Spin confusion: Stern-Gerlach experiment

[WWCY]

I have some serious issues trying to understand the idea of the spin in the context of the Stern-Gerlach experiment and would appreciate some assistance!

Assuming that a homogenous magnetic field $B$ in the "North-wards" $z$-direction, assume that there is a magnetic dipole moment $\mu$ pointing $3\pi /4$ away from the "North" (i.e. South-East).

My text states that there will be a torque that comes in the form of
$$\vec{\tau} = \vec{\mu} \times \vec{B}$$
which aligns the moment to the B-field, according to the right-hand rule.

It later states: "But the intrinsic angular momentum, the spin, of the atom is proportional to $\vec{\mu}$, and the rate change of angular momentum is just the torque, we have
$$\frac{d \vec{\mu}}{dt} \propto \vec{\mu} \times \vec{B}$$
which is to say the dipole moment precesses about $\vec{B}$"

It then goes on to say that the potential energy $E = - \vec{\mu} \cdot \vec{B}$ doesn't change, which means that the angle between $\vec{\mu}$ and $\vec{B}$ doesn't as well.

My questions are:

1) Applying the right-hand rule to $\frac{d \vec{\mu}}{dt}$ (for this example), I get a vector pointing out of this page, is this the direction in which the magnetic moment precesses?

2) It was stated that a magnetic moment in a $\vec{B}$ field experiences a torque aligning the moment to the magnetic field, yet $E = - \vec{\mu} \cdot \vec{B}$ was stated to be constant. Is this statement not contadictory? One states that the angle between $\mu$ and $B$ changes, while the other states that they stay constant.

The text later states that the magnetic moments are only measured to be either parallel or antiparallel to a given magnetic field. Which seems to throw the idea of precession out of the window and further adds to my confusion.

I must note that this is my first encounter with spins, so explanations that aren't too technical are greatly appreciated!

 

[DrClaude]

1) Applying the right-hand rule to $\frac{d \vec{\mu}}{dt}$ (for this example), I get a vector pointing out of this page, is this the direction in which the magnetic moment precesses?

The magnetic moment precesses around $\vec{B}$.

2) It was stated that a magnetic moment in a $\vec{B}$ field experiences a torque aligning the moment to the magnetic field, yet $E = - \vec{\mu} \cdot \vec{B}$ was stated to be constant. Is this statement not contadictory? One states that the angle between $\mu$ and $B$ changes, while the other states that they stay constant.

The first statement is classical, the second quantum mechanical. In QM, if the field is static, then the Hamiltonian is independent of time, so energy is conserved, as is the projection along $\vec{B}$.

The text later states that the magnetic moments are only measured to be either parallel or antiparallel to a given magnetic field. Which seems to throw the idea of precession out of the window and further adds to my confusion.

As I stated above, the precession is around $\vec{B}$, so it takes place in the plane perpendicular to $\vec{B}$. If you take the field to be aligned along z, that means that the probability of measuring spin-up or spin-down in z is constant, but not the probability of measuring spin-up or spin-down with respect to the x or y axes.

 

[WWCY]

Thank you for your response, however I'm afraid I don't quite follow,

a) How does this statement

But the intrinsic angular momentum, the spin, of the atom is proportional to $\vec{\mu}$, and the rate change of angular momentum is just the torque, we have

lead to this

$\frac{d \vec{\mu}}{dt} \propto \vec{\mu} \times \vec{B}$

and how does this expression lead to a conclusion about the precession of $\vec{\mu}$?

b)

As I stated above, the precession is around $\vec{B}$, so it takes place in the plane perpendicular to $\vec{B}$. If you take the field to be aligned along z, that means that the probability of measuring spin-up or spin-down in z is constant, but not the probability of measuring spin-up or spin-down with respect to the x or y axes.

But was it not the case that the magnetic moments were found to be either parallel or anti-parallel (and no in-betweens)? Does this not mean that there wasn't any precession, since this would mean that the atoms should end up spread out on the screen, since they would experience a variety of magnitudes of force?

c)

The first statement is classical, the second quantum mechanical. In QM, if the field is static, then the Hamiltonian is independent of time, so energy is conserved, as is the projection along $\vec{B}$.

My text seems to imply that the conserved projection on $\vec{B}$ was a classical statement, since it predicts that atoms land on the screen in all sorts of directions. It states "Rather, we have to accept that the classical prediction of a spread-out beam is inconsistent with experimental observation". What am I missing?

d) I'm getting the impression that if $\vec{\mu}$ points up, then the atom is also spin-up. However after trawling the internet for some equation relating the two, I find
$$\vec{\mu _s} = -\frac{e}{2m_e} g \ \vec{S}$$
which seems to suggest that a magnetic moment pointing up is actually spin-down. Again, I can't seem make sense of this.Attached below are the pages of the text I've been referencing. Also, apologies if my line of questioning is confused as I am too. Thank you for your patience.

 

[blue_leaf77]

how does this expression lead to a conclusion about the precession of ⃗μ\vec{\mu}?

