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When an electron is placed in a magnetic field, its spin precesses. Why doesn't this reveal the spin via angular momentum being transferred into the environment?


Impractical Thought Experiment

Suppose we have an electron with a spin that's either exactly $|\text{left}\rangle$ or exactly $|\text{right}\rangle$, but we don't know which. We apply a magnetic field pointing upward just long enough for the spin to precess by a half turn. In effect, we negated the direction of the electron's spin.

But doesn't just negating the spin violate conservation of angular momentum? If the electron's spin was $|\text{right}\rangle$ then it started with +1/2 angular momentum along the X axis but now it is -1/2. There's a missing +1-along-X bit of angular momentum; presumably it went into the surrounding system?

So, in principle, if we could know the total angular momentum of the relevant-parts-of-system-except-electron before and after precessing by a half turn, we would have determined the spin of the electron. Meaning it was measured, or at least entangled.

But that clearly can't be right, or we wouldn't describe what-an-eletron's-spin-does-when-hit-by-a-magnetic-field as precession. A partially measured/entangled spin doesn't act like a partially rotated spin.

I realize that I must be overlooking a very basic piece of information.

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  • $\begingroup$ IMHO, a typical example for what you are looking for would be spin-spin coupling in atoms and molecular systems. There the interaction between two (or more) spins causes exactly the kind of correlations that you are looking for. The answers below about macroscopic experiments are, of course, spot on. $\endgroup$ – CuriousOne May 6 '16 at 19:09
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You are right that the magnetic field, or rather the apparatus used to generate it, absorbs some angular momentum from the quantum spin. This apparatus includes, for example, some coils of wire containing more than $10^{23}$ electrons, as well as many other macroscopic components. The upshot is that increasing the angular momentum of this apparatus by an amount of order $\hbar$ leaves the apparatus in an almost indistinguishable state. However, in order for this process to constitute a decent measurement, the possible final states of the apparatus (depending on whether the spin was initially $\lvert \mathrm{left}\rangle$ or $\lvert \mathrm{right}\rangle$) must be distinguishable, i.e. almost orthogonal. If you want to think in terms of entanglement, this means that the entanglement generated between the magnetic field and the spin is negligibly small. This is why, so long as we only care about the dynamics of the quantum spin itself, we can reduce all the complexity of the field and the apparatus used to generate it into a single external, classical parameter entering the spin's Hamiltonian.

PS. You could ask essentially the same question, with essentially the same answer, about many different experimental setups, e.g. why doesn't the recoil momentum of a photon bouncing off the mirrors reveal the which-way information in a Mach-Zehnder inteferometer?

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  • $\begingroup$ Could you go into more detail about why the possible final states aren't orthogonal? It would of course be impractical to distinguish these states, but is the trace distance of the underlying kets actually near 0? I thought being able to distinguish even in principle would result in a mixed state instead of a pure state. $\endgroup$ – Craig Gidney May 6 '16 at 17:13
  • $\begingroup$ @CraigGidney I can't really go into more detail in the sense that I think you mean, and I doubt that any physicist could, because that would require solving an outrageously complicated many-body open quantum systems problem. Perhaps you could look at it in the following way. You are right that being able to perfectly distinguish even in principle would lead to a completely mixed state. Thus, the very fact that spins precess in a magnetic field shows you that the possible final states of the magnetic field + apparatus really do have vanishing trace distance. $\endgroup$ – Mark Mitchison May 6 '16 at 17:21
  • $\begingroup$ There's no over-simplified system that exhibits the same effect, or perhaps a reference? I want to understand the actual details of how this can happen, because it's very counter-intuitive to me. $\endgroup$ – Craig Gidney May 6 '16 at 17:26
  • $\begingroup$ @CraigGidney For a simplified (or at least simpler) system, maybe you could look at the work of Mollow on resonance flourescence. Here you have a two-level atom driven by light in a coherent state. For this particular state of the light field, you can show by a unitary transformation that the effect of the quantum field on the atom is equivalent to a classical plane-wave field. The intuition is the same, that the back-action of the atom on the field is essentially negligible (neglecting spontaneous emission into the laser mode) and thus there is no atom-field entanglement. $\endgroup$ – Mark Mitchison May 6 '16 at 17:30
  • $\begingroup$ The original reference is B. R. Mollow, Phys. Rev. A 12, 1919 (1975). $\endgroup$ – Mark Mitchison May 6 '16 at 17:38
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The surrounding system is in a very large very mixed state, and that washes out the distinction in angular momentum.

