Undergrad The significance of the Dirac notation

[entropy1]

Each spin state can be written as a superposition of these eigenstates,
$|\psi \rangle=\psi_{\uparrow} |\uparrow \rangle + \psi_{\downarrow} |\downarrow \rangle.$

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

 

[PeroK]

Right. So these states refer to the measured state? Or, as I understand, the eigenvectors of the operators?

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 $|\downarrow \uparrow \rangle$.

 

[PeroK]

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

That's a complex number. It's a way to represent the coefficient, instead of $c_1$ or $c_+$.

 

[entropy1]

and the state becomes (after measurement) either $|\uparrow \downarrow \rangle$ or $|\downarrow \uparrow \rangle$.

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?

 

[vanhees71]

Neither single particle has a state when the two-particle state is entangled. That is, more or less, the definition of entanglement.

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.

To understand this we must be a bit more precise. The state of a quantum system is described by a statistical operator $\hat{\rho}$, i.e., a positive semidefinite self-adjoint operator with $\mathrm{Tr} \hat{\rho}=1$. The state is a pure state if and only if there's a normalized vector $\Psi$ rangle such that $\hat{\rho}=|\Psi \rangle \langle \Psi |.$
Now take an arbitrary state $\hat{\rho}$ of a two-particle system. Then the state of one of the particles is defined as the socalled reduced state. To define it let $|a,b \rangle$ be an arbitrary product basis of the two-particle Hilbert space. Then the reduced state of particle 1 is
$$\hat{\rho}_1 = \sum_{a,a',b} |a \rangle \langle a,b|\hat{\rho}|a',b \rangle \langle a'|.$$
Now take as an example two spins as above and consider the entangled state defined as $\hat{\rho}=|\Psi \rangle \langle \Psi |$
$$|\Psi \rangle =\frac{1}{\sqrt{2}} (|\uparrow,\uparrow \rangle + |\downarrow,\downarrow \rangle).$$
Then
$$\hat{\rho}_1 = \sum_{a,a',b} |a \rangle \langle a'| \langle a,b|\Psi \rangle \langle \Psi|a',b \rangle.$$
Now
$$\langle a,b|\Psi \rangle=\frac{1}{\sqrt{2}} (\delta_{a,\uparrow} \delta_{b,\uparrow} + \delta_{a,\downarrow} \delta_{b,\downarrow}.$$
This leads to
$$\sum_b \langle a,b|\Psi \rangle \langle \Psi|a',b \rangle = \frac{1}{2} (\delta_{a,\uparrow} \delta_{a',\uparrow} + \delta_{a,\downarrow}{a',\downarrow})),$$
and thus finally
$$\hat{\rho}_1=\frac{1}{2} (|\uparrow \rangle \langle \uparrow | + |\downarrow \rangle \langle \downarrow|=\frac{1}{2} \hat{1}.$$
The spin state of particle 1 is thus completely undetermined, i.e., it's an unpolarized particle.

 

[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.