The significance of the Dirac notation

[PeroK]

Yes, it seems to me that the notation $|\uparrow\downarrow\rangle$ implies identical axis' to measure along in case of entanglement.

No. By convention it means the z-axis specifically. It has nothing to do with entanglement.

 

[entropy1]

No. By convention it means the z-axis specifically. It has nothing to do with entanglement.

Ok, so can the z-axis of Alice differ from that of Bob?

 

[kith]

Ok, but it is not obvious to me if the Hilbert Spaces used for a and b are identical or different.

It can be two copies of the same Hilbert space or two different Hilbert spaces.

Simply writing down the state $|ab \rangle$ is in no way enough to actually specify the physical situation. You need to specify what quantum system you are considering (An atom? The universe? The spin degrees of freedom of two electrons? The spin degrees of freedom of an electron and a proton?) and what the labels mean (An eigenstate of a certain observable which is named $AB$? The quantum state at a time $t_{ab}$? A product state of the spin degrees of freedom of two electrons witch chosen $z$-axes?). Dirac notation really is this flexible.

So let me give an example which includes all the information you need to specify (in principle, in practise many conventions apply):
We have two electrons. We consider only their spin degrees of freedom. We introduce a coordinate system with $x, y$ and $z$ axes for the first electron and a coordinate system with $x', y'$ and $z'$ axes for the second electron*. A valid state for this system then would be that the first electron is in the spin up state with respect to the $z$ axis and the second electron is in the spin down state with respect to the $z'$ axis:
$$| \uparrow_{z} \rangle \otimes |\downarrow_{z'} \rangle =: | \uparrow_{z} \downarrow_{z'} \rangle =: |ab \rangle$$
Now, $| ab \rangle$ has a well-defined meaning. Note that we don't need to specify how we realize the axes in the lab (By putting yard sticks on the wall? By using SG devices?) or how the state is reached (By performing a measurement? By letting a well-prepared system time evolve with a known Hamiltonian?). This only gets relevant if we want to actually prepare this state in the lab not if we want to reason about it hypothetically.

_____
*: This presupposes that we can distinguish the electrons somehow.

 

[PeroK]

Ok, so can the z-axis of Alice differ from that of Bob?

No. That would be silly. You must assume A and B have agreed a common coordinate system.

Imagine you were trying to do air traffic control and you and the aircraft were using different definitions of North and South? It doesn't really change anything. It just makes communications a mess.

 

[entropy1]

No. That would be silly. You must assume A and B have agreed a common coordinate system.

I agree. If we do that, then Alice and Bob could still have different basis', right? But then, if Alice happens to get outcome $|\uparrow_A\rangle$, then following Dirac in this way, the other particle must have state $|\downarrow_A\rangle$, but that won't be the outcome Bob gets, right? He will get $|\downarrow_B\rangle$, that doesn't have to be the same as $|\downarrow_A\rangle$. He will get it with probability $\langle\downarrow_A|\downarrow_B\rangle$, right?

 

[PeroK]

I agree. If we do that, then Alice and Bob could still have different basis', right? But then, if Alice happens to have outcome $|\uparrow_A\rangle$, then following Dirac, the other particle must have state $|\downarrow_A\rangle$, but that won't be the outcome Bob gets, right? He will get $|\downarrow_B\rangle$, that doesn't have to be the same as $|\downarrow_A\rangle$.

This is just so confused now. You need to get a grip on this. Alice and Bob can measure whatever they want. But, if Alice isn't measuring about the z-axis she can't get $|\uparrow_A\rangle$. I think you are trying to use the $A$ to denote both that it's Alice's particle and Alice's orientation. That won't do.

I suspect this is one of your problems: you are using $|\uparrow\rangle$ to mean "spin up in whatever direction was measured". That is too sloppy. All the up and down arrows refer to the common z direction, unless otherwise indicated.

I also suspect this is why you are misunderstanding what $|\uparrow \downarrow\rangle$ means in the first place. It means: the state composed of the first particle in the Z-spin-up state and the second particle in the Z-spin-down state.