Not very distinguishable


Trivial Example

Consider the trace distances between the following nearly-maximally-mixed density matrices:

$$\begin{align} D \left( \frac{1}{2} \begin{bmatrix} 1& & \\ &1& \\ & &0 \end{bmatrix}, \frac{1}{2} \begin{bmatrix} 0& & \\ &1& \\ & &1 \end{bmatrix} \right) &= \frac{1}{2} \\ D \left( \frac{1}{3} \begin{bmatrix} 1& & & \\ &1& & \\ & &1& \\ & & &0 \end{bmatrix}, \frac{1}{3} \begin{bmatrix} 0& & & \\ &1& & \\ & &1& \\ & & &1 \end{bmatrix} \right) &= \frac{1}{3} \\ D \left( \frac{1}{4} \begin{bmatrix} 1& & & & \\ &1& & & \\ & &1& & \\ & & &1& \\ & & & &0 \end{bmatrix}, \frac{1}{4} \begin{bmatrix} 0& & & & \\ &1& & & \\ & &1& & \\ & & &1& \\ & & & &1 \end{bmatrix} \right) &= \frac{1}{4} \\ \vdots \\ D \left( \frac{1}{n-1} (I_n - |0\rangle\langle0|), \frac{1}{n-1} (I_n - |n-1\rangle\langle n-1|) \right) &= \frac{1}{n-1} \end{align}$$

Notice that the second matrix is what you would get if you applied a "-1" operation to the state represented by the first matrix, yet the trace distances are getting closer to 0 as our uncertainty over the exact count increases. All of the 'middle' states cancel out, so that only the increasingly-unlikely-as-$n$-increases $|0\rangle\langle0|$ and $|n-1\rangle\langle n-1|$ states contribute to the trace distance.

Since $n$ is extremely large in practice (like... $10^{23}$ large), the trace distance is negligible and might as well be zero. The same effect happens when you have more plausible distributions of the angular momentum. This "hides" any difference caused by the electron, almost-exactly preserving its pure state.


Slightly Less Trivial Example

Suppose that we have a decent sized system-except-electron in a binomial-distributed pure state with $n=10^{24}$ components:

$$\psi_{\text{initial}} = \frac{1}{\sqrt{2^{n}}} \sum_{k=0}^n \sqrt{n \choose k}|k_{\text{apparatus}}\rangle|k_{\text{environment}}\rangle$$

The environment states are not accessible, so we trace them out to get the density matrix of the apparatus:

$$\rho_{\text{initial}} = \frac{1}{2^{n}} \sum_{k=0}^{n} {n \choose k}|k_{\text{apparatus}}\rangle\langle k_{\text{apparatus}}|$$

Now we perform the spin-negation, which either increments or decrements the states:

$$\rho_{\text{decrement}} = \frac{1}{2^{n}} \sum_{k=0}^{n} {n \choose k}|(k-1)_{\text{apparatus}}\rangle\langle (k-1)_{\text{apparatus}}|$$

$$\rho_{\text{increment}} = \frac{1}{2^{n}} \sum_{k=0}^{n} {n \choose k}|(k+1)_{\text{apparatus}}\rangle\langle (k+1)_{\text{apparatus}}|$$

The trace distance between these two states is:

$$\begin{align} d &= \frac{1}{2} \text{Tr} \left[ \text{abs} (\rho_{\text{increment}} - \rho_{\text{decrement}}) \right] \\ &= \frac{1}{2} \text{Tr} \left[ \text{abs}\left( \frac{1}{2^{n}} \sum_{k=0}^{n} {n \choose k}|k-1\rangle\langle k-1| - \frac{1}{2^{n}} \sum_{k=0}^{n} {n \choose k}|k+1\rangle\langle k+1|\right) \right] \\ &= \frac{1}{2} \frac{1}{2^{n}} \text{Tr} \left[ \text{abs}\left( \sum_{k=-1}^{n-1} {n \choose k+1}|k\rangle\langle k| - \sum_{k=+1}^{n+1} {n \choose k-1}|k\rangle\langle k|\right) \right] \\ &= \frac{1}{2^n 2} \text{Tr} \left[ \text{abs}\left( \begin{array}{ll} + & |{-1}\rangle\langle{-1}| \\ + &n |{0}\rangle\langle{0}| \\ - &n |n\rangle\langle n| \\ - &|n+1\rangle\langle n+1| \\ + &\sum_{k=1}^{n-1} |k\rangle\langle k| \left({n \choose k+1} - {n \choose k-1} \right) \end{array} \right) \right] \\ &= \frac{1}{2^n 2} \left[ 1 + n + n + 1 + \sum_{k=1}^{n-1} \left| {n \choose k+1} - {n \choose k-1} \right| \right] \\ &\approx \frac{1}{2^n 2} \left[ 2 + 2n + 2 \sum_{k=1}^{n/2} {n \choose k+1} - {n \choose k-1} \right] \\ &= \frac{1}{2^n} \left[ 1 + n + {n \choose n/2+1} + {n \choose n/2} - {n \choose 1} - {n \choose 0} \right] \\ &= \frac{1}{2^n} \left[ {n \choose n/2+1} + {n \choose n/2} \right] \\ &\approx \frac{2}{2^n} {n \choose n/2} \\ &\approx \frac{2}{2^n} \frac{2^n}{\sqrt{\pi n}} \\ &= \frac{2}{\sqrt{\pi n}} \\ &= \frac{2}{\sqrt{\pi 10^{24}}} \\ &\approx 10^{-12} \end{align} $$

So the trace distance is extremely close to zero, basically meaning that no process can distinguish between the two states at a rate better than chance. Therefore the information leakage from the electron's spin is negligible (after you trace out the inaccessible environment states, which is what unavoidably happens in practice).

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