A How can a spin moment align parallel to an external field?

JanSpintronics

Hello everybody,

Lets say we have an atom with an electron that have a spin. That spin can only pointing in 2 directions, in $s_1=\hbar*1/2$ and $s_1=-\hbar*1/2$ and therefore the magnetic moment of an electron is pointing on the opposite site. My Problem here is appearing when whe apply an magnetic field:
From experimental observation you know that the magnetic moment of a whole probe is going to be parallel to the external field, when its stron enough. Why is there such a thing? the direction of the Moment of a Spin is quantized and cant point in the direction of an external field, that is random applied in the room.
thanks for helping

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DrClaude

Mentor
I'm sorry, but your question is hard to understand.

Is it that you have a problem with when the quantization axis (defining spin up and spin down) is not the same as the direction of the external field?

JanSpintronics

I'm sorry, but your question is hard to understand.

Is it that you have a problem with when the quantization axis (defining spin up and spin down) is not the same as the direction of the external field?
Yeah kinda that, to be more precisely i meant that the magnetic moment of an electron isnt colinear to the direction of the external field. Because i saw a picture where an electron spinvector, or should better say his magnetic moment vector, is rotating around the magnetic field vector. i dont see why the spin vector can be parallel to the magnetic field.

Or should i see that in that way: the x and y components can be zero or whatever and are not quantized, so the vector can point on the direction of a field which is in z-direction only, when both components of x and y are zero?

DrClaude

Mentor
Because of the uncertainty principle, the spin cannot lie exactly along the z axis, because it would then be possible to know its spin along all 3 axes.

Representing it as a classical vector, a spin-1/2 would have a length of $\sqrt{s(s+1)} \hbar = \sqrt{\frac{3}{4}} \hbar$, while its projection on the z axis (in the spin-up case) would be $\hbar/2$. So you see that there is some angle between the axis and the spin vector. This is often represented by drawing a cone around the axis, meaning that the spin is somewhere on that cone (delocalized).

MisterX

The spin projection in any direction won't be as much as the total spin as mentioned by Dr. Claude. There are quantum effects which make this a bit different than a classical object pointing somewhere. A spin pointing up along some axis when measured on some different axis will measure some combination of both up and down. Although the average measured along some axis may be zero spin, you will always get either "up" or "down" on a single experiment regardless of what the initial spin state is and what axis was chosen to measure. You won't get an individual 0 measurement.

The axis a spin points up or down along can point in any direction. There's nothing special about the z axis.
We can use these vectors to represent spin up and spin down along the z axis:
$$\begin{bmatrix}1 \\ 0\end{bmatrix}, \, \begin{bmatrix}0 \\1\end{bmatrix}$$
and these represent spin up and spin down along the x axis:
$$\begin{bmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{bmatrix}, \, \begin{bmatrix}\frac{1}{\sqrt{2}} \\-\frac{1}{\sqrt{2}}\end{bmatrix}$$
and these vectors represent spin up and spin down along an axis in the $\phi,\, \theta$ direction:
$$\begin{bmatrix}\cos\frac{\theta}{2} \\ \sin\frac{\theta}{2} e^{i\phi} \end{bmatrix}, \, \begin{bmatrix}\sin\frac{\theta}{2} \\-\cos\frac{\theta}{2} e^{i\phi}\end{bmatrix}$$

The spin of a charged particle will naturally "relax" so that it's corresponding magnetic moment is in the direction of the applied field. So basically the spin can still point in any direction and be rotated continuously, although there are uniquely quantum properties like uncertainty.

ZapperZ

Staff Emeritus
2018 Award
Yeah kinda that, to be more precisely i meant that the magnetic moment of an electron isnt colinear to the direction of the external field. Because i saw a picture where an electron spinvector, or should better say his magnetic moment vector, is rotating around the magnetic field vector. i dont see why the spin vector can be parallel to the magnetic field.

Or should i see that in that way: the x and y components can be zero or whatever and are not quantized, so the vector can point on the direction of a field which is in z-direction only, when both components of x and y are zero?
I think you are confusing the alignment of the individual spin magnetic moment with the BULK magnetization of the material. These are two different things.

The population of each magnetic spin states depend on not just the external field, but also the temperature of the system. So there is a statistical factor involved in whether a spin magnetic moment is aligned with the external field.

The bulk magnetization is the total sum of ALL the magnetic spin moments of the entire material. In an RF field where the external magnetic field can change, the bulk magnetization can precess around a second fixed magnetic field (such as in a typical pulse NMR). This bulk magnetization is not as trivial as simply looking at the individual spin-state population.

Zz.

DrClaude

Mentor
A spin pointing up along some axis when measured on some different axis will measure some combination of both up and down. Although the average measured along some axis may be zero spin, you will always get either "up" or "down" on a single experiment regardless of what the initial spin state is and what axis was chosen to measure. You won't get an individual 0 measurement.
That first sentence in bold is incorrect and contradicted by the second. To be clear, a system can only be measured to be in an eigenstate of the observable corresponding to that measurement. If the system is spin-up along z, it is in a superposition of spin-up and spin.down along x, be a measurement along x will result in either spin-up along x or spin-down along x.

