A spin is created in a superposition of up and down states. A magnet is moved very slowly, towards the spin. What is the work done by the magnet. It may be helpful to imagine that the magnet is connected to a spring that expands or contracts depending on work done by the magnet.
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The work is of course equal to the potential energy of the dipole $\vec m$ in the magnetic field, $$ U = -\vec m \cdot \vec B.$$ Now, your wording indicates that the magnetic field $\vec B$ is macroscopic so it may be treated as a classical parameter. However, $\vec m$ from a spinning particle is quantum mechanical. For example, for an electron, $$ \vec m = -g_S \mu_B \frac{\vec S}{\hbar} $$ where the spin $g$-factor is $g_S\sim 2.0023$, $\mu_B=e\hbar/(2m_e)$ is the Bohr magneton, and $\vec S$ is the operator (or triplet of operators) of the electron's angular momentum – that act as $\hbar/2$ times the Pauli matrices on the spin-up and spin-down states. So up to a multiplicative constant $\gamma$, $$ U = \gamma \vec \sigma\cdot \vec B .$$ The work done on the electron is a $q$-number, an operator, which acts on the spin-up and spin-down states. For the up and down states relatively to the direction of the external magnetic field $\vec B$, the work done has a sharply defined value, an eigenvalue of $U$. For general linear superpositions, the work done on the electron has a certain probability amplitude to be positive and a certain probability amplitude to be negative. The squared absolute values of these probability amplitudes determine the probabilities that the work will be seen to be one (positive) value or the other (negative) value, just like everywhere in quantum mechanics. It's however very interesting to consider superpositions of states in this context, especially for $j=1/2$. For spin-1/2 particles, each superposition of up and down states is equivalent to "up" with respect to a certain axis in space. If we deal with general superpositions, this axis on which the spin is "up" will generally not be aligned with the direction of the magnetic field. If that's the case, the presence of the magnetic field will have the effect of causing "precession" of the axis defining the spin up. The axis at which the particle is polarized "up" will be "turning around" the direction of $\vec B$ like a ninny. This precession results from the time-dependent change of the phases of the wave function – which is opposite for the up and down states, so the relative phase is changing as well – and the fact that the relative phases of the wave functions always matter (for something) in quantum mechanics. The classic experiment measuring this spin-dependent magnetic work and exhibiting lots of the quantum properties is called the Stern-Gerlach experiment. With some probabilities, the particle behaves in one way or another. |
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