Wave function orthogonal components

[Alfred Cann]

1. The photon wave function, an EM wave, has orthogonal electric and magnetic components. I have gathered the impression that the electron wave function has only one. Is this correct?
2. By analogy with EM waves, can the electron's spin rate be identified with the frequency of its wave function?
3. Similarly, can the + and - 1/2 spins be identified with right and left hand circular polarizations of the wave?

 

[Chopin]

1. The electric and magnetic "components" are really sort of a misnomer...they're not actually independent components, they're both effects of the single underlying EM field. We only refer to them as two different fields because their effects were discovered in different ways, and act differently in day-to-day experiences. If you perform Lorentz boosts (the things that take you from one relativistic frame of reference to another), electric fields turn into magnetic fields and vice versa. So in modern physics you usually just think in terms of the single EM field, and extract out the electrical and magnetic effects at the end based on whatever frame of reference you're working in. In any case, there's no analogous effect with spin 1/2 fields like the electron.
2. The spin of an electron is not a real physical spin--nothing is rotating around. So there isn't a rotational frequency or a wavelength based on the spin. Electrons do however, like all particles, have a de Broglie wavelength, which is the wavelength of their wavefunction. Is that what you were thinking of?
3. Yes, or the vertical/horizontal polarizations if you want to work in that basis instead. In both cases they're the different "kinds" of particles you can have, or "configurations", or whatever word you want to use.

 

[WannabeNewton]

1. The photon wave function, an EM wave, has orthogonal electric and magnetic components. I have gathered the impression that the electron wave function has only one. Is this correct?

You're comparing apples to oranges. You should be careful in bringing quantum mechanical or even semi-classical concepts into a purely classical framework (in this case classical EM). A vacuum solution to Maxwell's equations (which reduce to two decoupled wave equations-one for $\vec{E}$ and one for $\vec{B}$) is simply a Fourier transform (i.e. a continuous superposition of plane waves). A single plane wave is described by a 4-vector (the 4-potential) $A_{\mu} = \epsilon_{\mu}e^{ik_{\nu}x^{\nu}}$ where $\epsilon_{\mu}$ is called the polarization vector and $k_{\mu}$ is the usual wave 4-vector.

From it we can get $\vec{E}$ and $\vec{B}$ in the usual way after choosing an inertial frame and splitting space-time into space and time relative to said frame. All the familiar properties of vacuum EM plane waves such as transverse polarization and orthogonality of $\vec{E}$ and $\vec{B}$ can be derived by plugging in the plane wave solutions into Maxwell's equations and using the radiation gauge; the point is that all of this comes from a 4-vector $A_{\mu}$

So you can't compare $A_{\mu}$ to a state vector $| \psi \rangle$ describing the state of an electron as the two are completely different objects. Furthermore the concept of a "photon wave function" is contentious. If you try to quantize $A_{\mu}$ by promoting it to a dynamical state vector $|A_{\mu} \rangle$ then you will end up getting negative energy eigenstates which is obviously problematic. Instead we promote $A_{\mu}$ to an operator field whose coefficients are creation and annihilation operators that act on vacuum to create photon states. Yet again $A_{\mu}$ in this QED context is quite a different object from $|\psi \rangle$ in regular QM.

2. By analogy with EM waves, can the electron's spin rate be identified with the frequency of its wave function?

Angular frequency is related to energy.

3. Similarly, can the + and - 1/2 spins be identified with right and left hand circular polarizations of the wave?

Don't confuse circular polarization states of an EM wave with spin! There is a loose analogy between the two that Sakurai discusses in section 1.1 of his QM text but you have to keep in mind that this is but an analogy. We do deal with left-handed and right-handed spinors but again don't take the analogy between spin and polarization states literally; for one thing in a circularly polarized EM wave $A_{\mu}$ has a rotate rate sensitive to angular frequency whereas spin is insensitive to angular frequency. Anyways, see here: http://farside.ph.utexas.edu/teaching/qm/lectures/node45.html

 

[Alfred Cann]

Thanks, Wannabe,
While I'm digesting that, here's another question.
An electron has a magnetic dipole moment associated with its spin. Suppose we have an electron in a 50/50 superposition of spin 1/2 and spin -1/2. Does it have zero magnetic moment?

 

[WannabeNewton]

Spin and magnetic moment are both operators so it doesn't make sense to say the operators themselves vanish. What we can talk about in a meaningful manner however are the averages of these operators.

For example, a general spin state can be written in terms of the $S_{z}$ spin eigenstates $|\uparrow \rangle , |\downarrow \rangle$ as $|\psi \rangle = c_1 |\uparrow \rangle + c_2|\downarrow \rangle$ where $S_z$ is the $z$ component of the spin operator relative to our chosen coordinate system and $|c_1|^2 + |c_2|^2 = 1$ due to the normalization constraint. If we imagine having infinite copies of the given system all prepared in the state $|\psi \rangle$ above and make measurements of $S_z$ for each copy then we can eventually write down the average or mean value associated with this state: $\langle S_z \rangle = \langle \psi | S_z | \psi \rangle = \frac{1}{2}\hbar(|c_1|^2 - |c_2|^2)$.

So yes if you take $c_1 = c_2 = \frac{1}{\sqrt{2}}$ (this choice is unique up to phase) you will get $\langle S_z \rangle = 0$ which is of course not surprising in the least bit. For an electron the intrinsic magnetic moment is proportional to the spin so the result carries over.

 

[Chopin]

So yes if you take $c_1 = c_2 = \frac{1}{\sqrt{2}}$ (this choice is unique up to phase) you will get $\langle S_z \rangle = 0$ which is of course not surprising in the least bit. For an electron the intrinsic magnetic moment is proportional to the spin so the result carries over.

It's important to note, though, that this is just the spin along the Z axis. Spin is a vector operator, so if the expectation value is 0 along one axis, then it will be non-zero along a different axis. More rigorously, an X +1/2 eigenstate is a superposition of Z +1/2 and Z -1/2 eigenstates with appropriate phases, as are X -1/2, Y +1/2 and Y -1/2. So if you have a superposition of Z +1/2 and Z -1/2 states, you're going to land in some combination of X and Y states, based on the phases of the Z states. Classically, you can think about this as meaning that the magnetic moment of that electron is just pointing in some direction perpendicular to the Z axis.

 

[Alfred Cann]

Dear Chopin and Wannabe,
It's clear I need a course in QM before I can understand you guys. I started Leonard Susskind's Stanford video lectures. Do you prefer something else?

 

[MisterX]

Dear Chopin and Wannabe,
It's clear I need a course in QM before I can understand you guys. I started Leonard Susskind's Stanford video lectures. Do you prefer something else?

I think this would be preferable:
http://www.physics.ox.ac.uk/users/Cruickshank/
accompanying book