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Next: Many-electron atoms Up: lecture_9 Previous: Electron spin

Spin wave functions

So far, we have not explicitly considered the spin of the electrons. For the next type of approximation we will consider, the so-called valene bond approximation, we will need to consider spin explicitly. As we have already seen in the case of the hydrogen atom, the wave functions of an electron depend on the coordinates ${\bf r}= (x,y,z)$ or $(r,\phi,\theta)$ and on the $z$-component of spin spin $S_z$ (the total spin $\vert{\bf S}\vert$ is fixed). While the coordinates can be anything, the spin $S_z$ can only take on two values $\hbar/2$ and $-\hbar/2$. Recall that these lead to the two values of the spin quantum number $m_s = \pm 1/2$. The value $m_s=1/2$ is what we call the ``spin-up'' state and $m_s=-1/2$ is what we call ``spin-down''.



We now define the spin wave functions. The spin-up wave function is denoted

\begin{displaymath}
\psi_{1/2}(S_z) = \psi_{\uparrow}(S_z)
\end{displaymath}

Since $S_z$ can take on only two values $\pm \hbar/2$, the spin wave function only have two values:

\begin{displaymath}
\psi_{\uparrow}(\hbar/2) = 1\;\;\;\;\;\;\;\psi_{\uparrow}(-\hbar/2) = 0
\end{displaymath}

The meaning of this wave function is that when the electron is in the spin-up state, the probability that a measurement of $S_z$ will yield the value $\hbar/2$ is $P(S_z = \hbar/2) = \vert\psi_{\uparrow}(\hbar/2)\vert^2 = 1$ and the probability that is value will be $-\hbar/2$ is 0. Similarly, the spin-down wave function is

\begin{displaymath}
\psi_{-1/2}(S_z) = \psi_{\downarrow}(S_z)
\end{displaymath}

where

\begin{displaymath}
\psi_{\downarrow}(\hbar/2) = 0\;\;\;\;\;\;\;\;\;\;
\psi_{\downarrow}(-\hbar/2) = 1
\end{displaymath}

so that the probability that a measurement of $S_z$ yields the value $-\hbar/2$ is 1 and that its value is $\hbar/2$ is 0. Note that the spin wave functions are normalized, meaning that they satisfy

\begin{displaymath}
\sum_{S_z = -\hbar/2}^{\hbar/2} \vert\psi_{m_s}(S_z)\vert^2 = 1
\end{displaymath}

and they are orthogonal, meaning that they satisfy

\begin{displaymath}
\sum_{S_z = -\hbar/2}^{\hbar/2}\psi_{\uparrow}(S_z)\psi_{\downarrow}(S_z) = 0
\end{displaymath}



Now, for a hydrogen atom, there are four quantum numbers, $n,l,m,m_s$, and the wave function depends on four coordinates, $r,\theta,\phi,s_z$. While $r,\theta,\phi$ are continuous, $s_z$ takes on only two values. The wave function can be expressed as a simple product

\begin{displaymath}
\psi_{nlmm_s}(r,\theta,\phi,s_z) = \psi_{nlm}(r,\theta,\phi)\psi_{m_s}(s_z)
\end{displaymath}

As a shorthand notation, we can generally represent the complete set of spatial and spin coordinates with a vector ${\bf x}$. The spatial coordinates can be $r,\theta,\phi$ or $x,y,z$ or any other set of spatial coordinates useful for a given problem. For the hydrogen atom, using this notation, we would write

\begin{displaymath}
\psi_{nlmm_s}({\bf x}) = \psi_{nlm}(r,\theta,\phi)\psi_{m_s}(s_z)
\end{displaymath}


next up previous
Next: Many-electron atoms Up: lecture_9 Previous: Electron spin
Mark E. Tuckerman 2011-11-03