To find out why precisely we need to specify a Z-direction here I suggest you review Susskind. These arrows do not refer to any other than the common Z-direction.

 

[entropy1]

Ok. So I feel I know enough in principle, but don't let it scare off other members if they want to post. If I feel I have to respond, I will.

 

[PeroK]

I agree. If we do that, then Alice and Bob could still have different basis', right? But then, if Alice happens to get outcome $|\uparrow_A\rangle$, then following Dirac in this way, the other particle must have state $|\downarrow_A\rangle$, but that won't be the outcome Bob gets, right? He will get $|\downarrow_B\rangle$, that doesn't have to be the same as $|\downarrow_A\rangle$. He will get it with probability $\langle\downarrow_A|\downarrow_B\rangle$, right?

No, because we started with a state $|\uparrow \downarrow \rangle$. If you want different bases for $A$ and $B$ you need to rewrite that state first.

Your notation is overloaded here to the point where no one knows what you mean.

 

[entropy1]

No, because we started with a state $|\uparrow \downarrow \rangle$. If you want different bases for $A$ and $B$ you need to rewrite that state first.

That makes sense to me. More than you expected perhaps. Because we seem to agree.

 

[entropy1]

This is just so confused now. You need to get a grip on this. Alice and Bob can measure whatever they want. But, if Alice isn't measuring about the z-axis she can't get $|\uparrow_A\rangle$. I think you are trying to use the $A$ to denote both that it's Alice's particle and Alice's orientation. That won't do.

I'm not confusing them. By $|\uparrow_A\rangle$ I mean spin-up in Alice's basis.

I suspect this is one of your problems: you are using $|\uparrow\rangle$ to mean "spin up in whatever direction was measured". That is too sloppy. All the up and down arrows refer to the common z direction, unless otherwise indicated.

The state will collapse along the direction of an eigenvector of the operator of the first measurement, to my knowledge.

I also suspect this is why you are misunderstanding what $|\uparrow \downarrow\rangle$ means in the first place. It means: the state composed of the first particle in the Z-spin-up state and the second particle in the Z-spin-down state.

Isn't the state that results from measurement along the direction of one of the eigenvectors of one of the operators (in fact the operator that measures first)? (collapse)

 

[PeterDonis]

The state will collapse along the direction of an eigenvector of the operator of the first measurement

The measurements are spacelike separated, so there is no invariant "first" measurement--in some frames, one measurement is first, and in some frames, the other is first. (And there will be a frame in which both measurements occur at exactly the same time, so neither one is first.)

In fact, the probabilities for results of the measurements do not depend at all on the order in which they are done.

 

[entropy1]

The measurements are spacelike separated, so there is no invariant "first" measurement--in some frames, one measurement is first, and in some frames, the other is first. (And there will be a frame in which both measurements occur at exactly the same time, so neither one is first.)

In fact, the probabilities for results of the measurements do not depend at all on the order in which they are done.

I know. It was trying to suggest an example. I ment something like: if we assume Alice's state collapsed to $e_A$ (eigenvector), then Bob's state will collapse to $e_A$ also. But for the same matter, Bob's state will collapse to $e_B$, and Alice's state will collapse to $e_B$ also. Dirac notation could be $|\uparrow\uparrow\rangle$. The result is the same. But the two ways you can look at it may be different. So if there are two ways to look at it, the notation may be confusing. That is what I mean.

 

[PeroK]

I know. It was trying to suggest an example. I ment something like: if we assume Alice's state collapsed to $e_A$ (eigenvector), then Bob's state will collapse to $e_A$ also. But for the same matter, Bob's state will collapse to $e_B$, and Alice's state will collapse to $e_B$ also. Dirac notation could be $|\uparrow\uparrow\rangle$. The result is the same. But the two ways you can look at it may be different. So if there are two ways to look at it, the notation may be confusing. That is what I mean.

Are you sure you're not treating the state $|\uparrow\uparrow\rangle$ as some sort of universal up-up state? Like you get up-up whatever the axes of measurement?

