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In other words, does there exist, for an electron, a definite direction in physical space such that a measurement of its spin along that direction always give spin up, 100% of the time?
For a single free electron that is not entangled with anything else, yes, your statement is correct.does there exist, for an electron, a definite direction in physical space such that a measurement of its spin along that direction always give spin up, 100% of the time?
I would say that if you have an electron and measure its spin then "100% of the time" makes no sense.In other words, does there exist, for an electron, a definite direction in physical space such that a measurement of its spin along that direction always give spin up, 100% of the time?
There's always a direction along which the spin is definitively up or down for a spin-1/2 particle. As @PeterDonis said this fails once the particle is entangled.In other words, does there exist, for an electron, a definite direction in physical space such that a measurement of its spin along that direction always give spin up, 100% of the time?
What about an electron in a mixed state?There's always a direction along which the spin is definitively up or down for a spin-1/2 particle. As @PeterDonis said this fails once the particle is entangled.
It's also special to spin-1/2. For spin-1 particles there are states where the particle is not definitively aligned in any direction.
Also doesn't have a definite alignment in any direction.What about an electron in a mixed state?
Spin 1/2 is described by an irreducible non-trivial representation of su(2). The Lie algebra su(2) has only one Casimir operator, ##\hat{\vec{s}}^2## with eigenvalues ##s(s+1)## (##s \in \{0,1/2,\ldots \}##). Thus each representation of su(2) is characterized by the eigenvalues of this Casimir operator.In other words, does there exist, for an electron, a definite direction in physical space such that a measurement of its spin along that direction always give spin up, 100% of the time?
How does entanglement cause a loss of the definite direction? Suppose two free electrons are both spin up vertically. They come together and become entangled. Why aren’t they still remain as both spin up?For a single free electron that is not entangled with anything else, yes, your statement is correct.
But that is pretty much the only case where it is. If the electron is entangled with anything else, in general it won't have a definite spin state, so your statement will not be true. And actual electrons are almost always entangled with something else. For example, an electron in a hydrogen atom is entangled with the proton.
You are getting confused between a state and the description of that state in a given basis.How does entanglement cause a loss of the definite direction? Suppose two free electrons are both spin up vertically. They come together and become entangled. Why aren’t they still remain as both spin up?
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Consider a system of two electrons in the state |11> (4.177). Both are spin up. But if I turn my measurement device upside down, then would both electrons be measured spin down? But that state is a different state, |1-1>, implying that I could just change the state by turning my measurement device, an absurd conclusion. But if we take the view that turning the measurement device does not change the measured spin to down, but still spin up, then that spin up is up with respect to what, if we can’t take reference to the measurement device?
And for the state |10>, how can the total spin be 1, when it is guaranteed that the electrons always have opposite spins? Is it that before measurement, both electrons’ spin are in the same direction, hence the total spin adds up to one, but during measurement, the measurement process flips one of the electron spin? But how do you decide which electron to flip?
In an entangled two-electron state, the single electron also has a definite state, which is given by the partial trace of the stat. op. over the other electron. This reduced state for a subsystem is in the case that the complete system is in a spin-entangled pure state, a mixed state. That's one of the special properties of entanglement!How does entanglement cause a loss of the definite direction? Suppose two free electrons are both spin up vertically. They come together and become entangled. Why aren’t they still remain as both spin up?
View attachment 248035
Consider a system of two electrons in the state |11> (4.177). Both are spin up. But if I turn my measurement device upside down, then would both electrons be measured spin down? But that state is a different state, |1-1>, implying that I could just change the state by turning my measurement device, an absurd conclusion. But if we take the view that turning the measurement device does not change the measured spin to down, but still spin up, then that spin up is “up” with respect to what, if we can’t take reference to the measurement device?
And for the state |10>, how can the total spin be 1, when it is guaranteed that the electrons always have opposite spins? Is it that before measurement, both electrons’ spin are in the same direction, hence the total spin adds up to 1, but during measurement, the measurement process flips one of the electron spin? But how do you decide which electron to flip? But flipping any one electron will still seem to make the total spin after measurement become 0 instead. So it’s puzzling how the total spin could be 1.
I don’t really understand this. But are you trying to say that each electron in a two-electron system still has a definite state?In an entangled two-electron state, the single electron also has a definite state, which is given by the partial trace of the stat. op. over the other electron. This reduced state for a subsystem is in the case that the complete system is in a spin-entangled pure state, a mixed state. That's one of the special properties of entanglement!
