# Position representation of spin states and spin operators

How can we represent a spin states $$\lvert S_x:+\rangle, \lvert S_y:+\rangle,\lvert S_z:+\rangle ,\lvert S_x:-\rangle, \lvert S_y:-\rangle$$ and $$\lvert S_z:-\rangle$$ in position representation like $$\langle x\lvert S_z:+\rangle$$ ?

Is there any position representation for spin operators $$S_z, S_y, S_z$$?

Zettili says in his book of Quantum mechanics that "Spin cannot be described by a differential operator". Does it mean spin cannot have a position representation? He does not talk about it more.

Unlike the angular momentum operators which inherit their position representation from the position representation of $$\hat x$$ and $$\hat p$$: $$\hat L_z\to -i\hbar (y\partial_x -x\partial_y)$$ it is not possible to write $$\hat S_z$$ in terms of the classical position and momenta because spin is an "intrinsic" rather than spatial degree of freedom: dimensional analysis shows that only products like $$yp_x$$ have units of angular momentum so, after accounting for the commutation relations to be respected, a position representation of spin would be more or less identical to the position representation of "ordinary" angular momentum.

Moreover, quantizing angular momentum necessarily leads to integer values of $$\ell$$ through the use of (spherical) coordinates see:

Gatland, Ian R. "Integer versus half-integer angular momentum." American Journal of Physics 74, no. 3 (2006): 191-192.

Thus a position-based representation for spin could not accommodate half-integer values.

• all the observables in-universe correspond to operators and all operators form a complete basis for expressing any wave function as a superposition of independent waves correspond to operators, can it possible to express a wave function corresponds to a momentum eigenvalue p in spin basis of $S_z$ operator? – ROBIN RAJ Jun 29 '19 at 10:19
• Your statement is true of “quantum states”, not “wavefunctions’, precisely because some internal degrees of freedom have quantum states but no wavefunctions, i.e. no states expressible in terms of position or momentum variables. – ZeroTheHero Jun 29 '19 at 11:17

Indeed, there is no position space representation of the spin states. That is because its Hilbert space is simply different from the Hilbert space of something like particle in a box. A very crude classical example would be think of a seesaw hinged at a place. Its configuration is given by two possible configurations: left-side-up or right-side-up (for simplicity assume you can't see an intermediate position).

If your seesaw wasn't fixed at a place but can move around in the park, then you will have something that needs to be described by both its position in the park as well as whether left or right side is up. In that case you will still have position as an additional information and you really can't write the configuration in terms its position (it is not like the seesaw being right-side-up is same as it being in the right-hand corner of the park). Such a description is what you will call a direct product of position space with spin space. An example of that would be a gas of spinning particles (say, electrons).

Alternatively, you can have an array of seesaws in the park. In that case you have to tell me which seesaw you are talking about in the park (described by its location in the park) and what's it configuration (left/right-side-up). Once again you don't have a position space representation but the position now becomes a label for the seesaws. Equivalent of that would be something like Heisenberg model where you have spins at different sites of a lattice.

• all the observables in-universe correspond to operators and all operators form a complete basis for expressing any wave function as a superposition of independent waves correspond to operators, can it possible to express a wave function corresponds to a momentum eigenvalue p in the spin basis of $S_z$operator? – ROBIN RAJ Jun 29 '19 at 10:25
• @ROBINRAJ Once again, momentum and position space representations are conjugate to each other, they describe the same Hilbert space. So the all my comments from above apply. Basically as emphasized by me and others here, one can't use the basis vectors of one Hilbert space to describe a totally different Hilbert space. Let me try to say it in another way: an analogous question to what you are asking would be whether it is possible to use the basis vectors of one space, say $\mathbb{R}^2$ to write an arbitrary vector in $\mathbb{R}^3$. Of course it is not possible. – nGlacTOwnS Jun 29 '19 at 17:02
• Now the example doesn't do complete justice because $\mathbb{R}^2$ is a subset of $\mathbb{R}^3$ (which might lead to additional question) but in general that is not the case and you are simply talking about different and distinct Hilbert spaces. – nGlacTOwnS Jun 29 '19 at 17:03
• can all classical observables describe same Hilbert space? – ROBIN RAJ Jun 29 '19 at 17:45
• I don't understand your question, can you elaborate with an example? – nGlacTOwnS Jun 29 '19 at 17:54

I'll try to give a simple and rough answer.

There is no position representation for the spin.

Roughly, that is because spin is a different space. Spins are things that live in a different space, which is independent of Hilbert's space.

On the one hand, you've got wavefunctions, which live in Hilbert's space.

$$\psi(x) \in \mathcal{H}$$

You can use Dirac notation: $$|\psi \rangle$$.

On the other hand, spins are objects living in $$\mathbb{C}^n$$. With $$s=½$$, we've got $$n=2$$, so ½-spins live in $$\mathbb{C}^2$$. We use a basis that is

$$\lbrace \left( \begin{array}{c} 1 \\ 0 \end{array} \right); \ \left( \begin{array}{c} 0 \\ 1 \end{array} \right) \rbrace$$

The first one corresponds to spin-up and the second one corresponds to spin down.

