The significance of the Dirac notation

[entropy1]

An example would be a two particle system in which $A$ measures the spin of the one particle on some axis, and $B$ measures the spin of the other particle

So in case of $|ab\rangle$, a signifies measurement in basis A, and b in basis B (so to say)?

I think I misunderstand you. The states of particles A and B are opposite in case of entanglement.

 

[vanhees71]

No, $a$ signifies a state of particle 1 and $b$ a state of particle 2. The state the of two particles together is described by the tensor-product state $|ab \rangle=|a \rangle \otimes |b \rangle$.

 

[Nugatory]

Because if the basises of A an B differ and A measures outcome $|\uparrow \rangle$, then B can't measure outcome $|\downarrow \rangle$.

Before I answer this, I’ll start with what might be a quibble but probably isn’t: the phrase “A measures outcome $|\uparrow \rangle$“ is complete nonsense. $|\uparrow \rangle$ isn’t a measurement result, it’s a vector in a Hilbert space and measurement results are numbers. And furthermore, it’s not even a vector in the Hilbert space of states of our two-particle system.
This might be a quibble because I know you meant “A’s detector measures spin-up”... but I also strongly suspect that your confused wording is the result of an underlying confusion about what a ket represents.

But with that said, we can look at what happens if A and B use different bases to label vectors in their respective two dimensional Hilbert spaces, a natural choice if their measuring devices are aligned along different axes. The two Hilbert spaces are the same, the choice of basis only affects the labels we write down inside the kets. The four-dimensional Hilbert space of states of the two-particle system is of course still the tensor product of these two two-dimensional Hilbert spaces, and we can still choose $|\uparrow\uparrow\rangle$, $|\uparrow\downarrow\rangle$, $|\downarrow\uparrow\rangle$, $|\downarrow\downarrow\rangle$ as a basis within this Hilbert space (using the common convention that, for example, $|\uparrow\uparrow \rangle$ means $|\uparrow_A\rangle\otimes|\uparrow_B\rangle$ and the subscripts remove the ambiguity about which vector belongs to which Hilbert space).

Now after A measures spin-up, the state of the two particle system will be something like $\alpha|\uparrow\uparrow \rangle+ \beta|\uparrow\downarrow \rangle$. (This is the same state that we’ve been writing as $|\uparrow\downarrow \rangle$ when we’ve chosen to align the detectors and base vectors on the same axis, just written in a different basis).

It’s clear from the Born rule that B can measure either up or down.

 

[entropy1]

No, $a$ signifies a state of particle 1 and $b$ a state of particle 2. The state the of two particles together is described by the tensor-product state $|ab \rangle=|a \rangle \otimes |b \rangle$.

Of course. But in the case of entanglement, if measurement yields state a (of A), then state b is relative to state a, right?

 

[entropy1]

Before I answer this, I’ll start with what might be a quibble but probably isn’t: the phrase “A measures outcome $|\uparrow \rangle$“ is complete nonsense. $|\uparrow \rangle$ isn’t a measurement result, it’s a vector in a Hilbert space and measurement results are numbers. And furthermore, it’s not even a vector in the Hilbert space of states of our two-particle system.

Yes, I'm a bit confused but that is my problem asking questions here without much preceding study. And yes, I come a bit sloppy, but I mean to do that to reduce the confusion. And I prefer to identify the eigenvector with the measurement outcome instead of the eigenvalue, of which the latter seems trivial to me.

Concerning the rest of your post, I'm going to study it before placing more posts.

 

[Nugatory]

Of course. But in the case of entanglement, if measurement yields state a (of A), then state b is relative to state a, right?

No. There is no “state of A” and “state of B”, only the state of the single quantum system with observables that are measured at different places. And there’s no “b is relative to a” here either; when we write $|ab\rangle$ the meanings of $a$ and $b$ are determined by whatever conventions we’ve used to label vectors in the two Hilbert spaces whose tensor product forms the Hilbert space that we’re interested in; there’s no particular relationship between these conventions.

(And you do understand that ##|ab\rangle is not an entangled state.)

 

[entropy1]

No. There is no “state of A” and “state of B”, only the state of the single quantum system with observables that are measured at different places. And there’s no “b is relative to a” here either; when we write $|ab\rangle$ the meanings of $a$ and $b$ are determined by whatever conventions we’ve used to label vectors in the two Hilbert spaces whose tensor product forms the Hilbert space that we’re interested in; there’s no particular relationship between these conventions.

My example here is the orientations of SG magnets or the orientations of polarization filters.

(And you do understand that ##|ab\rangle is not an entangled state.)

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

 

[vanhees71]

Of course. But in the case of entanglement, if measurement yields state a (of A), then state b is relative to state a, right?

Take the example from posting #63,
$$|\Psi \rangle=\alpha |\uparrow \uparrow \rangle + \beta |\downarrow \downarrow \rangle$$
and measure for simplicity the spin $z$ component of both particles. The probabilities are given by
$$P(a,b)=|\langle ab|7\Psi \rangle|^2.$$
For the four possible outcomes you simply get the probabilities
$$P(\uparrow,\uparrow)=|\alpha|^2, \quad P(\downarrow,\downarrow)=|\beta|^2, \quad P(\uparrow,\downarrow)=P(\downarrow,\uparrow)=0.$$
This just tells you: It's impossible that you find opposite spins then measureing the $s_z$ components of both particles and the probability that both spins are up is $|\alpha|^2$ and for both spin components down is $|\beta|^2$. That's all. There's not more you can say about the outcome of measurements when measuring the $s_z$-components of both particles. In this minimal interpretation there's nothing very surprising about this result. The one-to-one correlation of the outcomes for the $s_z$ components of the two particles is due to the preparation of the two-particle spin state in this specific entangled state.

 

[PeroK]

Ok, so what I mean is, if we get $|\uparrow \downarrow \rangle$ as an outcome, do we interpret that as $|\uparrow \rangle$ is what is measured by A, and $|\downarrow \rangle$ is the state of B, or as $|\downarrow \rangle$ is what is measured by B, and $|\uparrow \rangle$ is the state of A?

I wonder whether you are confused by the notation here. Usually, by default, $|\uparrow \rangle$ and $|\downarrow \rangle$ mean the state of spin-up in the z-direction and spin-down in the z-direction respectively. If you start measuring spin about different axes, then you need notation to denote spin-up and spin-down about other axes.

You could use $|\uparrow_z \rangle$, $|\uparrow_x \rangle$ and $|\uparrow_y \rangle$, for example. In this case, if $A$ is measured about the z-axis, then the resulting state of particle $A$ is either $|\uparrow_z \rangle$ or $|\downarrow_z \rangle$. And, if $B$ is measured about the x-axis, then the resulting state of particle $B$ is $|\uparrow_x \rangle$ or $|\downarrow_x \rangle$.

 

[entropy1]

I wonder whether you are confused by the notation here. Usually, by default, $|\uparrow \rangle$ and $|\downarrow \rangle$ mean the state of spin-up in the z-direction and spin-down in the z-direction respectively. If you start measuring spin about different axes, then you need notation to denote spin-up and spin-down about other axes.

You could use $|\uparrow_z \rangle$, $|\uparrow_x \rangle$ and $|\uparrow_y \rangle$, for example. In this case, if $A$ is measured about the z-axis, then the resulting state of particle $A$ is either $|\uparrow_z \rangle$ or $|\downarrow_z \rangle$. And, if $B$ is measured about the x-axis, then the resulting state of particle $B$ is $|\uparrow_x \rangle$ or $|\downarrow_x \rangle$.

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

 

[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.