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Can electron spin change? If so how does this happen?

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Can electron spin change? If so how does this happen?

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When people talk about spin then may mean two different things. One is the absolute value of spin (the length of the vector). For the electron this value is [tex] \hbar/2[/tex], and it never changes, i.e., this is a fixed property of the electron, like its mass or charge.Can electron spin change? If so how does this happen?

Another thing is spin projection on a given axis (a vector component). This projection may be either [tex] +\hbar/2[/tex] or [tex] -\hbar/2[/tex], with probability weight assigned to each value. These probabilities may change in electron interactions, collisions, etc.

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Vanadium 50

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It is incorrect to describe a single electron as paramagnetic or diamagnetic. These are properties of bulk materials, not individual electrons.

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Vanadium 50

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First, as I said before, it is incorrect to describe a single electron as paramagnetic or diamagnetic. These are properties of bulk materials, not individual electrons. Second, the only person discussing applied magnetic fields is you. Third, the terms describing the orientation of spins with respect to external fields is not para- and dia-, but rather para- and ortho-.

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So the magnitude of the vector doesn't change but it's components do in a conservative way so as to keep a constant spin magnitude of h/4pi.When people talk about spin then may mean two different things. One is the absolute value of spin (the length of the vector). For the electron this value is [tex] \hbar/2[/tex], and it never changes, i.e., this is a fixed property of the electron, like its mass or charge.

Another thing is spin projection on a given axis (a vector component). This projection may be either [tex] +\hbar/2[/tex] or [tex] -\hbar/2[/tex], with probability weight assigned to each value. These probabilities may change in electron interactions, collisions, etc.

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That's right.So the magnitude of the vector doesn't change but it's components do in a conservative way so as to keep a constant spin magnitude of h/4pi.

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Pure physicists may not be aware of it, but the only real practical manifestation of spins is exposed by applying some equivalent of an external magnetic field to the device.

First, as I said before, it is incorrect to describe a single electron as paramagnetic or diamagnetic. These are properties of bulk materials, not individual electrons. Second, the only person discussing applied magnetic fields is you. Third, the terms describing the orientation of spins with respect to external fields is not para- and dia-, but rather para- and ortho-.

That basically takes spin (and all the entailing theoretical discussion) out of the Hilbert space and shows that it's real and it could be used.

The fact that Grampa's referring to - precession of spin - is the basis of the first proposed spinFET in 1989 by Datta and Das.

In fact, spin and magnetic fields are so entangled that the entire field of Spintronics (crowned by its first Nobel prize in 2007) is founded upon those two.

I'd go easy with Grampa if you haven't read a sentence involving spins and applied magnetic fields. Because that's your fault.

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Great answer and insight.When people talk about spin then may mean two different things. One is the absolute value of spin (the length of the vector). For the electron this value is [tex] \hbar/2[/tex], and it never changes, i.e., this is a fixed property of the electron, like its mass or charge.

Another thing is spin projection on a given axis (a vector component). This projection may be either [tex] +\hbar/2[/tex] or [tex] -\hbar/2[/tex], with probability weight assigned to each value. These probabilities may change in electron interactions, collisions, etc.

This must be a sticky post to every spin question. first understand what you are talking about:

i) Is it the spin vector (projection)?

ii) Or is it simply the magnitude of that?

The first one will be important when magnetization and EXTERNAL magnetic fields are present

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Vanadium 50

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Isn't it [itex]\sqrt{3}\hbar/2[/itex]? It's actually in an eigenstate of SOne is the absolute value of spin (the length of the vector). For the electron this value is [tex] \hbar/2[/tex], and it never changes, i.e., this is a fixed property of the electron, like its mass or charge.

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Matterwave

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It is in an eigenstate of [tex]S^2[/tex] with eigenvalue [tex]\frac{\hbar^2}{4}[/tex]. Take the square root of that and you get the correct answer.

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Vanadium 50

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Why isn't the eigenvalue [itex]\sqrt{S(S+1)}[/itex]?

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jtbell

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No, the "z-component" (actually the component along any direction) of the spin angular momentum vector has that value (either + or -).No, the electron's spin is [itex]\frac{\hbar}{2}[/itex]

The

http://hyperphysics.phy-astr.gsu.edu/Hbase/spin.html

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Matterwave

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Indeed, I had forgotten that the spin can never point directly in the +/- z direction.

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so what are the allowed z-component values for a delta baryon? (spin 3/2)

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Matterwave

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3/2, 1/2, -1/2, -3/2

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thank you. :-)

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jtbell

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Don't forget to multiply by [itex]\hbar[/itex] if you're talking about the physical quantity (angular momentum) and not the quantum number.3/2, 1/2, -1/2, -3/2

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Matterwave

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[tex]\frac{3\hbar}{2}, \frac{\hbar}{2}, \frac{-\hbar}{2}, \frac{-3\hbar}{2}[/tex]

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