Representing states in the full Hilbert space

Given a representation of the states that span the spin Hilbert space, we now need to consider the problem of representing the the states the span the full Hilbert space:

$${\cal H} = {\cal H}_r \bigotimes {\cal H}_s$$

We will work with the following complete set of commuting observables (CSCO): $\{X,Y,Z,S^2,S_z\}$, which means that the basis vectors which span the full Hilbert space must be simultaneous eigenvectors of these five operators. These will be represented as

$$\vert{\bf r}\;s\;m_s\rangle = \vert{\bf r}\rangle\bigotimes \vert s\;m_s\rangle$$

that is, they will be a tensor product of the usual coordinate eigenvector and the simultaneous eigenvector of $S^2$ and $S_z$. Thus, they will satisfy the eigenvalue equations:

$$X\vert{\bf r}\;\;s\;m_s\rangle = x \vert{\bf r}\;\;s\;m_s\rangle$$

$$Y\vert{\bf r}\;\;s\;m_s\rangle = y \vert{\bf r}\;\;s\;m_s\rangle$$

$$Z\vert{\bf r}\;\;s\;m_s\rangle = z \vert{\bf r}\;\;s\;m_s\rangle$$

$$S^2\vert{\bf r}\;\;s\;m_s\rangle = s(s+1)\hbar^2 \vert{\bf r}\;\;s\;m_s\rangle$$

$$S_z\vert{\bf r}\;\;s\;m_s\rangle = m_s\hbar \vert{\bf r}\;\;s\;m_s\rangle$$

The basis vectors will also satisfy an orthogonality relation:

$$\rangle {\bf r}\;\;s\;m_s\vert{\bf r}'\;s\;m_s'\rangle = \delta_{m_s m_s'}\delta^{(3)} ({\bf r}-{\bf r}')$$

Any arbitrary vector $\vert\phi\rangle$ in the Hilbert space can be expanded in terms of these basis vectors:

$$\vert\phi\rangle = \sum_{m_s=-s}^s \int\;d{\bf r}\;\; \vert{\bf r}\;\;s\;m_s\rangle\langle{\bf r}\;\;s\;m_s\vert\phi\rangle$$

The expansion coefficients can, as usual, be designated as functions of ${\bf r}$:

$$\langle {\bf r}\;\;s\;m_s\vert\phi\rangle = \phi_{s,m_s}({\bf r})$$

For the case of spin-1/2, the expansion takes the form

$$\vert\phi\rangle = \sum_{m_s=-1/2}^{1/2} \int\;d{\bf r}\;\;\left\vert{\bf r}\;\;{1 \over 2}\;m_s\right> \left<{\bf r}\;\;{1 \over 2}\;m_s\right\vert\phi\rangle$$

$$= \int\;d{\bf r}\;\;\left( \left\vert{\bf r}\;\;{1 \over 2}\;-{1 \o... ...}\right> \left<{\bf r}\;\;{1 \over 2}\;{1 \over 2}\right\vert\phi\rangle\right)$$

The coefficients are designated by

$$\left<{\bf r}\;\;{1 \over 2}\;{1 \over 2}\right\vert\phi\rangle =... ...\phi_{{1 \over 2}}({\bf r})\;\;\;\;\;{\rm or}\;\;\;\;\;\phi_{\uparrow}({\bf r})$$

$$\left<{\bf r}\;\;{1 \over 2}\;-{1 \over 2}\right\vert\phi\rangle ... ...i_{-{1 \over 2}}({\bf r})\;\;\;\;\;{\rm or}\;\;\;\;\;\phi_{\downarrow}({\bf r})$$

Then, since the basis vectors are:

$$\left\vert{\bf r}\;\;{1 \over 2}\;{1 \over 2}\right> = \vert{\bf r}\rangle \bigotimes\;\left\vert{1 \over 2}\;{1 \over 2}\right> = \vert{\bf r}\rangle \bigotimes {\left(\matrix{1 \cr 0}\right)}$$

$$\left\vert{\bf r}\;\;{1 \over 2}\;-{1 \over 2}\right> = \vert{\bf r}\rangle \bigotimes\;\left\vert{1 \over 2}\;-{1 \over 2}\right> = \vert{\bf r}\rangle \bigotimes {\left(\matrix{0 \cr 1}\right)}$$

the expansion can be written as

$$\vert\phi\rangle = \int\;d{\bf r}\; \left(\vert{\bf r}\rangle \bigotimes {\left(\mat... ...le \bigotimes {\left(\matrix{1 \cr 0}\right)}\phi_{{1 \over 2}}({\bf r})\right)$$

$$= \int\;d{\bf r}\;\vert{\bf r}\rangle \bigotimes \left[{\left(\matr... ...}}({\bf r}) + {\left(\matrix{1 \cr 0}\right)}\phi_{{1 \over 2}}({\bf r})\right]$$

$$= \int\;d{\bf r}\;\vert{\bf r}\rangle \bigotimes \left(\matrix{\phi_{{1 \over 2}}({\bf r}) \cr \phi_{-{1 \over 2}}({\bf r})}\right)$$

The vector

$$\left(\matrix{\phi_{{1 \over 2}}({\bf r}) \cr \phi_{-{1 \over 2}}({\bf r})}\right)$$

is called a two-component spinor. Note that

$$\langle \phi\vert\phi\rangle = \int\;d{\bf r}\;\int\;d{\bf r}'\; \left(\matrix{\phi_{{1 \over 2}}^*({\bf r}')$$

$$= \int\;d{\bf r}\;\int\;d{\bf r}'\; \left[\phi_{{1 \over 2}}({\bf r... ...r 2}}({\bf r})\phi_{-{1 \over 2}}({\bf r})\right]\delta^{(3)}({\bf r}-{\bf r}')$$

$$= \int\;d{\bf r}\;\left(\vert\phi_{{1 \over 2}}({\bf r})\vert^2 + \vert\phi_{-{1 \over 2}}({\bf r})\vert^2\right)$$

Example: If we have a spin-independent Hamiltonian that is also spherically symmetric, then the quantum numbers that characterize the states will be $n,l,m,s,m_s$. Thus, for the hydrogen atom,

$$H = \left[-{\hbar^2 \over 2\mu}{1 \over r}{\partial^2 \over ... ...l r^2}r + {l(l+1)\hbar^2 \over 2\mu r^2}-{e^2 \over r}\right]$$

which is spin independent. The ground state will, therefore, be twofold degenerate with the two eigenfunctions being:

$$\psi_{100{1 \over 2}\;{1 \over 2}}(r,\theta,\varphi) = \left({1 \over \pi a_0^3}\right)^{1/2}e^{-r/a_0}{\left(\matrix{1 \cr 0}\right)}$$

$$\psi_{100{1 \over 2}\;-{1 \over 2}}(r,\theta,\varphi) = \left({1 \over \pi a_0^3}\right)^{1/2}e^{-r/a_0}{\left(\matrix{0 \cr 1}\right)}$$