Apart from that, you appear to be placing way, way too much weight on ordinary language descriptions. Physics is not done in ordinary language. It is done in math. You keep trying to make hairsplitting distinctions about ordinary language that is already known to be inadequate to describe the physics anyway. You would be far better served by forgetting all about ordinary language and learning how to describe the physics with math.
You are saying shutup and calculate. Does that mean that the "and,or" is not a question in QM( math) in the same sense that we do not ask what are virtual particle that are represented by math in QFT.
 
  • #115
selfsimilar said:
You are saying shutup and calculate.
In this forum, yes, because in this forum we just discuss the basic math of QM, without adopting any particular interpretation.

selfsimilar said:
Does that mean that the "and,or" is not a question in QM( math) in the same sense that we do not ask what are virtual particle that are represented by math in QFT.
I don't understand what you mean by this.
 
  • #116
PeterDonis said:
I don't understand what you mean by this.
I mean the math is silent on how to interpret the probabilities as "and" or "or".
 
  • #117
selfsimilar said:
I mean the math is silent on how to interpret the probabilities as "and" or "or".
The math is not silent on what probabilities mean operationally: you can test them experimentally.

The math is "silent" on "and" or "or" because those words are not math.
 
  • #118
HighPhy said:
Apologies to everyone for insisting on this thread. I would like to shed some light on this point.

Most widely circulated scientific articles present Schroedinger's Cat paradox in the following way:

"Suppose a perfectly closed lead box, that is, in such a way that you cannot understand in any way what is inside it.
The cat until the act of observation is both alive and dead, when you check collapses the wave function and the cat is either alive or dead."

This seems to me a misrepresentation of this thought experiment, typical of bad pop-science. Otherwise I am the one who has missed the point.

How can the phrase "both dead and alive" be a synonym for "a superposition of two states, dead or alive"? Is the inherent formulation of this thought experiment correct? And what role does superposition play?
Let me answer this as a quantum information theorist might. The difference between a classical bit of information (cbit) and a quantum bit of information (qubit) is that you can get from a pure state to another pure state continuously through other pure states for a qubit, while you are only passing through mixed states between pure states for the cbit.

For example, suppose your cbit is a box and a measurement of the box (opening it) reveals one of two outcomes: a ball (yes) or no ball (no). The probability space has two axes, one represents "yes" and the other "no". Those are pure states, i.e., they represent actual measurement outcomes of a single trial of the experiment. Any state between those pure states, e.g., 80% yes, 20% no, does not represent the outcome of some new measurement, it represents a distribution of the yes/no outcomes of the original measurement, i.e., it's a mixed state. But, if the ball-box combo was a qubit, then that 80-20 state would have to correspond to the outcome of some other measurement with 100% probability.

For example, the x-spin state $$\frac{|\text{x+}\rangle + |\text{x-}\rangle}{\sqrt{2}}$$ means you will get 50% "up" results and 50% "down" results when you make an x-spin measurement of electrons in this state. If the x direction is horizontal, then your Stern-Gerlach magnets are oriented horizontally and electrons in this state are deflected in equal degree to the right and left in equal numbers. Since this is a qubit, your electron state must also be a pure state for some measurement corresponding to an outcome with 100% probability. What is that measurement and its outcome in this case? A z-spin "up" state works. In other words, you pass electrons through vertically oriented (z direction) SG magnets and those that are deflected up (towards North magnetic pole) are selected for future measurement. Those electrons, if subjected to a z-spin measurement, will produce the outcome z-spin "up" with 100% certainty, but if subjected to an x-spin measurement will produce "up-down" (physically, right-left) outcomes in 50-50 fashion, i.e., $$|\text{z+}\rangle = \frac{|\text{x+}\rangle + |\text{x-}\rangle}{\sqrt{2}}$$. Now you understand the physical difference between a cbit and a qubit.

