The significance of the Dirac notation

In summary: 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.Ok.
  • #1
entropy1
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If we have the wavefunction ##|ab \rangle##, what do the a and b stand for? In particular, do a and b signify an outcome of some pending or possible measurement, or do they signify some aspect of the wavefunction, and if so, which aspect?
 
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  • #2
entropy1 said:
If we have the wavefunction ##|ab \rangle##, what do the a and b stand for? In particular, do a and b signify an outcome of some pending or possible measurement, or do they signify some aspect of the wavefunction, and if so, which aspect?
Where are you learning Dirac notation?
 
  • #3
PeroK said:
Where are you learning Dirac notation?
I learned it from Susskind. Perhaps he explained it in his book but I can't really remember.
 
  • #4
entropy1 said:
I learned it from Susskind.
What does Susskind say is meant by ##|ab \rangle##?
 
  • #5
PeroK said:
What does Susskind say is meant by ##|ab \rangle##?
I figured it was equivalent with ##|↑↓\rangle##.
 
  • #6
entropy1 said:
I figured it was equivalent with ##|↑↓\rangle##.
##a## and ##b## are variables and ##\uparrow## and ##\downarrow## represent specific states. Do you understand the concept of a state?
 
  • #7
PeroK said:
Do you understand the concept of a state?
No, not really. I am confused about it. I suppose it is a vector in Hilbert Space.
 
  • #9
If I understand you correctly, a and b represent states. My question is: which states do they represent? The states of an unmeasured wavefunction or the states of measured outcomes (eigenvectors)?
 
  • #10
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].
 
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  • #11
entropy1 said:
If I understand you correctly, a and b represent states. My question is: which states do they represent?
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.
 
  • #12
Maybe to be more precise: if we have ##c_1|↑↓\rangle + c_2|↓↑\rangle## representing an entangled state, what do each of the four states signify?
 
  • #13
entropy1 said:
Maybe to be more precise: if we have ##c_1|↑↓\rangle + c_2|↓↑\rangle## representing an entangled state, what do each of the four states signify?
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.
 
  • #14
PeroK said:
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.
They can't however mean the states of the particles before measurement, for, say, the spin has no value. So it must be the state after measurement, then, or, the eigenvector that represents the measurement outcome?
 
  • #15
entropy1 said:
They can't however mean the states of the particles before measurement, for, say, the spin has no value. So it must be the state after measurement, then, or, the eigenvector that represents the measurement outcome?
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.

You could say that the system has been prepared in that entangled state.
 
  • #16
PeroK said:
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.

You could say that the system has been prepared in that entangled state.
But the states (the four arrows) must represent something right? 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.
 
  • #17
entropy1 said:
But the states (the four arrows) must represent something right? 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.
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.
 
  • #18
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.
 
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  • #19
entropy1 said:
So in this matter, what is really going on?

What do you mean by that? This is the way QM works.
 
  • #20
entropy1 said:
If we have the wavefunction ##|ab \rangle##, what do the a and b stand for? In particular, do a and b signify an outcome of some pending or possible measurement, or do they signify some aspect of the wavefunction, and if so, which aspect?
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.
 
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  • #21
PeroK said:
What do you mean by that? This is the way QM works.
I just mean the four arrows in the entangled state are signifying different things depending on your point of view.
 
  • #22
entropy1 said:
I just mean the four arrows in the entangled state are signifying different things depending on your point of view.
There's no point of view involved. There is a well-defined (entangled) state for a composite system of two particles.
 
  • #23
PeroK said:
There's no point of view involved. There is a well-defined (entangled) state for a composite system of two particles.
Ok so we have a different POV :oldbiggrin:

If you feel like it, read #18 :wink:

I don't know why I said I didn't understand the state concept. Well, at least not fully.
 
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  • #24
entropy1 said:
I just mean the four arrows in the entangled state are signifying different things depending on your point of view.
No, they mean whatever they mean in whatever notation we decided to use.
 
  • #25
Nugatory said:
No, they mean whatever they mean in whatever notation we decided to use.
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.
 
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  • #26
entropy1 said:
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 ↓ a state, not a measurement outcome.
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.
 
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  • #27
entropy1 said:
But the states (the four arrows) must represent something right? 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.
Well, it's a bit vaguely formulated, but I guess what's meant is that you consider two spin-1/2 particles.

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.
 
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  • #28
Thanks. I guess I need to know, if we take ##| \uparrow \rangle## as the state of one of the particles of the entangled pair, whether it is the state before the (spin-)measurement, or the state after the measurement.
 
  • #29
entropy1 said:
Thanks. I guess I need to know, if we take ##| \uparrow \rangle## as the state of one of the particles of the entangled pair, whether it is the state before the (spin-)measurement, or the state after the measurement.
Neither single particle has a state when the two-particle state is entangled. That is, more or less, the definition of entanglement.
 
  • #30
PeroK said:
Neither single particle has a state when the two-particle state is entangled. That is, more or less, the definition of entanglement.
Right. So these states refer to the measured state? Or, as I understand, the eigenvectors of the operators?
 
  • #31
vanhees71 said:
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} ##?
 
  • #32
entropy1 said:
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##.
 
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  • #33
entropy1 said:
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_+##.
 
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  • #34
PeroK said:
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?
 
  • #35
PeroK said:
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.
 
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