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## PHY456H1F: Quantum Mechanics II. Lecture 16 (Taught by Prof J.E. Sipe). Hydrogen atom with spin, and two spin systems.

Posted by peeterjoot on November 2, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# The hydrogen atom with spin.

READING: what chapter of [1] ?

For a spinless hydrogen atom, the Hamiltonian was

\begin{aligned}H = H_{\text{CM}} \otimes H_{\text{rel}}\end{aligned} \hspace{\stretch{1}}(2.1)

where we have independent Hamiltonian’s for the motion of the center of mass and the relative motion of the electron to the proton.

The basis kets for these could be designated ${\left\lvert {\mathbf{p}_\text{CM}} \right\rangle}$ and ${\left\lvert {\mathbf{p}_\text{rel}} \right\rangle}$ respectively.

Now we want to augment this, treating

\begin{aligned}H = H_{\text{CM}} \otimes H_{\text{rel}} \otimes H_{\text{s}}\end{aligned} \hspace{\stretch{1}}(2.2)

where $H_{\text{s}}$ is the Hamiltonian for the spin of the electron. We are neglecting the spin of the proton, but that could also be included (this turns out to be a lesser effect).

We’ll introduce a Hamiltonian including the dynamics of the relative motion and the electron spin

\begin{aligned}H_{\text{rel}} \otimes H_{\text{s}}\end{aligned} \hspace{\stretch{1}}(2.3)

Covering the Hilbert space for this system we’ll use basis kets

\begin{aligned}{\left\lvert {nlm\pm} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}\begin{aligned}{\left\lvert {nlm+} \right\rangle} &\rightarrow \begin{bmatrix}\left\langle{{\mathbf{r}+}} \vert {{nlm+}}\right\rangle \\ \left\langle{{\mathbf{r}-}} \vert {{nlm+}}\right\rangle \\ \end{bmatrix}=\begin{bmatrix}\Phi_{nlm}(\mathbf{r}) \\ 0\end{bmatrix} \\ {\left\lvert {nlm-} \right\rangle} &\rightarrow \begin{bmatrix}\left\langle{{\mathbf{r}+}} \vert {{nlm-}}\right\rangle \\ \left\langle{{\mathbf{r}-}} \vert {{nlm-}}\right\rangle \\ \end{bmatrix}=\begin{bmatrix}0 \\ \Phi_{nlm}(\mathbf{r}) \end{bmatrix}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.5)

Here $\mathbf{r}$ should be understood to really mean $\mathbf{r}_\text{rel}$. Our full Hamiltonian, after introducing a magnetic pertubation is

\begin{aligned}H = \frac{P_\text{CM}^2}{2M} + \left(\frac{P_\text{rel}^2}{2\mu}-\frac{e^2}{R_\text{rel}}\right)- \boldsymbol{\mu}_0 \cdot \mathbf{B}- \boldsymbol{\mu}_s \cdot \mathbf{B}\end{aligned} \hspace{\stretch{1}}(2.6)

where

\begin{aligned}M = m_\text{proton} + m_\text{electron},\end{aligned} \hspace{\stretch{1}}(2.7)

and

\begin{aligned}\frac{1}{{\mu}} = \frac{1}{{m_\text{proton}}} + \frac{1}{{m_\text{electron}}}.\end{aligned} \hspace{\stretch{1}}(2.8)

For a uniform magnetic field

\begin{aligned}\boldsymbol{\mu}_0 &= \left( -\frac{e}{2 m c} \right) \mathbf{L} \\ \boldsymbol{\mu}_s &= g \left( -\frac{e}{2 m c} \right) \mathbf{S}\end{aligned} \hspace{\stretch{1}}(2.9)

We also have higher order terms (higher order multipoles) and relativistic corrections (like spin orbit coupling [2]).

# Two spins.

Example: Consider two electrons, 1 in each of 2 quantum dots.

\begin{aligned}H = H_{1} \otimes H_{2}\end{aligned} \hspace{\stretch{1}}(3.11)

where $H_1$ and $H_2$ are both spin Hamiltonian’s for respective 2D Hilbert spaces. Our complete Hilbert space is thus a 4D space.

