Larmor Precession and Force on a Magnetic Dipole

[Rubiss]

Homework Statement

I have two questions. Both involve derivations from textbooks, not end-of-the-chapter problems. The two textbooks are Quantum Mechanics by Griffiths, and A Modern Approach to Quantum Mechanics by John Townsend.

My first question is about the discussion of Larmor precession found on page 180 of Griffiths. On that page, he calculates
$$\langle S_{x} \rangle = \frac{\hbar}{2}\mbox{sin}(\alpha)\mbox{cos}(\gamma B_{0}t)$$
$$\langle S_{y} \rangle = -\frac{\hbar}{2}\mbox{sin}(\alpha)\mbox{sin}(\gamma B_{0}t)$$
$$\langle S_{z} \rangle = \frac{\hbar}{2}\mbox{cos}(\alpha)$$
and then states, "Evidently
$$\langle \bf{S} \rangle$$ is tilted at a constant angle $$\alpha$$ to the z-axis, and precesses about the field the field at the Larmor frequency $$\omega = \gamma B_{0}$$..."
My question is how is does one know something about the expectation value of S from the expectation values of the components of S? Isn't it correct that
$$\langle {\bf S} \rangle \neq \langle S_{x} \rangle + \langle S_{y} \rangle + \langle S_{z} \rangle$$

Is one supposed to just "see/know" that because the expectation values of the components of the spin have a trigonometric function of alpha, the expectation value of the spin is tilted at an angle alpha to the z-axis? Or, is this understanding similar to transforming/decomposing a vector from spherical coordinates to Cartesian coordinates (looking at figure 4.10 and decomposing S)? If that is the case, then I understand.


My second question involves the force on a magnetic dipole. On page 3 of Townsend, he states that, "If we call the direction in which the inhomogeneous magnetic field is large the z direction, we see that
$$F_{z} = \vec{\mu}\cdot \frac{\partial \vec{B}}{\partial z}\simeq \mu_{z}\frac{\partial B_{z}}{\partial z}"$$

Here's my question. I know that $$\vec{F}=\nabla(\vec{\mu}\cdot\vec{B})$$
How does he pull the partial derivative into the dot product before first applying the dot product? Is it because the magnetic dipole moment is a constant in the z direction?

Homework Equations

$$\vec{F}=\nabla(\vec{\mu}\cdot\vec{B})$$

$$\vec{S}=S_{x}\hat{x}+S_{y}\hat{y}+S_{z}\hat{z}$$

$$\langle S^{2} \rangle = \langle S_{x}^{2} \rangle + \langle S_{y}^{2} \rangle + \langle S_{z}^{2} \rangle$$

The Attempt at a Solution

Refer to the problem description. Thank you for any help.

 

[rude man]

Homework Statement

My second question involves the force on a magnetic dipole. On page 3 of Townsend, he states that, "If we call the direction in which the inhomogeneous magnetic field is large the z direction, we see that
$$F_{z} = \vec{\mu}\cdot \frac{\partial \vec{B}}{\partial z}\simeq \mu_{z}\frac{\partial B_{z}}{\partial z}"$$

Here's my question. I know that $$\vec{F}=\nabla(\vec{\mu}\cdot\vec{B})$$
How does he pull the partial derivative into the dot product before first applying the dot product? Is it because the magnetic dipole moment is a constant in the z direction?

Homework Equations

$$\vec{F}=\nabla(\vec{\mu}\cdot\vec{B})$$

$$\vec{S}=S_{x}\hat{x}+S_{y}\hat{y}+S_{z}\hat{z}$$

$$\langle S^{2} \rangle = \langle S_{x}^{2} \rangle + \langle S_{y}^{2} \rangle + \langle S_{z}^{2} \rangle$$

The Attempt at a Solution

(b):
Yes.
The magnetic dipole moment is not a function of distance. It's really defined by the torque exerted on the dipole by a homogeneous magnetic field B: τ = μ x B. Under a homogeneous (grad(B) = 0) B field there is however no force acting on the dipole.

So given μ = μk, k = unit vector in z direction, and dk/dz = 0, , so μ can indeed be put outside the partial derivative.