So, the full wavefunction is the tensor product of those tow things, that is

$$\psi(x)\otimes \left( \begin{array}{c} a \\ b \end{array} \right)$$

The symbol $$\otimes$$ is usually ommited. So basically, a spinor is

$$\left( \begin{array}{c} \phi(x) \\ \varphi(x) \end{array} \right)$$

and it is the result of doign the tensor product of a wavefunction and a spin vector.

Only things in Hilbert space can have a position representation.

On the other hand, spins belong to an independent complex space $$\mathbb{C}^n$$.

• all the observables in-universe correspond to operators and all operators form a complete basis for expressing any wave function as a superposition of independent waves correspond to operators, can it possible to express a wave function corresponds to a momentum eigenvalue p in spin basis of $Sz$ operator? – ROBIN RAJ Jun 29 '19 at 10:30
• @ROBINRAJ sorry, but I don't understand the second part of your comment. As for the first one, indeed, spins are observables, and thus they have their corresponding operators $(S_x, S_y,S_z, S^2...$, but those operators act on the spin-space. That's a different space than that of the wavefunctions. Spin is a "new thing", which is independent of the spatial part (they can be related, but they are separate things). I don't know if this is what you were asking... – FGSUZ Jun 29 '19 at 13:13
• My second part says We know that $S_z$ operator form complete basics, so can we express a plane wave corresponds to a momentum value p in the complete basics of $S_z$ operator? – ROBIN RAJ Jun 29 '19 at 17:46
• Okay, I think I get you. No, you can't express any plane wave. Because the eigenvectors of $S_z$ form a complete basis for the space, yes, but what space? The spin space. The vectors (1 0) and (0 1) form a basis for the spin ½ space. But that's the spin part. It has nothing to do with the spatial part. Momentum $p$ lives in $\mathcal{H}$. The spin is a different part of the puzzle. – FGSUZ Jun 29 '19 at 23:17
• Can we have any theoretical proof exists that give quantized spin angular momentum value for an electron is $\hbar/2$ or $- \hbar/2$ – ROBIN RAJ Jun 30 '19 at 6:32

Here's the fifth answer, in a more mathematical fashion.

Using the standard terminology, for a massive non-specially relativistic particle with spin in three dimensions, the set of operators $$\{x,y,z,p_x,p_y,p_z, S_x,S_y, S_z\}$$ is irreducible, which means there is no linear or non linear functional dependence between them. This is a consequence of the representation theory of the Galilei algebra. We are then forced to the algebraic commutation relations which derive from the representation theory of the Galilei algebra (ignore i and hbar):

$$[x_i,p_j] = \delta_{ij}, [x_i, S_j] =0, [p_k, S_l] =0, [S_m,S_n]=\epsilon_{mnq}S_q .$$

Let us call the vector space of this representation $$V$$. That $$x$$ and $$p_x$$ are irreducible and do not commute, it means that there are conditions to apply the famous uniqueness theorem of Stone and von Neumann to conclude that $$V$$ must have $$L^2 (\mathbb R^3)$$ as a subspace, so it is necessarily infinite-dimensional.

Let us now consider that the spin operators S form a representation of $$\text{so}(3)$$ on $$V$$ by (essentially) self-adjoint operators. We know that $$\text{so}(3)$$ is the compact Lie algebra of the compact group $$\text{SO(3)}$$, so, by the theorem of Peter-Weyl, the representation space of the commutation relation must be finite-dimensional. We are to conclude that the commutation relation for the spin operators should take place on a finite-dimensional subspace of $$V$$, we call $$\mathcal S_{fin}$$. Of course, $$\mathcal S_{fin}$$ is isomorphic as a vector space to $$\mathbb {C}^n$$ for a particular finite $$n$$.

There are two options. The first:

$$V = L^2 (\mathbb R^3) \otimes \mathcal S_{fin},$$

which basically answers the question (the spin states do not have a position representation), because they live in different (rigged) Hilbert spaces and, by convention/definition, the scalar product between a spin state and a position (eigen)state is nil. The same applies to momentum and spin.

The second option would be $$V= L^2 (\mathbb R^3), \mathcal S_{fin} < L^2 (\mathbb R^3).$$

One can easily show that the three essentially self-adjoint operators $$\epsilon_{ijk}x_j p_k$$ satisfy the $$\text{so(3)}$$ commutation relations on $$L^2 (S_2) whose dimension is necessarily an even integer. We have that, due to a coordinate change implemented unitarily, $$L^2 (\mathbb R^3) = L^2 (S^2) \otimes L^2 ((0,\infty), r^2 dr)$$. The dimension of $$\mathcal S_{fin}$$, however, can be any natural number. So, for integer spins, we have that the representation spaces of the same algebra are isomorphic. It follows that the generators are one the multiple of the other, hence the position, momentum, spin cannot be irreducible anymore.

The existence of spin comes purely from enforcing a projective unitary representation of the Galilei group on a (rigged) Hilbert space. Group theory shows that we are actually representing the universal covering group of which has a subgroup isomorphic to $$\text{SU(2)}$$. Had we not been forced to use projection representations, but only vector ones, we would have had no way to exhibit a spin operator. The same goes for the specially relativistic case.

• all the observables in-universe correspond to operators and all operators form a complete basis for expressing any wave function as a superposition of independent waves correspond to operators, can it possible to express a wave function corresponds to a momentum eigenvalue p in the spin basis of $S_z$operator? – ROBIN RAJ Jun 29 '19 at 10:30