Wave function of spin 1/2 under parity

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1. The problem statement, all variables and given/known data

How does the wave function of spin 1/2 change under parity?

2. Relevant equations



3. The attempt at a solution

The behavior of the eigenfunctions of orbital angular momentum L is easily seen from their explicit form, namely spherical function Y^m_l is multiplied by (-1)^l, where l=0,1,2,...

Spin operator S commutes with the parity, so the eigenket of S is expected to be a parity eigenket.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

arkajad

"Spin operator S commutes with the parity"

How do you know it? Where from? Either you know exactly what is the parity transformation, then you can check it, but also you can compute the transformation on spin eigenstates, or someone told it to you without any proof - a thing to believe in?

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Parity=space inversion, i.e. a 3x3 matrix with -1 on the diagonal. The spin operator S is a unitary operator of rotation ~ 1-i S/h. (you can look in Sakurai JJ, page 254)

arkajad

But this is not an answer to the question: how do you know that Parity (3x3 matrix) commutes with S - the spin operator for spin 1/2 particle. You can't commute 3x3 matrix with 2x2 matrix. It does not make sense.

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Parity P commutes with the infinitesimal operator of rotation D from which the spin operator S is derived. All group properties and commutation relations are the same for D and S. So if D (3x3 matrix) commutes with P, then S also commutes. At this moment we are not saying anything about the spin value. The lowest dimension for which the commutation relations are hold is 2 and one can then construct Pauli matrices for S=1/2. However the act in different space.

For the spin 1 the matrix you mentioned is 3x3, for S=3/2 4x4, ... ;-)

arkajad

Spin is not derived from the same infinitesimal operator that 3D space rotations are derived. Space rotation operators are 3x3 matrices, spin rotations are 2x2 matrices. They are not the same. What is true, is that both satisfy the same commutation relations between themselves, but that alone is not enough to deduce some definite conclusion about parity.

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Clearly, rotations affect the properties of physical systems, and to see how we do have to consider how we construct the spin operators. Angular momentum operator (it does not matter orbital of spin) is defined so that the _operator_ of an infinitesimal rotation by d\phi around axis n is 1- i(nJ)/h d\phi. This is by analogy with the momentum operator. The finite rotation is then given by the operator D(\phi)=exp(- i(nJ)/h \phi)
For every rotation R represented by 3x3 matrix in 3D <==> operator in D(R) acting in angular momentum space. The dimension of this space is 2J+1.

Actually from the above definition it follows that by 2\pi rotation the sign of the wave function is changed by (-1)^(2J), implying that e.g. for the neutron (S=1/2) to get back to the initial phase one needs to rotate by 4\pi, which had been indeed observed experimentally ( http://prl.aps.org/abstract/PRL/v35/i16/p1053_1 ).

So, my initial question was: how does the the wave function phase of S=1/2 change under the space inversion?

arkajad

You don't know it, because space inversion has determinant -1 and you cannot get -1 from exponenensial of infinetisimal rotations, because these exponentials always give you proper rotations of SO(3) of determinant +1.

More precisely: for instance you are looking for a linear operator ([tex]2\times 2[/tex] complex matrix) [tex]\mathcal{P}[/tex] such that

[tex]\mathcal{P}\sigma_i\mathcal{P}^{-1}=-\sigma_i,\quad (i=1,2,3[/tex]

or

[tex]\mathcal{P}\sigma_i+\sigma_i\mathcal{P}=0, \quad (i=1,2,3).[/tex]

It is an easy exercise in the matrix algebra to see that the only solution of the last equation is [tex]\mathcal{P}=0.[/tex]

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Then, it must be somehow related to the dimension of the ket states, because we know that for the angular momentum L the wavefunction changes to [tex](-1)^l[/tex], where l=0,1,2,3... The matrix algebra for S=1 or l=1 is the same.

arkajad

Indeed it is. The situation is different when you change from 2x2 Pauli matrices to 4x4 Dirac matrices. The point is that there are two inequivalent representations of the Pauli anticommutation algebra [tex]\{\sigma_i,\sigma_j\}=2\delta_{ij}[/tex], one given by [tex]\sigma_i[/tex], the other one given by [tex]-\sigma_i[/tex].