The state ##|\Psi \rangle=|\uparrow,\uparrow \rangle## is not a spin-entangled state since it's a product state. The pure state of the two-spin systems is ##\hat{\rho}=|\Psi \rangle \langle \Psi|##. In this case the partial trace is of course a pure state,
$$\hat{\rho}_2=\mathrm{Tr}_1 \hat{\rho} =\sum_{i,j,k \in \{\uparrow,\downarrow \}} |j \rangle \langle ij|\hat{\rho}|ik \rangle \langle k|=|\uparrow \rangle \langle \uparrow |.$$
For the electron in a ground-state hydrogen atom, since there is no definite direction for its spin, does it mean that the probability of measuring spin up and down are both half, regardless of the direction of measurement?For a single free electron that is not entangled with anything else, yes, your statement is correct.
But that is pretty much the only case where it is. If the electron is entangled with anything else, in general it won't have a definite spin state, so your statement will not be true. And actual electrons are almost always entangled with something else. For example, an electron in a hydrogen atom is entangled with the proton.
The usual representation for a spin entangled pair of electrons is:I’m trying to figure out when two electrons come together and get entangled, which state, |11>, |10>, |1-1> or |00>, the resultant system would be in.
It depends on how they are entangled. There are four basis states for the Hilbert space of a two-electron system if we are only looking at the spin part of the state, three called "triplet" and one called "singlet". None of the ones you wrote down are among them; you forgot that electrons are indistinguishable so the spin part of the two-electron wave function has to be either symmetric (if the position/momentum part is antisymmetric) or antisymmetric (if the position/momentum part is symmetric).I’m trying to figure out when two electrons come together and get entangled, which state, |11>, |10>, |1-1> or |00>, the resultant system would be in.
Theoretically, yes. Practically speaking, we have no way of making such a measurement.For the electron in a ground-state hydrogen atom, since there is no definite direction for its spin, does it mean that the probability of measuring spin up and down are both half, regardless of the direction of measurement?
No. Measurement involves interaction with an external apparatus, so if you measure a system you can't just think of the system by itself, with regard to conservation laws or anything else; you have to think of the system plus apparatus.if there is no definite direction, does this violate the conservation of information?
Making a hydrogen atom doesn't just mean "bring an electron and a proton together" with nothing else being involved. The electron-proton system has to give up energy to become bound. The process of giving up energy involves emitting something, most likely photons, and the emitted photons carry away whatever angular momentum they need to to put the hydrogen atom in its final state, as well as whatever energy they need to to do that.Consider two identical protons A and B. A is approached by a "spin up" electron from the bottom, while B is approached by a "spin down" electron from the right. Clearly, these two hydrogen atoms are formed from different initial states. If they end up having the same final state, both with no definite spin direction, then where does the information about the initial spin go?
Another way to put this would be that even if you have an electron whose spin in the ##z## direction, say, you have just measured to be "up", that does not mean it's the same as a little spinning ball with angular momentum ##\hbar / 2## about the ##z## axis, as would be the case for a classical object that had just given that measurement result.A spin 1/2 particle has total angular momentum √3/2. At most 1/2 points in any given direction, so its spin is most assuredly not in some definite direction.
If you haven't already done so, you might want to google for "Clebsch-Gordan coefficients" - spin doesn't combine the way you'd intuitively expect.I suspect when two up electrons come together, the system will be in....
I'm using the same notation as 4.177 and 4.178. So according to them, |11>, |10> and |1-1> are the triplet, |00> is the singlet state, and the symmetric or antisymmetric requirement is already taken care of, right?It depends on how they are entangled. There are four basis states for the Hilbert space of a two-electron system if we are only looking at the spin part of the state, three called "triplet" and one called "singlet". None of the ones you wrote down are among them; you forgot that electrons are indistinguishable so the spin part of the two-electron wave function has to be either symmetric (if the position/momentum part is antisymmetric) or antisymmetric (if the position/momentum part is symmetric).
Ah, ok. Yes, with that notation, all of those states are either symmetric or antisymmetric.I'm using the same notation as 4.177 and 4.178. So according to them, |11>, |10> and |1-1> are the triplet, |00> is the singlet state, and the symmetric or antisymmetric requirement is already taken care of, right?
My bad, corrected. Thanks.No, this is one of the triplet states. The singlet-state ket is
$$\frac{1}{\sqrt{2}} (|10 \rangle-|01 \rangle).$$