The significance of the Dirac notation

[PeroK]

Ok. Suppose the state becomes $|\uparrow \downarrow \rangle$. Suppose the first state $\uparrow$ belongs to particle A. So A has state $\uparrow$. What will be what we measure at the side of particle B?

By definition particle $B$ is in the state $\downarrow \rangle$.

 

[entropy1]

By definition particle $B$ is in the state $\downarrow \rangle$.

Which eigenvector on side B would that be?

Or: the state of B is defined before it is measured, right?

 

[vanhees71]

What do you mean by $\psi_{\uparrow}$?

It's the corresponding component of $|\psi \rangle$ in the expansion in the $\sigma_z$-basis with eigenvalue $\uparrow$ (as I said, I find the notation with the arrows a bit ideosyncratic, but what can one do if it's so widely used)?

 

[PeroK]

This is also very misleading popular-science talk. Right is that the single-particle state in a two-particle system that is in an entangled pure state is in a mixed state.

Yes, I've changed "defined" to "pure".

 

[entropy1]

if you measure either particle you break the entanglement and the state becomes (after measurement) either $|\uparrow \downarrow \rangle$ or $|\downarrow \uparrow \rangle$.

Ok, so, the measurement consolidates the states, it assigns this states to the particles?

 

[PeroK]

Ok, so, the measurement consolidates the states, it assigns this states to the particles?

That's an strange way to describe it!

 

[entropy1]

That's an strange way to describe it!

But it is correct? How would you say it?

 

[PeroK]

But it is correct? How would you say it?

We generally talk about the outcome of a measurement and the state collapsing to the appropriate eigenstate. That's the usual terminology.

 

[entropy1]

We generally talk about the outcome of a measurement and the state collapsing to the appropriate eigenstate. That's the usual terminology.

Agreed. But the particles are in their own point in spacetime when measured, so are not measured simulaneously. So when particle A is measured, does (due to the breaking of the entanglement) particle B get assigned its state? Or is it more like "as if" it gets that state assigned?

 

[PeroK]

Agreed. But the particles are in their own point in spacetime when measured, so are not measured simulaneously. So when particle A is measured, does (due to the breaking of the entanglement) particle B get assigned its state?

Yes, see here (for example):

https://en.wikipedia.org/wiki/Quantum_entanglement#Quantum_mechanical_framework

 

[entropy1]

Yes, see here (for example):

https://en.wikipedia.org/wiki/Quantum_entanglement#Quantum_mechanical_framework

They speak of Bob getting the identical (reciprocal) result as Alice has measured, when measured in the same basis as Alice. But if the basises differ, the figures only agree when we assume the reciprocal state from Alice gets assigned to the particle of Bob. Is that correct?

 

[PeroK]

They speak of Bob getting the identical (reciprocal) result as Alice has measured, when measured in the same basis as Alice. But if the basises differ, the figures only agree when we assume the reciprocal state from Alice gets assigned to the particle of Bob. Is that correct?

It gets more complicated if measurements of spin are made about different axes. That's true in general for spin 1/2 particles.

 

[entropy1]

It gets more complicated if measurements of spin are made about different axes. That's true in general for spin 1/2 particles.

Suppose Alice measures outcome $\overrightarrow{A}$, then Bob measures outcome $\overrightarrow{B}$ with probability $\overrightarrow{\overline{A}}\cdot \overrightarrow{B}$, right?

 

[PeroK]

Suppose Alice measures $\overrightarrow{A}$, then Bob measures $\overrightarrow{\overline{A}}\cdot \overrightarrow{B}$, right?

I'm not familiar with that notation.

 

[entropy1]

I'm not familiar with that notation.

A=$\varphi_A$, B=$\varphi_B$, I suppose. And the inproduct. The projection of the state on the eigenvector. Nevermind. Thanks for so far.

 

[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.

 

[vanhees71]

Ok, so, the measurement consolidates the states, it assigns this states to the particles?

First of all the measurement measures an observable. The state describes how the quantum system is prepared and the probabilities to find any possible value (an eigenvalue of the corresponding self-adjoint operator) when you (accurately) measure this observable.

What happens to the system when measured depends on the details of the measurement device, i.e., in the quantum mechanical description on the detailed interactions between the measurement device and the measured quantum system. In many cases the quantum system gets destroyed. E.g., when detecting a photon the usual way is through the photoelectric effect with the detector material, and thus the photon gets absorbed.

In some rare cases you can realize what's known as a "von Neumann filter measurement". Then you can use the measurement device to filter out the system according to the measurement result. E.g., using a typical Stern-Gerlach experiment for measuring a spin component, i.e., an inhomogeneous magnetic field with the right properties (large homogeneous part in the direction of the spin component and sufficiently large field gradient) you split a beam of particles in several partial beams, each having a certain value of the so measured spin component (i.e., the SG magnet leads to a (nearly) perfect entanglement between the spin component and the position of the particles at the end of the magnet). To prepare particles with a certain value of the measured spin component you just put absorbers in front of all the partial beams with the "unwanted" spin component.

In many textbooks this is described as the "collapse of the state", but one should be aware that this is only an effective description of a von Neumann filter measurement, which is described by the quantum dynamical formalism as any other dynamics of the quantum system. Particularly it's due to local interactions with the measurement/preparation equipment (in the above example the magnetic field and the absorbers for the unwanted spin states), and thus the "collapse" in reality is not some esoteric mechanism outside of quantum mechanics, but it's just an effective description of the filter measurement.

For a thorough summary of the foundations of QM in terms of the Dirac notation/Hilbert-space formalism, see

https://www.physicsforums.com/insights/the-7-basic-rules-of-quantum-mechanics/

 

[entropy1]

I'm not sure what you mean by a "measured" state. You have an entangled state. E.g:
$$\frac 1 {\sqrt 2}(|\uparrow \downarrow \rangle + |\downarrow \uparrow \rangle)$$
In this state neither particle has a (edit) pure single-particle state; the system has not been "measured" (although you could say it must have been prepared in this state); if you measure either particle you break the entanglement and the state becomes (after measurement) either $|\uparrow \downarrow \rangle$ or $|\downarrow \uparrow \rangle$.

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?

 

[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?

How do you get an outcome without doing a measurement?

 

[entropy1]

How do you get an outcome without doing a measurement?

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

 

[PeroK]

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

If A and B measure spin about the different axes, then the final state will not be up or down in the z-basis, but up or down in the directions of measurement (with the appropriate statistical correlation).

Note that the initial state could have equally been described in any basis.

 

[DrClaude]

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?

There is no difference. The state after measurement is $|\uparrow \downarrow \rangle$ without regards to who performed the measurement first. This is also compatible with special relativity, in that different observers will not agree on who of A and B measured their particle first, but the result is still the same.

 

[entropy1]

There is no difference. The state after measurement is $|\uparrow \downarrow \rangle$ without regards to who performed the measurement first. This is also compatible with special relativity, in that different observers will not agree on who of A and B measured their particle first, but the result is still the same.

Ok, so to me that appears like not really precisely signified.

 

[DrClaude]

The state after measurement is $|\uparrow \downarrow \rangle$ without regards to who performed the measurement first.

I forgot to add that this extends to the case where one performs a measurement but not the other.

 

[Nugatory]

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?

Look at post #20 again... the answer to this exact question is there in the last two sentences.

 

[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.