2.1: Representing states in the full Hilbert space
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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:
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
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:
\(\displaystyle X\vert{\bf r}\;\;s\;m_s\rangle\) | \(\textstyle =\) | \(\displaystyle x \vert{\bf r}\;\;s\;m_s\rangle\) | |
\(\displaystyle Y\vert{\bf r}\;\;s\;m_s\rangle\) | \(\textstyle =\) | \(\displaystyle y \vert{\bf r}\;\;s\;m_s\rangle\) | |
\(\displaystyle Z\vert{\bf r}\;\;s\;m_s\rangle\) | \(\textstyle =\) | \(\displaystyle z \vert{\bf r}\;\;s\;m_s\rangle\) | |
\(\displaystyle S^2\vert{\bf r}\;\;s\;m_s\rangle\) | \(\textstyle =\) | \(\displaystyle s(s+1)\hbar^2 \vert{\bf r}\;\;s\;m_s\rangle\) | |
\(\displaystyle S_z\vert{\bf r}\;\;s\;m_s\rangle\) | \(\textstyle =\) | \(\displaystyle m_s\hbar \vert{\bf r}\;\;s\;m_s\rangle\) |
The basis vectors will also satisfy an orthogonality relation:
Any arbitrary vector \(\vert\phi\rangle\) in the Hilbert space can be expanded in terms of these basis vectors:
The expansion coefficients can, as usual, be designated as functions of \({\bf r}\):
For the case of spin-1/2, the expansion takes the form
\(\displaystyle \vert\phi\rangle\) | \(\textstyle =\) | ||
\(\textstyle =\) |
The coefficients are designated by
Then, since the basis vectors are:
\(\displaystyle \left\vert{\bf r}\;\;{1 \over 2}\;{1 \over 2}\right>\) | \(\textstyle =\) | ||
\(\displaystyle \left\vert{\bf r}\;\;{1 \over 2}\;-{1 \over 2}\right>\) | \(\textstyle =\) |
the expansion can be written as
\(\displaystyle \vert\phi\rangle\) | \(\textstyle =\) | ||
\(\textstyle =\) | |||
\(\textstyle =\) |
The vector
is called a two-component spinor. Note that
\(\displaystyle \langle \phi\vert\phi\rangle\) | \(\textstyle =\) | ||
\(\textstyle =\) | |||
\(\textstyle =\) |
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,
which is spin independent. The ground state will, therefore, be twofold degenerate with the two eigenfunctions being:
\(\displaystyle \psi_{100{1 \over 2}\;{1 \over 2}}(r,\theta,\varphi)\) | \(\textstyle =\) | \(\displaystyle \left({1 \over \pi a_0^3}\right)^{1/2}e^{-r/a_0}{\left(\matrix{1 \cr 0}\right)}\) | |
\(\displaystyle \psi_{100{1 \over 2}\;-{1 \over 2}}(r,\theta,\varphi)\) | \(\textstyle =\) | \(\displaystyle \left({1 \over \pi a_0^3}\right)^{1/2}e^{-r/a_0}{\left(\matrix{0 \cr 1}\right)}\) |