The rate of change of angular momentum is equal to the torque, that's what you get in classical mechanics. It's just the rotational analog of the rate of change of momentum being equal to force.
Assume your homogenous magnetic field points in positive z direction, $\mathbf B = B\mathbf k$ and the dipole moment $\mathbf \mu = \mu_x \mathbf i + \mu_y \mathbf j + \mu_z \mathbf k$, then
$$
\mathbf \mu \times \mathbf B = \mathbf i B \mu_y - \mathbf j B \mu_x
$$
Putting this into the equation of motion of the dipole
$$
\frac{d\mathbf \mu}{dt} = C(\mathbf \mu \times \mathbf B) \\
\dot{\mu}_x \mathbf i + \dot{\mu}_y \mathbf j + \dot{\mu}_z \mathbf k = \mathbf i C \mu_y - \mathbf j C \mu_x
$$
where $C$ in the second line is different from the one in the first line, the latter is proportional in $B$.
Now you have a set of differential equations to solve
$$
\dot\mu_z = 0 \\
\dot\mu_x = C^2\mu_y \\
\dot\mu_y = -C^2\mu_x \\$$
Solving this system of equations (e.g. by differentiating the second equation w.r.t. time once and use the third equation to eliminate $\dot\mu_y$) you get the time dependent dipole moment
$$
\mu_x(t) = \mu_y(0) \sin(\omega t) + \mu_x(0) \cos(\omega t) \\
\mu_y(t) = \mu_y(0) \cos(\omega t) - \mu_y(0) \sin(\omega t) \\
\mu_z(t) =\mu_z(0)
$$
with $\omega$ proportional to $B$. The z component is constant over time but the x and y components undergo a circular motion, this is by definition called precession.
Up to now, the derivation is completely classical. If you resort to QM by considering a region of constant magnetic field in the z direction and calculating the time-dependent expectation value of spin components you will get a very similar expressions.
In a constant magnetic field $B$, the Hamiltonian of a spin-only system is $H = CBS_z$ which results in the time-evolution operator of the form $U = \exp(-iCBS_zt)$ (with $C$ another constant). The expectation values of spin components turns out to be given by
$$
\langle S_x(t) \rangle = \langle U^\dagger S_x U\rangle = \langle S_x(0) \rangle \cos(\omega t) - \langle S_y(0) \rangle \sin(\omega t) \\
\langle S_x(t) \rangle = \langle U^\dagger S_y U\rangle = \langle S_y(0) \rangle \cos(\omega t) + \langle S_x(0) \rangle \sin(\omega t) \\
\langle S_z(t) \rangle = \langle U^\dagger S_z U\rangle = \langle S_z(0)
$$
Which closely resembles the classical analogue. As a trivia it might be a good exercise to derive $U^\dagger S_i U$ ($i=x,y,z$).

But was it not the case that the magnetic moments were found to be either parallel or anti-parallel (and no in-betweens)? Does this not mean that there wasn't any precession

In the above scheme if you measure the z component of spin you will indeed observe two spots on the screen aligned in z direction but recording the x and y components is not easy. In typical experiment they send the atoms in a continuous beam (i.e. not pulsed or bunched), therefore what is recorded in the screen is an accumulation of many measurements from different times obscuring the oscillating movement of the spot in x or y direction.

 

[WWCY]

The rate of change of angular momentum is equal to the torque, that's what you get in classical mechanics. It's just the rotational analog of the rate of change of momentum being equal to force.
Assume your homogenous magnetic field points in positive z direction, $\mathbf B = B\mathbf k$ and the dipole moment $\mathbf \mu = \mu_x \mathbf i + \mu_y \mathbf j + \mu_z \mathbf k$, then
$$
\mathbf \mu \times \mathbf B = \mathbf i B \mu_y - \mathbf j B \mu_x
$$
Putting this into the equation of motion of the dipole
$$
\frac{d\mathbf \mu}{dt} = C(\mathbf \mu \times \mathbf B) \\
\dot{\mu}_x \mathbf i + \dot{\mu}_y \mathbf j + \dot{\mu}_z \mathbf k = \mathbf i C \mu_y - \mathbf j C \mu_x
$$
where $C$ in the second line is different from the one in the first line, the former is proportional in $B$.
Now you have a set of differential equations to solve
$$
\dot\mu_z = 0 \\
\dot\mu_x = C^2\mu_y \\
\dot\mu_y = -C^2\mu_x \\$$
Solving this system of equations (e.g. by differentiating the second equation w.r.t. time once and use the third equation to eliminate $\dot\mu_y$) you get the time dependent dipole moment
$$
\mu_x(t) = \mu_y(0) \sin(\omega t) + \mu_x(0) \cos(\omega t) \\
\mu_y(t) = \mu_y(0) \cos(\omega t) - \mu_y(0) \sin(\omega t) \\
\mu_z(t) =\mu_z(0)
$$
with $\omega$ proportional to $B$. The z component is constant over time but the x and y components undergo a circular motion, this is by definition called precession.
Up to now, the derivation is completely classical. If you resort to QM by considering a region of constant magnetic field in the z direction and calculating the time-dependent expectation value of spin components you will get a very similar expressions.
In a constant magnetic field $B$, the Hamiltonian of a spin-only system is $H = CBS_z$ which results in the time-evolution operator of the form $U = \exp(-iCBS_zt)$ (with $C$ another constant). The expectation values of spin components turns out to be given by
$$
\langle S_x(t) \rangle = \langle U^\dagger S_x U\rangle = \langle S_x(0) \rangle \cos(\omega t) - \langle S_y(0) \rangle \sin(\omega t) \\
\langle S_x(t) \rangle = \langle U^\dagger S_y U\rangle = \langle S_y(0) \rangle \cos(\omega t) + \langle S_x(0) \rangle \sin(\omega t) \\
\langle S_z(t) \rangle = \langle U^\dagger S_z U\rangle = \langle S_z(0)
$$
Which closely resembles the classical analogue. As a trivia it might be a good exercise to derive $U^\dagger S_i U$ ($i=x,y,z$).

In the above scheme if you measure the z component of spin you will indeed observe two spots on the screen aligned in z direction but recording the x and y components is not easy. In typical experiment they send the atoms in a continuous beam (i.e. not pulsed or bunched), therefore what is recorded in the screen is an accumulation of many measurements from different times obscuring the oscillating movement of the spot in x or y direction.

The equations of motion helped a great deal, thank you for your time!

 

[blue_leaf77]

The equations of motion helped a great deal, thank you for your time!

You're welcome.