 

[entropy1]

Are you sure you're not treating the state $|\uparrow\uparrow\rangle$ as some sort of universal up-up state? Like you get up-up whatever the axes of measurement?

No. To the contrary.

 

[PeroK]

No. To the contrary.

Which is ...?

 

[entropy1]

Which is ...?

I know. It was trying to suggest an example. I ment something like: if we assume Alice's state collapsed to $e_A$ (eigenvector), then Bob's state will collapse to $e_A$ also. But for the same matter, Bob's state will collapse to $e_B$, and Alice's state will collapse to $e_B$ also. Dirac notation could be $|\uparrow\uparrow\rangle$. The result is the same.

But in the first option, both up arrows signify state $e_A$. In the second option both up arrows signify state $e_B$. That's not te same thing. At least, that is how I can not help but see it.

 

[PeterDonis]

if we assume Alice's state collapsed to $e_A$ (eigenvector), then Bob's state will collapse to $e_A$ also. But for the same matter, Bob's state will collapse to $e_B$, and Alice's state will collapse to $e_B$ also.

Neither of these are correct unless both Alice and Bob are measuring spin about the same axis direction. If they are measuring spin about different axis directions, you cannot know what state either Alice's or Bob's qubit collapses to until that qubit is measured; you can't assign it a state based solely on the other qubit's measurement result.

 

[entropy1]

Neither of these are correct unless both Alice and Bob are measuring spin about the same axis direction. If they are measuring spin about different axis directions, you cannot know what state either Alice's or Bob's qubit collapses to until that qubit is measured; you can't assign it a state based solely on the other qubit's measurement result.

$|\uparrow\uparrow\rangle$ shows two identical states. At least one of them signifies a measured state. The other must then be the same state. With different basis' we don't measure two identical states of course. You could say that the first one measured (for example Alice) collapses and we call the state $e_A = |\uparrow\rangle$. Then Bob won't measure the same state. He will measure the projection of $e_A$ along one of his eigenvectors. The converse holds for if Bob measures first.

 

[PeterDonis]

$|\uparrow\uparrow\rangle$ shows two identical states.

No. It shows a single state of a two-qubit system in which the spins of both qubits are correlated, so if they are both measured in the same spin axis direction, the results will be the same.

At least one of them signifies a measured state.

Not necessarily.

The other must then be the same state.

No, both spins are correlated in the way I described above.

With different basis' we don't measure two identical states of course.

You say "of course" but I'm not sure you realize why: it's because neither of the states that are eigenstates in the new basis are "the same" as either of the states that are eigenstates in the old basis.

He will measure the projection of $e_A$ along one of his eigenvectors.

No, he won't. He will measure one of his eigenvalues.

 

[Vanadium 50]

This thread is ostensibly about notation, but it seems that the real problem is more fundamental. It seems that the OP is unclear about what exactly is being notated. Maybe a step back would be useful.

I am more convinced of this than ever.

You've moved from asking us about notation to arguing about it with @PeroK .

I took a look at your posting history. You've posted around one hundred threads over a period of years, and I would characterize many of them as "trying to understand QM by getting the words right". The fact that it's gone on for years means that it's not working for you. Have you thought about a different approach? Take a class. Buy a textbook. Otherwise, we will be right back in the same spot in a year. And two. And five.

 

[PeterDonis]

in the first option, both up arrows signify state $e_A$. In the second option both up arrows signify state $e_B$.

Then you should make this clear by using the subscripts and writing these two distinct states as $| \uparrow_A \uparrow_A \rangle$ and $| \uparrow_B \uparrow_B \rangle$. Otherwise you'll just confuse people, including yourself.

 

[entropy1]

Then you should make this clear by using the subscripts and writing these two distinct states as $| \uparrow_A \uparrow_A \rangle$ and $| \uparrow_B \uparrow_B \rangle$. Otherwise you'll just confuse people, including yourself.

Yes, I was thinking along this line myself. I thought perhaps $| \uparrow_A \uparrow_A \rangle + | \uparrow_B \uparrow_B \rangle$ might do the trick. Or perhaps it should be $| \uparrow_A \uparrow_A \rangle + | \uparrow_B \uparrow_B \rangle + | \downarrow_A \downarrow_A \rangle + | \downarrow_B \downarrow_B \rangle$.

