How does spin work in the position representation?

In summary, when dealing with a particle with spin, its quantum state will live in the tensor product of two Hilbert spaces representing position and spin states. The wavefunction for this particle will be a scalar-valued function of position and a spinor-valued function of position. This allows for a more comprehensive understanding of the particle's properties, including probabilities for both position and spin states. This concept can be extended to particles with arbitrary spin, with the number of components in the wavefunction corresponding to the spin value. The important aspect of these multi-component objects is that they allow for meaningful rotations, just like real rotations in 3-dimensional space.
  • #1
lugita15
1,554
15
When you have an ordinary (read: spin zero) particle with a quantum state |ψ>, the corresponding position-space wave function <x|ψ> is an ordinary scalar function of position. How do things work when the particle has spin? In that case its quantum state will live in the tensor product of two Hilbert spaces: an infinite-dimensional (rigged) Hilbert space for position, and presumably a finite dimensional Hilbert space for its spin states. So its state will be of the form |ψ>=|ρ>|χ>, where |ρ> is the positional state and |χ> is the spin state, and the amplitude to find it in a particular position and spin is given by <x|<s||ρ>|χ>. So then the position space wave function should be written as ψ(x)=ρ(x)χ(s), where ρ(x)=<x|ρ> and χ(s)=<s|χ> are scalar-valued wave functions representing the positional and spin states.

But instead I think I've seen people write ψ(x)=ρ(x)χ(x), where χ(x)=<x|χ> is called a "spinor-valued wavefunction" representing the spin state. What's going on here? Is this correct, and what does this mean? I don't know too much about spinors, other than the fact that they can be represented in terms of matrices. Are spinors supposed to be the irreducible representations of the group SU(2) in the positional Hilbert space?

Any help would be greatly appreciated.

Thank You in Advance.
 
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  • #2
The way to deal with this is to extend the list of parameters to the wavefunction. It's no longer enough to ask "what is the probability that the particle is at position x?". Instead, you have to ask "what is the probability that the particle is at position x and with spin state s?". To use your notation, if you have a state [itex]|\psi\rangle = |\rho\rangle | \chi\rangle[/itex], then converting to the position representation will result in a wavefunction with an extra spin state parameter: [itex]\psi(x, s) = (\langle x| \langle s|)(|\rho\rangle | \chi\rangle)[/itex].

Another equivalent way to write this is to say that we'll still just have a function of position [itex]\psi(x)[/itex], but that instead of being a simple complex number at each point in space, it's a pair of complex numbers, one for the probability that it's spin-up, and the other for the probability that it's spin-down. This method ends up being useful for a number of reasons as one gets further into the mathematics of spin, so it's very common. These two-component objects are spinors.
 
  • #3
Chopin said:
The way to deal with this is to extend the list of parameters to the wavefunction. It's no longer enough to ask "what is the probability that the particle is at position x?". Instead, you have to ask "what is the probability that the particle is at position x and with spin state s?". To use your notation, if you have a state [itex]|\psi\rangle = |\rho\rangle | \chi\rangle[/itex], then converting to the position representation will result in a wavefunction with an extra spin state parameter: [itex]\psi(x, s) = (\langle x| \langle s|)(|\rho\rangle | \chi\rangle)[/itex].
But is it valid to factor [itex]\psi(x, s)[/itex] into a scalar-valued function of position and a "spinor-valued" function of position?
Chopin said:
Another equivalent way to write this is to say that we'll still just have a function of position [itex]\psi(x)[/itex], but that instead of being a simple complex number at each point in space, it's a pair of complex numbers, one for the probability that it's spin-up, and the other for the probability that it's spin-down. This method ends up being useful for a number of reasons as one gets further into the mathematics of spin, so it's very common. These two-component objects are spinors.
So which is the spinor, the ordered pair of complex numbers or the wavefunction which outputs such pairs? Also, can this notion be generalized to arbitrary spin, not just for 1/2, by using ordered n-tuples of complex numbers. And do the complex numbers represent probabilities or just amplitudes?
 
  • #5
You can use some of those isomorphisms as [tex]V^*\otimes W \cong Hom(V,W)[/tex] and if you identify $V^*$ and $V$, you can represent the states as elements in the tensor product or as vector valued functions. For example for a single particle you can think of [tex]L^2(\mathbb R^3)\otimes \mathbb C^n[/tex] as [tex]Hom(L^2(\mathbb R^3), \mathbb C^n)[/tex].
 
  • #6
lugita15 said:
But is it valid to factor [itex]\psi(x, s)[/itex] into a scalar-valued function of position and a "spinor-valued" function of position?
Well, I guess it depends on what you mean by "valid". Mathematically it's a thing you can do, but probably not a useful thing. You usually either write the wavefunction as a spinor-valued function of position, or a complex-valued function of position and spin.

lugita15 said:
So which is the spinor, the ordered pair of complex numbers or the wavefunction which outputs such pairs?
The ordered pair. The wavefunction which outputs such pairs is a spinor-valued function.

lugita15 said:
And do the complex numbers represent probabilities or just amplitudes?
Sorry, I meant amplitude. You have to square them just like usual to get a probability.

lugita15 said:
Also, can this notion be generalized to arbitrary spin, not just for 1/2, by using ordered n-tuples of complex numbers.
Sure. For spin 1, it's a 3-component object, aka a vector. In general, for spin [itex]n[/itex], it's a [itex]2n+1[/itex] component object, so you can go as high as you want.

The important thing about these multi-component objects, though, is that there's a meaningful concept of "rotation" on them. When I say meaningful, I mean that if you apply two rotation operations to one of them, the resulting object is consistent with what you'd get if you'd applied the single combined rotation to it--the rotation operations compose just like real rotations do. For the vector (spin 1) case, we already know what operations will do this--it's just the set of 3-dimensional rotation matrices. What's novel is that in the spinor case, we can find a set of 2-dimensional rotation matrices which do the same thing. For instance, a rotation of [itex]\theta[/itex] degrees about the Z axis is represented by [itex]e^{i \theta\sigma_z}[/itex], where [itex]\sigma_z[/itex] is the Z Pauli spin matrix.

To use the fancy terminology, these matrices form a two-dimensional representation of SU(2). It's necessary to use objects which have a representation of the rotation group so that you can meaningfully talk about taking a system and rotating it in space--if there were no such representation, then it would be impossible to solve the Schrodinger equation and then change to a rotated basis, because there would be no way to transform the spinors.
 

What is spin in the position representation?

Spin in the position representation refers to the intrinsic angular momentum of a particle in quantum mechanics. It is a quantum mechanical property that cannot be directly observed, but its effects can be seen in experiments.

How is spin represented in the position representation?

In the position representation, spin is represented by a mathematical object called the spinor. This is a complex vector with two components that describes the spin state of a particle. The spinor is used to calculate the probability of finding a particle in a particular spin state at a given position.

What are the possible values of spin in the position representation?

In the position representation, spin can have two possible values: up spin and down spin. These correspond to the two components of the spinor and are denoted by the symbols ↑ and ↓. In quantum mechanics, spin is quantized, meaning it can only have discrete values.

How does spin affect the position of a particle?

Spin does not directly affect the position of a particle in the position representation. Instead, it is a property that describes the rotational symmetry of a particle. However, the spin of a particle can affect its energy, momentum, and interactions with other particles, which can indirectly influence its position.

What experiments have been done to study spin in the position representation?

There have been numerous experiments to study spin in the position representation, including the Stern-Gerlach experiment and the double-slit experiment. These experiments have provided evidence for the quantized nature of spin and its role in determining the behavior of particles at the quantum level.

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