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2.1: Representing states in the full Hilbert space

  • Page ID
    20878
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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:

    \begin{displaymath}
{\cal H} = {\cal H}_r \bigotimes {\cal H}_s
\end{displaymath}

    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

    \begin{displaymath}
\vert{\bf r}\;s\;m_s\rangle = \vert{\bf r}\rangle\bigotimes \vert s\;m_s\rangle
\end{displaymath}

    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:

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

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

    \begin{displaymath}
\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
\end{displaymath}

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

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

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

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

    The coefficients are designated by

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

    Then, since the basis vectors are:

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

    the expansion can be written as

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

    The vector

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

    is called a two-component spinor. Note that

    \(\displaystyle \langle \phi\vert\phi\rangle\) \(\textstyle =\) $\displaystyle \int\;d{\bf r}\;\int\;d{\bf r}'\;
\left(\matrix{\phi_{{1 \over 2}}^*({\bf r}')$
    \(\textstyle =\) $\displaystyle \int\;d{\bf r}\;\int\;d{\bf r}'\;
\left[\phi_{{1 \over 2}}({\bf r...
...r 2}}({\bf r})\phi_{-{1 \over 2}}({\bf r})\right]\delta^{(3)}({\bf r}-{\bf r}')$
    \(\textstyle =\) $\displaystyle \int\;d{\bf r}\;\left(\vert\phi_{{1 \over 2}}({\bf r})\vert^2 +
\vert\phi_{-{1 \over 2}}({\bf r})\vert^2\right)$

    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,

    \begin{displaymath}
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]
\end{displaymath}

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

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

    This page titled 2.1: Representing states in the full Hilbert space is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark E. Tuckerman.

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