Connection between Pauli XYZ and spatial XYZ

nomadreid · Feb 7, 2015

If I understand correctly (no guarantee), the angles A and B in the generator of the three Pauli matrices (excluding the identity):

cos A...................exp(-iB) sin A
exp(iB) sin A.................-cos A
refer to angles in a Hilbert space, for example the angles of the Bloch sphere. However, are they at the same time the angles between the axes in which a magnetic field would be positioned in order to measure spin? Otherwise put, what is the connection between the Pauli matrices and spatial coordinates?

stevendaryl · Feb 8, 2015

nomadreid said: ↑

If I understand correctly (no guarantee), the angles A and B in the generator of the three Pauli matrices (excluding the identity):

cos A...................exp(-iB) sin A
exp(iB) sin A.................-cos A
refer to angles in a Hilbert space, for example the angles of the Bloch sphere. However, are they at the same time the angles between the axes in which a magnetic field would be positioned in order to measure spin? Otherwise put, what is the connection between the Pauli matrices and spatial coordinates?

Yes, they are definitely related. Let [itex]\hat{m}[/itex] be a unit vector with components [itex]m_x = sin(A) cos(B), m_y = sin(A) sin(B), m_z = cos(A)[/itex]. Then your matrix can be written as [itex]M = \hat{m} \cdot \vec{\sigma}[/itex]. That matrix represents rotation about the axis [itex]\vec{m}[/itex] in the following sense:

Let [itex]\chi[/itex] be the spinor representing a particle that has spin-up in the direction [itex]\vec{r}[/itex]. Then [itex]e^{i \frac{\epsilon}{2} M} \chi[/itex] is the spinor representing a particle that has spin-up in the direction [itex]\vec{r}'[/itex], where [itex]\vec{r}'[/itex] is obtained from [itex]\vec{r}[/itex] by rotating it through an angle [itex]\epsilon[/itex] about the axis [itex]\hat{m}[/itex]

A concrete example is to consider rotations about the y-axis. In that case, [itex]e^{i \frac{\epsilon}{2} M}[/itex] is the matrix [itex]\left( \begin{array}\\ cos(\epsilon/2) & sin(\epsilon/2) \\ -sin(\epsilon/2) & cos(\epsilon/2) \end{array} \right)[/itex]. If we let [itex]\chi[/itex] be spin-up in the z-direction, then it is represented by [itex]\chi = \left( \begin{array}\\ 1 \\ 0 \end{array} \right)[/itex]. Then [itex]e^{i \frac{\epsilon}{2} M} \chi = \left( \begin{array}\\ cos(\epsilon/2) \\ -sin(\epsilon/2) \end{array} \right)[/itex]. That represents spin-up in the direction that is in the z-x plane and that makes an angle of [itex]\epsilon[/itex] with the z-axis.

The factors of 2 in various places seem strange, but that's the origin of the weird fact that rotation about 360 degrees (or [itex]2 \pi[/itex] radians) doesn't return you to the same state, it returns you to the negative of the same state:[itex]\left( \begin{array}\\ cos((2 \pi)/2) \\ -sin((2 \pi)/2) \end{array} \right) = \left( \begin{array}\\ -1\\0\end{array} \right)[/itex].

nomadreid · Feb 9, 2015

Thanks very much, stevendaryl. Let me make sure I understand which angles are referring to those in physical space, and which ones in Hilbert space. The epsilon refers to the angle in physical space, and m_x, m_y, m_z and r^→ are referring to vectors in physical space, whereas χ is referring to a representation in Hilbert space.... is that correct?
If that is correct, this means that the rotation you are referring to in the last paragraph is also physical. As you say, then, that the 2π (physical) rotation gives you the negative of the state. But since the same direction can be considered at 0±2πn from itself, then unless the spinor has some kind of "memory", this would indicate that the two states are equivalent. Am I on the wrong track here?

stevendaryl · Feb 9, 2015

nomadreid said: ↑

Thanks very much, stevendaryl. Let me make sure I understand which angles are referring to those in physical space, and which ones in Hilbert space. The epsilon refers to the angle in physical space, and m_x, m_y, m_z and r^→ are referring to vectors in physical space, whereas χ is referring to a representation in Hilbert space.... is that correct?
If that is correct, this means that the rotation you are referring to in the last paragraph is also physical. As you say, then, that the 2π (physical) rotation gives you the negative of the state. But since the same direction can be considered at 0±2πn from itself, then unless the spinor has some kind of "memory", this would indicate that the two states are equivalent. Am I on the wrong track here?

Yes, [itex]m_x, m_y, m_z[/itex] is a physical vector. And yes, multiplying a quantum state by a real (or complex) number leaves it the same physical state.

nomadreid · Feb 9, 2015

Thanks again, stevendaryl. The ideas are starting to clear up. While I am on the subject, you mentioned that multiplying a state by a complex number leaves it in the same physical state. I have always been puzzled by this: as I understand it when two states are simultaneously multiplied by the same scalar, then the inner product of |e^iθ u> with |e^iθ v> = <u*e^-iθ |e^iθ v> = <u*|v>, but when only one of them is, say v, then <u*|e^iθ v> = e^iθ<u*| v>, and since <u*| v> is itself a complex number, then since I can choose θ to get any (unit) complex number I wish, this would seem to indicate that all inner products are equivalent, which is of course absolute nonsense. I would be grateful if you could clear this point up as well for me.

stevendaryl · Feb 9, 2015

nomadreid said: ↑

Thanks again, stevendaryl. The ideas are starting to clear up. While I am on the subject, you mentioned that multiplying a state by a complex number leaves it in the same physical state. I have always been puzzled by this: as I understand it when two states are simultaneously multiplied by the same scalar, then the inner product of |e^iθ u> with |e^iθ v> = <u*e^-iθ |e^iθ v> = <u*|v>, but when only one of them is, say v, then <u*|e^iθ v> = e^iθ<u*| v>, and since <u*| v> is itself a complex number, then since I can choose θ to get any (unit) complex number I wish, this would seem to indicate that all inner products are equivalent, which is of course absolute nonsense. I would be grateful if you could clear this point up as well for me.

Well, what's physically meaningful in quantum mechanics is not amplitudes, but squared amplitudes. Or the square of the sum of a bunch of amplitudes.

So if you have an amplitude [itex]A[/itex] that is the sum of component amplitudes: [itex]A = A_1 + A_2 + ...[/itex], then

[itex]|A|^2 = \sum_{ij} |A_i| |A_j| cos(\theta_{ij})[/itex]

where [itex]\theta_{ij}[/itex] is the difference in phases between the complex numbers [itex]A_i[/itex] and [itex]A_j[/itex]

Multiplying all the amplitudes [itex]A_i[/itex] by the same phase factor [itex]e^{i\theta}[/itex] makes no difference to [itex]|A|^2[/itex]. However, if you change the phases of each term separately, then you'll get a different answer. So phases don't matter, but differences between phases do.

So getting back to spin-1/2 particles, if you rotate a single particle through [itex]2\pi[/itex], it makes no difference. But if you could somehow get interference between a state that is rotated and one that is not rotated, then you will get interference effects.

nomadreid · Feb 9, 2015

Many thanks, stevendaryl. I think what you are referring to is what my book terms global versus relative phases.

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Connection between Pauli XYZ and spatial XYZ

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