JanSpintronics

Oookaaaaay that confuses a bit more than help... Which one of you is right ?
Because of the uncertainty principle, the spin cannot lie exactly along the z axis, because it would then be possible to know its spin along all 3 axes.

Representing it as a classical vector, a spin-1/2 would have a length of $\sqrt{s(s+1)} \hbar = \sqrt{\frac{3}{4}} \hbar$, while its projection on the z axis (in the spin-up case) would be $\hbar/2$. So you see that there is some angle between the axis and the spin vector. This is often represented by drawing a cone around the axis, meaning that the spin is somewhere on that cone (delocalized).
Okkay if i get you right, it isnt possible to measure it cause...its just not possible?
Why i can see in every Paper or Books or whatever, that the authors say that the Spins are ALIGNED with the magnetization?

I think you are confusing the alignment of the individual spin magnetic moment with the BULK magnetization of the material. These are two different things.

The population of each magnetic spin states depend on not just the external field, but also the temperature of the system. So there is a statistical factor involved in whether a spin magnetic moment is aligned with the external field.

The bulk magnetization is the total sum of ALL the magnetic spin moments of the entire material. In an RF field where the external magnetic field can change, the bulk magnetization can precess around a second fixed magnetic field (such as in a typical pulse NMR). This bulk magnetization is not as trivial as simply looking at the individual spin-state population.
okaay, you meant its possible or i should say its often the case that you have some up spins to the left direction and some up spins the right direction and so they cancel out and you have a spin pointing just above?
But i just dont know how to get the quantized part of the spins in that picture... Does that quantized mean that this z-vector of the spin can have just special values?

DrClaude

Mentor
Okkay if i get you right, it isnt possible to measure it cause...its just not possible?
Why i can see in every Paper or Books or whatever, that the authors say that the Spins are ALIGNED with the magnetization?
Because that is what is is called when a spin has a maximum projection along the magnetization axis. Quantum is not classical, and the spin is not actually parallel in the classical sense of a vector being parallel to an axis, but is in a quantum state that is as close as possible to that classical situation.

JanSpintronics

Because that is what is is called when a spin has a maximum projection along the magnetization axis.
What do you mean with maximum projection?
So you mean that the z-component have his maximum, which is always aligned?
Quantum is not classical, and the spin is not actually parallel in the classical sense of a vector being parallel to an axis, but is in a quantum state that is as close as possible to that classical situation.
What do you mean with close as possible? Is it for an example you have an up magnetization and a spin of an electron is not aligned with it but nearest alignement it can have is the up spin position $+\hbar/2$ ?

And how does it fit in with the picture of Precission of a Spin moment? For example if you will take a spin in a magnetic field, you will get a precission of that spin which you meant in your 2. post. But this spin precession (the movement of that cone) will relax in some time and the spin will be "aligned" with the magnetization. So it will looks lie a cone with smaler and smaler radius until its zero. But this spin cant be aligned with the magnezation when it comes to this damping. So it just stop to precess or how it can be explain?

DrClaude

Mentor
Consider the following figure:

It is the classical representation of quantum angular momentum. The right panel shows the possible orientations for $l=3$, where you see that the angular momentum is never actually aligned (in the classical sense) with the $z$ axis, but there is a state that is considered to be aligned (in the quantum mechanical sense) with the $z$ axis, the one for which $m=l$, which is illustrated in the left panel. The position of the classical vector in that left panel is not fixed, since its projection is "delocalized" over the entire $xy$ plane.

This is not the same as precession. Precession comes about when the spin lies in the $xy$ plane and can be thought of as rotating in that $xy$ plane. As it relaxes, it starts to point towards the field ($z$ axis), until it ends up completely aligned. Note that this representation of the spin rotating in the $xy$ plane is a classical representation. In reality, it is the expectation values $\langle S_x \rangle$ and $\langle S_y \rangle$ that have a sinusoidal behavior (the spin itself never points exactly along the $x$ or the $y$ axis for the same reason that it never points along $z$).

JanSpintronics

So let me summarize and try to understand it:
When we apply a magnetic field in the z-axis on a spin in the z-axis he will NOT precess around it, because that quantum Alignment means that you have the closest and nearest position to the magnetization vector and thats what it called align with the magnetization. Thats the case when the spin has his maximum quantum position to the magnetization, but he is still delocalized in the that cone in the left pic.

So then we have a spin in the state below the one before, if that was the m=3 state, then we have a state of m=2. (Is that what you meant wih a spin in the x-axis?)
Then we have a precession in the xy plane till it points in the z-axis by a apllied field in the z axis like before.

By the way, you mean with the expectual values that the spin sometimes on the on one side and than around 180 degrees on the other side and the sum of this is the axis?

But just consider the case where m ist zero, so you have the spin in the xy-plane and it would be locatied, right?

And just for make sure: you meant the "complited aligned" in the quantum sense right?