We’ll write

\begin{aligned}\begin{aligned}{\left\lvert {+} \right\rangle}_1 \otimes {\left\lvert {+} \right\rangle}_2 &= {\left\lvert {++} \right\rangle} \\ {\left\lvert {+} \right\rangle}_1 \otimes {\left\lvert {-} \right\rangle}_2 &= {\left\lvert {+-} \right\rangle} \\ {\left\lvert {-} \right\rangle}_1 \otimes {\left\lvert {+} \right\rangle}_2 &= {\left\lvert {-+} \right\rangle} \\ {\left\lvert {-} \right\rangle}_1 \otimes {\left\lvert {-} \right\rangle}_2 &= {\left\lvert {--} \right\rangle} \end{aligned}\end{aligned} \hspace{\stretch{1}}(3.12)

Can introduce

\begin{aligned}\mathbf{S}_1 &= \mathbf{S}_1^{(1)} \otimes I^{(2)} \\ \mathbf{S}_2 &= I^{(1)} \otimes \mathbf{S}_2^{(2)}\end{aligned} \hspace{\stretch{1}}(3.13)

Here we “promote” each of the individual spin operators to spin operators in the complete Hilbert space.

We write

\begin{aligned}S_{1z}{\left\lvert {++} \right\rangle} &= \frac{\hbar}{2} {\left\lvert {++} \right\rangle} \\ S_{1z}{\left\lvert {+-} \right\rangle} &= \frac{\hbar}{2} {\left\lvert {+-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.15)

Write

\begin{aligned}\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2,\end{aligned} \hspace{\stretch{1}}(3.17)

for the full spin angular momentum operator. The $z$ component of this operator is

\begin{aligned}S_z = S_{1z} + S_{2z}\end{aligned} \hspace{\stretch{1}}(3.18)

\begin{aligned}S_z{\left\lvert {++} \right\rangle} &= (S_{1z} + S_{2z}) {\left\lvert {++} \right\rangle} = \left( \frac{\hbar}{2} +\frac{\hbar}{2} \right) {\left\lvert {++} \right\rangle} = \hbar {\left\lvert {++} \right\rangle} \\ S_z{\left\lvert {+-} \right\rangle} &= (S_{1z} + S_{2z}) {\left\lvert {+-} \right\rangle} = \left( \frac{\hbar}{2} -\frac{\hbar}{2} \right) {\left\lvert {+-} \right\rangle} = 0 \\ S_z{\left\lvert {-+} \right\rangle} &= (S_{1z} + S_{2z}) {\left\lvert {-+} \right\rangle} = \left( -\frac{\hbar}{2} +\frac{\hbar}{2} \right) {\left\lvert {-+} \right\rangle} = 0 \\ S_z{\left\lvert {--} \right\rangle} &= (S_{1z} + S_{2z}) {\left\lvert {--} \right\rangle} = \left( -\frac{\hbar}{2} -\frac{\hbar}{2} \right) {\left\lvert {--} \right\rangle} = -\hbar {\left\lvert {--} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.19)

So, we find that ${\left\lvert {x x} \right\rangle}$ are all eigenkets of $S_z$. These will also all be eigenkets of $\mathbf{S}_1^2 = S_{1x}^2 +S_{1y}^2 +S_{1z}^2$ since we have

\begin{aligned}S_1^2 {\left\lvert {x x} \right\rangle} &= \hbar^2 \left(\frac{1}{{2}}\right) \left(1 + \frac{1}{{2}}\right) {\left\lvert {x x} \right\rangle} = \frac{3}{4} \hbar^2 {\left\lvert {x x} \right\rangle} \\ S_2^2 {\left\lvert {x x} \right\rangle} &= \hbar^2 \left(\frac{1}{{2}}\right) \left(1 + \frac{1}{{2}}\right) {\left\lvert {x x} \right\rangle} = \frac{3}{4} \hbar^2 {\left\lvert {x x} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.23)

\begin{aligned}\begin{aligned}S^2 &= (\mathbf{S}_1^2+\mathbf{S}_2^2) \cdot(\mathbf{S}_1^2+\mathbf{S}_2^2) \\ &= S_1^2 + S_2^2 + 2 \mathbf{S}_1 \cdot \mathbf{S}_2\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.25)