 

[PeterDonis]

I thought perhaps $| \uparrow_A \uparrow_A \rangle + | \uparrow_B \uparrow_B \rangle$ might do the trick.

That is a different state of the two-qubit system than any of the ones we have been talking about up to now. Which state do you want to talk about?

 

[entropy1]

That is a different state of the two-qubit system than any of the ones we have been talking about up to now. Which state do you want to talk about?

Sorry, I mean an entangled state thereby.

It was just a hunch.

 

[PeterDonis]

I mean an entangled state thereby.

An entangled state would be something like $1 / \sqrt{2} \left( | \uparrow_A \uparrow_A \rangle + | \downarrow_A \downarrow_A \rangle \right)$.

 

[entropy1]

A different way to put my question is: Suppose we can write our entangled state as $|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle$ and observe that the first entries in both sub-states represents the possible outcome that we supposedly to measure. The physical manifestation of the outcome however may differ according to the orientation of the measurement basis. So the arrow of this first entry is not representing a known physical manifestation until we have measured an outcome and associate that outcome with one of the two possible arrows in the first entry. Then, after the measurement of the first entry yields either spin up or down, people seem to assert that the spin of the other particle (the second entry) is in fact either up or down too with respect to the orientation of the spin of the first particle. Thus, we claim to only know the spin of the first particle after having measured it, but know the spin of the other particle while not having measured it. And if the measurement basis of the second particle has a different orientation from the basis of the first particle, the arrow of this second particle cannot represent the actual outcome, while the first arrow is supposed to do just that! what is left of the meaning of up arrow or down arrow in the entangled state? Does the entangled state mean to represent measurement outcomes? It doesn't seem so at all! The the up arrow and the down arrow seem to represent nothing at all!

You can see it thus also: suppose the SGM's have a 45 degree angle from each other: SGM A 0 degrees and SGM B 45 degrees. If we assume the first entry is measured first, up arrow means 0 degrees and down arrow means 180 degrees. If we assume the second entry is measured first, up arrow means 45 degrees and down arrow means 225 degrees. So the meaning of the arrows vary depending on who is supposedly measuring first.

 

[PeroK]

Either you understand what a state represents or you don't. If you do, the meaning is clear. Your attempt to describe a state in these terms indicates that you have not grasped the mathematical formalism and relationship between a state and all possible measurement outcomes.

 

[entropy1]

Either you understand what a state represents or you don't. If you do, the meaning is clear. Your attempt to describe a state in these terms indicates that you have not grasped the mathematical formalism and relationship between a state and all possible measurement outcomes.

Could well be. You probably conclude that from the "muddyness" of my argument. Or just the math of QM prevails. I hope this is clearer?:

You can see it thus also: suppose the SGM's have a 45 degree angle from each other: SGM A 0 degrees and SGM B 45 degrees. If we assume the first entry is measured first, up arrow means 0 degrees and down arrow means 180 degrees. If we assume the second entry is measured first, up arrow means 45 degrees and down arrow means 225 degrees. So the meaning of the arrows vary depending on who is supposedly measuring first.

From this it should be clear what I mean.

I guess that Dirac notation is equivalent to other notations, so Dirac is probably not the problem here.

Me against the physics community lol

 

[PeroK]

Could well be. You probably conclude that from the "muddyness" of my argument. Or just the math of QM prevails. I hope this is clearer?:

It's not an argument, as such. It's just a muddy way of trying to say what can be said more simply.

So the meaning of the arrows vary depending on who is supposedly measuring first.

In any case, you've ended up at this patently false conclusion.

 

[PeroK]

Or just the math of QM prevails.

You could check out this thread for how to understand these states in terms of different measurement axes:

https://www.physicsforums.com/threa...s-in-two-different-basis.993252/#post-6388834

 

[PeterDonis]

Me against the physics community lol

And you will lose.

Enough is enough. Thread closed.