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entropy1

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- #1

entropy1

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- #2

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Where are you learning Dirac notation?

- #3

entropy1

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I learned it from Susskind. Perhaps he explained it in his book but I can't really remember.Where are you learning Dirac notation?

- #4

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What does Susskind say is meant by ##|ab \rangle##?I learned it from Susskind.

- #5

entropy1

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I figured it was equivalent with ##|↑↓\rangle##.What does Susskind say is meant by ##|ab \rangle##?

- #6

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##a## and ##b## are variables and ##\uparrow## and ##\downarrow## represent specificI figured it was equivalent with ##|↑↓\rangle##.

- #7

entropy1

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No, not really. I am confused about it. I suppose it is a vector in Hilbert Space.Do you understand the concept of a state?

- #8

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https://en.wikipedia.org/wiki/Quantum_stateNo, not really.

The answer to your original question is that it is a composite state:

https://en.wikipedia.org/wiki/Bra–ket_notation#Composite_bras_and_kets

- #9

entropy1

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- #10

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They are generic ones, even though in most situations are eigenstates of particular observables, case in which a and b are replaced by other letters or signs.

More precisely, mathematicians are fond of using the letter A for a generic (not necessarily self-adjoint) operator in a Hilbert space. In Dirac notation its spectral equation then reads [tex]|a\rangle =a |a\rangle [/tex].

More precisely, mathematicians are fond of using the letter A for a generic (not necessarily self-adjoint) operator in a Hilbert space. In Dirac notation its spectral equation then reads [tex]|a\rangle =a |a\rangle [/tex].

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- #11

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They are variables. They could be any states, given the context they are presented in. Similar to ##ax^2 + bx + c## being a general quadratic equation.If I understand you correctly, a and b represent states. My question is: which states do they represent?

- #12

entropy1

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- #13

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The state ##|\uparrow \downarrow \rangle## is a composite (two particle) state where (loosely) the first particle is in the up state and the second particle is in the down state. The state ##|\downarrow \uparrow \rangle## is similarly defined. The whole thing is a linear combination of these two states.entangledstate, what do each of the four states signify?

- #14

entropy1

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They can't however mean the states of the particlesThe state ##|\uparrow \downarrow \rangle## is a composite (two particle) state where (loosely) the first particle is in the up state and the second particle is in the down state. The state ##|\downarrow \uparrow \rangle## is similarly defined. The whole thing is a linear combination of these two states.

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I suspect you are missing something fundamental in your understanding of the nature of QM states. So, I'm not sure how to answer that. The state you quoted does not imply a measurement of spin on the system or on either particle.They can't however mean the states of the particlesbeforemeasurement, for, say, the spin has no value. So it must be the stateaftermeasurement, then, or, the eigenvector that represents the measurement outcome?

You could say that the system has been

- #16

entropy1

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But the states (the four arrows) must representI suspect you are missing something fundamental in your understanding of the nature of QM states. So, I'm not sure how to answer that. The state you quoted does not imply a measurement of spin on the system or on either particle.

You could say that the system has beenpreparedin that entangled state.

- #17

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You don't have to assume that the spins are not well-defined, the entangled composite state tells you that. Like all states it contains the information about the likelihood of any measurement outcome on the system.But the states (the four arrows) must representsomethingright? If the spins (let's assume) of the entangled pair are not defined before measurement, this entangled state does not represent those spins, right? Yet.

- #18

entropy1

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Ok. Suppose particle A measures spin up. The entangled state "collapses" to ##|↑↓\rangle##. So the ↓ state can now signify the state of particle B (*before* it is measured). So now ↑ signifies a measurement outcome and ↓ a state *before* measurement, *not* a measurement outcome.

But you could also measure particle B first, in which case ↓ signifies the measurement outcome (particle B), and ↑ the state of particle A (*before* measurement). So in this matter, what is *really* going on with respect to the ontology of the arrow states?

You*could* say that ##|↑↓\rangle## means that you can measure particle A up and particle B down. But if the measurement basises are not the same, we won't measure these states simultaneously.

But you could also measure particle B first, in which case ↓ signifies the measurement outcome (particle B), and ↑ the state of particle A (

You

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- #19

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So in this matter, what isreallygoing on?

What do you mean by that? This is the way QM works.

- #20

Nugatory

Mentor

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The symbols within a bra or a ket are just labels.

We start with some Hilbert space that is appropriate for whatever problem we're considering. Then we think of a convenient scheme for labeling the vectors of that Hilbert space, and use these labels as we please. A two-dimensional Hilbert space will be spanned by two basis vectors, and if our problem involves the z-axis spin of the particle it's natural to choose the spin-up-z and spin-down-z vectors as the basis. Once we've done that, we can call these vectors ##|\uparrow\rangle## and ##|\downarrow\rangle##, or ##|z_+\rangle## and ##|z_-\rangle##, or ##|+\rangle## and ##|-\rangle##, or ##|sheep\rangle## and ##|goat\rangle##, or ...