Are all the product kets also eigenkets of $S^2$? Calculate

\begin{aligned}S^2 {\left\lvert {+-} \right\rangle} &= (S_1^2 + S_2^2 + 2 \mathbf{S}_1 \cdot \mathbf{S}_2) {\left\lvert {+-} \right\rangle} \\ &=\left(\frac{3}{4}\hbar^2+\frac{3}{4}\hbar^2\right)+ 2 S_{1x} S_{2x} {\left\lvert {+-} \right\rangle} + 2 S_{1y} S_{2y} {\left\lvert {+-} \right\rangle} + 2 S_{1z} S_{2z} {\left\lvert {+-} \right\rangle} \end{aligned}

For the $z$ mixed terms, we have

\begin{aligned}2 S_{1z} S_{2z} {\left\lvert {+-} \right\rangle} = 2 \left(\frac{\hbar}{2}\right)\left(-\frac{\hbar}{2}\right){\left\lvert {+-} \right\rangle}\end{aligned} \hspace{\stretch{1}}(3.26)

So

\begin{aligned}S^2{\left\lvert {+-} \right\rangle} = \hbar^2 {\left\lvert {+-} \right\rangle} + 2 S_{1x} S_{2x} {\left\lvert {+-} \right\rangle} + 2 S_{1y} S_{2y} {\left\lvert {+-} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.27)

Since we have set our spin direction in the z direction with

\begin{aligned}{\left\lvert {+} \right\rangle} &\rightarrow \begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\left\lvert {-} \right\rangle} &\rightarrow \begin{bmatrix}0 \\ 1 \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.28)

We have

\begin{aligned}S_x{\left\lvert {+} \right\rangle} &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix} =\frac{\hbar}{2}\begin{bmatrix}0 \\ 1 \end{bmatrix}=\frac{\hbar}{2} {\left\lvert {-} \right\rangle} \\ S_x{\left\lvert {-} \right\rangle} &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix} =\frac{\hbar}{2}\begin{bmatrix}1 \\ 0 \end{bmatrix}=\frac{\hbar}{2} {\left\lvert {+} \right\rangle} \\ S_y{\left\lvert {+} \right\rangle} &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix} =\frac{i\hbar}{2}\begin{bmatrix}0 \\ 1 \end{bmatrix}=\frac{i\hbar}{2} {\left\lvert {-} \right\rangle} \\ S_y{\left\lvert {-} \right\rangle} &\rightarrow \frac{\hbar}{2} \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix} =\frac{-i\hbar}{2}\begin{bmatrix}1 \\ 0 \end{bmatrix}=-\frac{i\hbar}{2} {\left\lvert {+} \right\rangle} \\ \end{aligned}

And are able to arrive at the action of $S^2$ on our mixed composite state

\begin{aligned}S^2{\left\lvert {+-} \right\rangle} = \hbar^2 ({\left\lvert {+-} \right\rangle} + {\left\lvert {-+} \right\rangle} ).\end{aligned} \hspace{\stretch{1}}(3.30)

For the action on the ${\left\lvert {++} \right\rangle}$ state we have

\begin{aligned}S^2 {\left\lvert {++} \right\rangle} &=\left(\frac{3}{4}\hbar^2 +\frac{3}{4}\hbar^2\right){\left\lvert {++} \right\rangle} + 2 \frac{\hbar^2}{4} {\left\lvert {--} \right\rangle} + 2 i^2 \frac{\hbar^2}{4} {\left\lvert {--} \right\rangle} +2 \left(\frac{\hbar}{2}\right)\left(\frac{\hbar}{2}\right){\left\lvert {++} \right\rangle} \\ &=2 \hbar^2 {\left\lvert {++} \right\rangle} \\ \end{aligned}

