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## Summary:

- The project of spin on real space

## Main Question or Discussion Point

If the spin space is independent of the real space, what is the meaning of, for example, the z-component of the spin?

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## Summary:

- The project of spin on real space

If the spin space is independent of the real space, what is the meaning of, for example, the z-component of the spin?

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The z-component of spin is a measurable quantity. The abstract spin vectors describe the spin state, but the spin operators represent real observables.Summary::The project of spin on real space

If the spin space is independent of the real space, what is the meaning of, for example, the z-component of the spin?

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It isn't. The spinthe spin space is independent of the real space

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Do you mean that there is something of real-rotation entity in spins?It isn't. The spindegrees of freedomare independent of the configuration space (position and momentum) degrees of freedom. But the spin operators are still connected to directions in real space.

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The problem is that we have not a real intuition for the quantum world, so we have to use some more abstract thinking to get an intuition about it. Spin is most easily introduced as a field-theoretical concept, because then you have at least some intuition from classical field theory (though the only classical field theory except fluid dynamics, where you only deal with orbitarl angular momentum, is electromagnetism, and there you deal with a massless relativistic vector field, where the notion of spin is not as for massive fields, and there's no non-relativistic limit for it).

To get an idea, how to introduce angular momentum in a field theory we consider simply wave mechanics a la Schrödinger as a classical field theory though there is no classical physical interpretation to the Schrödinger wave function. To define what's angular momentum we have to remember that in classical mechanics (Hamilton formalism) angular momentum is the quantity that is conserved due to invariance under rotations (isotropy of Galilean spacetime) and thus angular momentum is the generator of rotations.

So let's start with the usual scalar Schrödinger field, ##\psi(t,\vec{x})##. Consider an infinitesimal rotation of the spatial coordinates, ##\vec{x}'=\vec{x}+\delta \vec{\varphi} \times \vec{x}##, where ##\delta \vec{\varphi}## is an infinitesimal vector pointing in the direction of the rotation axis and ##\delta \varphi=|\delta \vec{\varphi}|## is the infinitesimal rotation angle. Now since the Schrödinger field is a scalar field it transforms as

$$\psi'(t,\vec{x}')=\psi(t,\vec{x})=\psi(t,\vec{x}'-\delta \vec{\varphi} \times \vec{x}) = \psi(t,\vec{x}') - (\delta \vec{\varphi} \times \vec{x}) \cdot \vec{\nabla}' \psi(t,\vec{x}').$$

The infinitesimal change of the field thus is

$$\delta \psi(t,\vec{x})=-\delta \vec{\varphi} \cdot (\vec{x} \times \vec{\nabla}) \psi(t,\vec{x}).$$

Since (using units with ##\hbar=1## for simplicity)

$$\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}$$

you can write this is

$$\delta \psi(t,\vec{x})= -\mathrm{i} \delta \vec{\varphi} \cdot (\vec{x} \times \hat{\vec{p}}) \psi(t,\vec{x}).$$

Since (total) angular momentum generates by definition rotations you should have

$$\delta \psi(t,\vec{x})=-\mathrm{i} \delta \vec{\varphi} \cdot \hat{\vec{J}} \psi(t,\vec{x}).$$

Since this is true for any ##\delta \vec{\varphi}## this implies that for the usual scalar Schrödinger field

$$\hat{\vec{J}} = \hat{\vec{x}} \times \hat{\vec{p}}=\hat{\vec{L}},$$

i.e., for this case the total angular momentum is just the (quantized) familiar orbital angular momentum of classical mechanics.

However, there are more possibilities for a field theory. In quantum mechanics it turns out that from the commutation relations for angular momenta you can get some kind of angular momentum not familiar from classical physics, namely socalled spinor representations with half-integer spin quantum number. The most simple is spin 1/2, describing, e.g., an electron. In non-relativistic physics this spin is independent of ##\hat{\vec{x}}## and ##\hat{\vec{p}}## and thus you have a fourth independent variable which usually is chosen as ##\hat{s}_3=\frac{1}{2} \sigma_3## with the Pauli matrix ##\sigma_3=\mathrm{diag}(1,-1)##. Then the wave functions get (Weyl-)spinor valued, i.e., the wave function has two components ##\psi_{\sigma}(t,\vec{x})## with ##\sigma \in \{1/2,-1/2 \}##, and we write the wave function as a column vector,

$$\psi(t,\vec{x})=\begin{pmatrix} \psi_{1/2}(t,\vec{x}) \\ \psi_{-1/2}(t,\vec{x}) \end{pmatrix}.$$

As it turns out, the infinitesimal rotation of such an object is defined as

$$\psi'(t,\vec{x}')=(1- \mathrm{i} \delta \vec{\varphi} \cdot \hat{\vec{s}} \cdot \psi(t,\vec{x}'-\delta \vec{\varphi} \times \vec{x}').$$

The same calculation as above, neglecting all contributions with ##\delta \varphi^2## (linear approximation), you get

$$\delta \psi(t,\vec{x}) = -\mathrm{i} \delta \vec{\varphi} \cdot (\hat{\vec{L}}+\hat{\vec{s}}) \psi(t,\vec{x}),$$

which leads to

$$\hat{\vec{J}}=\hat{\vec{L}} + \hat{\vec{s}}.$$

So the spin is an additional intrinsic contribution to the angular momentum of the particle, independent of its orbital motion.

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No. I meant what I said. Read it again.Do you mean that there is something of real-rotation entity in spins?

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bobob

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