(a) EDIT
sorry, I'm deleting what I wrote here previously as being totally irrelevant ...
[/B]

 

[Rubiss]

(b):
Yes.
The magnetic dipole moment is not a function of distance. It's really defined by the torque exerted on the dipole by a homogeneous magnetic field B: τ = μ x B. Under a homogeneous (grad(B) = 0) B field there is however no force acting on the dipole.

So given μ = μk, k = unit vector in z direction, and dk/dz = 0, , so μ can indeed be put outside the partial derivative.

(a) EDIT
sorry, I'm deleting what I wrote here previously as being totally irrelevant ...
[/B]

Or, if the magnetic dipole moment doesn't point entirely in the z direction, we just use the component in the z direction, right? By the way, why do we ignore the other components of the dipole moment? Is it because we assume that all other components of the magnetic field are small compared to the z-component of the magnetic field? Is that what Townsend means when he says, "If we call the direction in which the inhomogeneous magnetic field is large the z direction..."?


Can anyone help with my first question about Larmor precession?

 

[rude man]

Or, if the magnetic dipole moment doesn't point entirely in the z direction, we just use the component in the z direction, right? By the way, why do we ignore the other components of the dipole moment? Is it because we assume that all other components of the magnetic field are small compared to the z-component of the magnetic field? Is that what Townsend means when he says, "If we call the direction in which the inhomogeneous magnetic field is large the z direction..."?


Can anyone help with my first question about Larmor precession?

Yes, that's why we take the gradient of the dot-product of μ and B. The only part of μ that counts is the part directed along B.

We ignore the components of μ along any other direction than B because the formula says to! Ha ha , how's that for a cop-out? Well, I don't want to re-derive the force formula here; I'm sure your textbook can do a better job than I.

I will say that even if the x or y components of μ were 100 times the size of the z component, assuming B is entirely along z, still that would not matter. Trust the ol' dot product! It wipes out all the orthogonal stuff.

What's interesting is that unless there is a gradient of B, there is no force.

Also note for future reference that the direction of a magnetic dipole points from S to N, unlike a B field which points from N to S. Don't ask me why, but you have to adhere to this convention or the signs come up wrong.

 

[alemsalem]

why isn't <S> the (vector) sum of its three components? it's a linear operator.

 

[Rubiss]

why isn't <S> the (vector) sum of its three components? it's a linear operator.

The vector S is the sum of its three components, but I don't think the expectation value of the vector S is the expectation value of its three components. That is, I don't think we can take
$$\vec{S}=S_{x}\hat{x}+S_{y}\hat{y}+S_{z}\hat{z}$$
and then take the expectation value of both sides to obtain
$$\langle \vec{S} \rangle \neq \langle S_{x}\hat{x}+S_{y}\hat{y}+S_{z}\hat{z} \rangle \neq \langle S_{x} \rangle + \langle S_{y} \rangle + \langle S_{z} \rangle$$
I thought that first we obtain
$$S^{2} = S_{x}^{2} + S_{y}^{2} + S_{z}^{2}$$
and then take the expectation value of both sides, which we can do because we are now dealing with scalars instead of vectors:
$$\langle S^{2} \rangle = \langle S_{x}^{2} \rangle + \langle S_{y}^{2} \rangle + \langle S_{z}^{2} \rangle$$

But I could be wrong... Can someone confirm this? Right or wrong, can someone help me with my first question about what Griffiths means concerning Larmor precession? Thanks again for everyone's help.

 

[alemsalem]

The vector S is the sum of its three components, but I don't think the expectation value of the vector S is the expectation value of its three components. That is, I don't think we can take
$$\vec{S}=S_{x}\hat{x}+S_{y}\hat{y}+S_{z}\hat{z}$$
and then take the expectation value of both sides to obtain
$$\langle \vec{S} \rangle \neq \langle S_{x}\hat{x}+S_{y}\hat{y}+S_{z}\hat{z} \rangle \neq \langle S_{x} \rangle + \langle S_{y} \rangle + \langle S_{z} \rangle$$

why not? isn't that what you do with the position and momentum vector?
I think what he meant (as you said) is that it looks like a transformation to spherical coordinates (h/2, α, Bγt).