You will always need the context (possibly implied) to know which vectors in which Hilbert space are represented by any particular kets.

There is one fairly common convention that you'll want to be aware of. Suppose the operator ##A## acts on the vectors of one Hilbert space, and we've decided to write the eigenvector of ##A## that has eigenvalue ##a## as ##|a\rangle##; and the operator ##B## acts on the vectors of another Hilbert space, and we've decided to write the eigenvector of ##B## that has eigenvalue ##b## as ##|b\rangle##. In this case the vector ##|a\rangle\otimes|b\rangle## is a vector in yet a third Hilbert space, the tensor product of the first two, and this will often be written as ##|ab\rangle##. (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 - that's what's going on with the ##|\uparrow\downarrow\rangle## notation).

This convention is common enough that you can assume that it is the answer to your question unless something else in the surrounding context says otherwise.

- #21

entropy1

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I just mean the four arrows in the entangled state are signifying different things depending on your point of view.What do you mean by that? This is the way QM works.

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There's no point of view involved. There is a well-defined (entangled) state for a composite system of two particles.I just mean the four arrows in the entangled state are signifying different things depending on your point of view.

- #23

entropy1

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Ok so we have a different POVThere's no point of view involved. There is a well-defined (entangled) state for a composite system of two particles.

If you feel like it, read #18

I don't know why I said I didn't understand the state concept. Well, at least not fully.

- #24

Nugatory

Mentor

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No, they mean whatever they mean in whatever notation we decided to use.I just mean the four arrows in the entangled state are signifying different things depending on your point of view.

- #25

entropy1

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Sure. And it works if you mean that. But I find it confusing. It appears to me as if it has no clearly defined ontology. Anyway, not everyone shares this concern of course.No, they mean whatever they mean in whatever notation we decided to use.

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- #26

vela

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You're making a distinction where there is none. The notation ##\lvert \uparrow \downarrow \rangle## means particle 1 is in the spin-up state and particle 2 is in the spin-down state, period.Ok. Suppose particle A measures spin up. The entangled state "collapses" to |↑↓⟩. So the ↓ state can now signify the state of particle B. So now ↑ signifies a measurement outcome and ↓ astate,nota measurement outcome.

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- #27

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Well, it's a bit vaguely formulated, but I guess what's meant is that you consider two spin-1/2 particles.But the states (the four arrows) must representsomethingright? If the spins (let's assume) of the entangled pair are not defined before measurement, this entangled state does not represent those spins, right? Yet.

Each particle's spin states are described in a 2D Hilbert space, spanned by the eigenvectors of one spin component (usually one chooses ##\hat{s}_z##) with eigenvalues ##+1/2## and ##-1/2##. The corresponding eigenstates are often denoted by ##|\uparrow \rangle## and ##|\downarrow \rangle##, which seems to be the notation used in your book. Each spin state can be written as a superposition of these eigenstates,

$$|\psi \rangle=\psi_{\uparrow} |\uparrow \rangle + \psi_{\downarrow} |\downarrow \rangle.$$

Now take two indepdendent particles. Then their spins are described by superpositions of product states ##|\psi_1 \rangle \otimes |\psi_2 \rangle \equiv |\psi_1,\psi_2 \rangle##. Now the general two-spin state can be written in terms of the four basis vectors

$$|\uparrow ,\uparrow \rangle, \quad |\uparrow,\downarrow \rangle, \quad |\downarrow,\uparrow \rangle, \quad |\downarrow,\downarrow \rangle,$$

i.e., the two-spin Hilbert space is four-dimensional. The most general two-spin state thus is given by

$$|\Psi \rangle = \sum_{a,b \in \{\uparrow,\downarrow\}} \Psi_{ab} |a,b \rangle.$$

I don't know, who has invented thus ##\uparrow##-##\downarrow## notation instead of simply using ##|\sigma_z \rangle## with ##\sigma_z \in \{-1/2,1/2 \}##, which is a much more intuitive notation, but that doesn't really matter.

- #28

entropy1

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- #29

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Neither single particle has a state when the two-particle state is entangled. That is, more or less, the definition of entanglement.beforethe (spin-)measurement, or the stateafterthe measurement.

- #30

entropy1

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Right. So these states refer to theNeither single particle has a state when the two-particle state is entangled. That is, more or less, the definition of entanglement.

- #31

entropy1

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What do you mean by ##\psi_{\uparrow} ##?Each spin state can be written as a superposition of these eigenstates,

##|\psi \rangle=\psi_{\uparrow} |\uparrow \rangle + \psi_{\downarrow} |\downarrow \rangle.##

- #32

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I'm not sure what you mean by a "measured" state. You have an entangled state. E.g:Right. So these states refer to themeasuredstate? Or, as I understand, the eigenvectors of the operators?

$$\frac 1 {\sqrt 2}(|\uparrow \downarrow \rangle + |\downarrow \uparrow \rangle)$$

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

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- #33

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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_+##.

- #34

entropy1

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

- #35

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

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.

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