The problem with the way most people present Schroedinger's Cat is that they only talk about a measurement with outcomes of Live Cat (LC) and Dead Cat (DC). With that information alone, we could have a cbit. The problem Schroedinger was pointing out is that quantum mechanics is supposedly applicable to anything. Therefore, it should be possible to render the Cat-Box system a qubit rather than a cbit in which case the state $$\frac{|\text{LC}\rangle + |\text{DC}\rangle}{\sqrt{2}}$$ must represent the outcome of some measurement with 100% certainty. What is that measurement? And, what does its outcome mean physically? Can you turn the Cat-Box system into a qubit simply by adding a quantum trigger mechanism for the deadly gas? Does that help answer the questions we need answered to understand the Cat-Box system as a qubit? We could answer those questions for the spin of an electron, but his point was we have no answers for the Cat-Box system. So, is quantum mechanics really applicable to any thing?

We have a book forthcoming with Oxford UP in June 2024 titled "Einstein's Entanglement: Bell Inequalities, Relativity, and the Qubit" in which we propose a principle justification for the existence of qubits (whence the Hilbert space of quantum mechanics). It was written so that someone with knowledge of intro physics alone could understand it. Since you're a physics student, you might be interested in that.
 
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  • #119
RUTA said:
We have a book forthcoming with Oxford UP in June 2024 titled "Einstein's Entanglement: Bell Inequalities, Relativity, and the Qubit" in which we propose a principle justification for the existence of qubits (whence the Hilbert space of quantum mechanics). It was written so that someone with knowledge of intro physics alone could understand it. Since you're a physics student, you might be interested in that.
I'm looking forward to it!
 
  • #120
RUTA said:
Let me answer this as a quantum information theorist might.
Can quantum information theorist say something different than the original theory? AFAIK the theory says that you cannot know the exact value of the observable before measurement, even worse you cannot know things like "and", "or" question which this thread is about, you can only know the value after measurement. So I am really surprised why there are endless threads about that. The theory just has nothing to say about that (before measurement) by the postulates, that is that. Everything else is a made up idea trying to reach a similar structure to quantum mechanics by providing a "mechanism" before measurement henceforth called interpretation.
 
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  • #121
I was re-reading the previous posts, but a question arose.
Some posts ago, I said:

HighPhy said:
When you observe the spin, this probability distribution collapses to a defined state, and then your measurement changes that probability distribution. Depending on the case, it can collapse it to a very simple one - probability 1 for a certain value and 0 for all others.

As answers, I obtained:

PeroK said:
Spin is a 3D vector quantity. In QM, only spin about one axis can be defined - spin about the other two axes remains undefined. If we measure about the z-axis, we get either ##\pm \frac \hbar 2## and the state collapses to z-spin-up or z-spin-down. Subsequent measurements of a free particle will always give the same outcome. But, measurements about the x or y-axis will give ##\pm \frac \hbar 2## with equal probability.

Moreover, if the particle is in a magnetic field, then the state will naturally evolve from the initial state or z-spin-up or z-spin-down. Look up Larmor Precession.
PeterDonis said:
For spin measurements, this will always be the case, since spin is a discrete observable.
What is the particular connection that allows these two responses to be viewed as related?
Sorry for not grasping it.
 
  • #122
HighPhy said:
What is the particular connection that allows these two responses to be viewed as related?
My statement that you quoted was in response to your statement about collapse. Look for the word "collapse" in what you quoted from @PeroK. What does it say about that?
 
  • #123
PeterDonis said:
My statement that you quoted was in response to your statement about collapse. Look for the word "collapse" in what you quoted from @PeroK. What does it say about that?
It says "if we measure about the z-axis, we get either ##\pm \frac{\hbar}{2}## and the state collapses to z-spin-up or z-spin-down."

I may have understood. I focused my attention on the rest of the response, but perhaps what @PeroK says is the mathematical description of your words in the comment I quoted. Correct?
 
  • #124
HighPhy said:
what @PeroK says is the mathematical description of your words in the comment I quoted. Correct?
Yes.
 
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  • #125
HighPhy said:
It says "if we measure about the z-axis, we get either ##\pm \frac{\hbar}{2}## and the state collapses to z-spin-up or z-spin-down."

I may have understood. I focused my attention on the rest of the response, but perhaps what @PeroK says is the mathematical description of your words in the comment I quoted. Correct?
The spin state is an abstract vector in a 2D complex vector space; and, the spin measurable is a vector in physical space - although its components cannot all be well-defined.
 

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