Can electron spin change?

  • #1

Main Question or Discussion Point

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

Answers and Replies

  • #2
2,257
7
the direction of its spin will precess around an externally applied magnetic field. also it can change from diamagnetic to paramagnetic.
 
  • #3
1,744
48
Can electron spin change? If so how does this happen?
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.
 
  • #4
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
23,839
6,281
It is correct that the orientation of the electron spin can be changed.

It is incorrect to describe a single electron as paramagnetic or diamagnetic. These are properties of bulk materials, not individual electrons.
 
  • #5
2,257
7
well yes. I was speaking loosely. but the electron will either align with or against the applied magnetic field. this is analogous to para and diamagnetism.
 
  • #6
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
23,839
6,281
Grampa, for heaven's sake, please turn your fount of misinformation down a notch.

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-.
 
  • #7
4,662
5
The electron cannot change the magnitude of its spin or its magnetic moment. In the hydrogen atom, it is in the field of the proton spin (much weaker). There are only two alignments permitted; same direction, and opposite direction, They differ by 1420 MHz (21 cm). This is perhaps the most dominant microwave emission (and absorption) line in the universe.
 
  • #8
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.
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.
 
  • #9
1,744
48
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.
That's right.
 
  • #10
483
2
Grampa, for heaven's sake, please turn your fount of misinformation down a notch.

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-.
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.

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.
 
Last edited:
  • #11
483
2
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.
Great answer and insight.

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
 
  • #12
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
23,839
6,281
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.
Isn't it [itex]\sqrt{3}\hbar/2[/itex]? It's actually in an eigenstate of S2, right?
 
  • #13
Matterwave
Science Advisor
Gold Member
3,965
326
No, the electron's spin is [tex]\frac{\hbar}{2}[/tex]

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.
 
  • #14
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
23,839
6,281
Why isn't the eigenvalue [itex]\sqrt{S(S+1)}[/itex]?
 
  • #15
jtbell
Mentor
15,500
3,296
No, the electron's spin is [itex]\frac{\hbar}{2}[/itex]
No, the "z-component" (actually the component along any direction) of the spin angular momentum vector has that value (either + or -).

The magnitude of the spin angular momentum vector is a fixed [itex]\sqrt{3} \hbar / 2[/itex].

http://hyperphysics.phy-astr.gsu.edu/Hbase/spin.html
 
  • #16
Matterwave
Science Advisor
Gold Member
3,965
326
Ah, you are indeed correct. I was wrong. Sorry.

Indeed, I had forgotten that the spin can never point directly in the +/- z direction.
 
  • #17
2,257
7
so what are the allowed z-component values for a delta baryon? (spin 3/2)
 
  • #18
Matterwave
Science Advisor
Gold Member
3,965
326
3/2, 1/2, -1/2, -3/2
 
  • #19
2,257
7
thank you. :-)
 
  • #20
jtbell
Mentor
15,500
3,296
3/2, 1/2, -1/2, -3/2
Don't forget to multiply by [itex]\hbar[/itex] if you're talking about the physical quantity (angular momentum) and not the quantum number. :smile:
 
  • #21
Matterwave
Science Advisor
Gold Member
3,965
326
I think the hbar is assumed. I mean, numbers don't have units...so...but if you want to be pedantic alright.

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

Related Threads for: Can electron spin change?

Replies
2
Views
802
Replies
19
Views
3K
Replies
6
Views
576
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
19
Views
931
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
11
Views
702
  • Last Post
Replies
4
Views
600
Top