and on the ${\left\lvert {--} \right\rangle}$ state we have

\begin{aligned}S^2 {\left\lvert {--} \right\rangle} &=\left(\frac{3}{4}\hbar^2 +\frac{3}{4}\hbar^2\right){\left\lvert {--} \right\rangle} + 2 \frac{(-\hbar)^2}{4} {\left\lvert {++} \right\rangle} + 2 i^2 \frac{\hbar^2}{4} {\left\lvert {++} \right\rangle} +2 \left(-\frac{\hbar}{2}\right)\left(-\frac{\hbar}{2}\right){\left\lvert {--} \right\rangle} \\ &=2 \hbar^2 {\left\lvert {--} \right\rangle} \end{aligned}

All of this can be assembled into a tidier matrix form

\begin{aligned}S^2\rightarrow \hbar^2\begin{bmatrix}2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.31)

where the matrix is taken with respect to the (ordered) basis

\begin{aligned}\{{\left\lvert {++} \right\rangle},{\left\lvert {+-} \right\rangle},{\left\lvert {-+} \right\rangle},{\left\lvert {--} \right\rangle}\}.\end{aligned} \hspace{\stretch{1}}(3.32)

However,

\begin{aligned}\left[{S^2},{S_z}\right] &= 0 \\ \left[{S_i},{S_j}\right] &= i \hbar \sum_k \epsilon_{ijk} S_k\end{aligned} \hspace{\stretch{1}}(3.33)

It should be possible to find eigenkets of $S^2$ and $S_z$

\begin{aligned}S^2 {\left\lvert {s m_s} \right\rangle} &= s(s+1)\hbar^2 {\left\lvert {s m_s} \right\rangle} \\ S_z {\left\lvert {s m_s} \right\rangle} &= \hbar m_s {\left\lvert {s m_s} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.35)

An orthonormal set of eigenkets of $S^2$ and $S_z$ is found to be

\begin{aligned}\begin{array}{l l}{\left\lvert {++} \right\rangle} & \mbox{latex s = 1and $m_s = 1$} \\ \frac{1}{{\sqrt{2}}} \left( {\left\lvert {+-} \right\rangle} + {\left\lvert {-+} \right\rangle} \right) & \mbox{$s = 1$ and $m_s = 0$} \\ {\left\lvert {–} \right\rangle} & \mbox{$s = 1$ and $m_s = -1$} \\ \frac{1}{{\sqrt{2}}} \left( {\left\lvert {+-} \right\rangle} – {\left\lvert {-+} \right\rangle} \right) & \mbox{$s = 0$ and $m_s = 0$}\end{array}\end{aligned} \hspace{\stretch{1}}(3.37)

The first three kets here can be grouped into a triplet in a 3D Hilbert space, whereas the last treated as a singlet in a 1D Hilbert space.

Form a grouping

\begin{aligned}H = H_1 \otimes H_2\end{aligned} \hspace{\stretch{1}}(3.38)

Can write

\begin{aligned}\frac{1}{{2}} \otimes \frac{1}{{2}} = 1 \oplus 0\end{aligned} \hspace{\stretch{1}}(3.39)

where the $1$ and $0$ here refer to the spin index $s$.

## Other examples

Consider, perhaps, the $l=5$ state of the hydrogen atom

\begin{aligned}J_1^2 {\left\lvert {j_1 m_1} \right\rangle} &= j_1(j_1+1)\hbar^2 {\left\lvert {j_1 m_1} \right\rangle} \\ J_{1z} {\left\lvert {j_1 m_1} \right\rangle} &= \hbar m_1 {\left\lvert {j_1 m_1} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.40)

\begin{aligned}J_2^2 {\left\lvert {j_2 m_2} \right\rangle} &= j_2(j_2+1)\hbar^2 {\left\lvert {j_2 m_2} \right\rangle} \\ J_{2z} {\left\lvert {j_2 m_2} \right\rangle} &= \hbar m_2 {\left\lvert {j_2 m_2} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.42)

Consider the Hilbert space spanned by ${\left\lvert {j_1 m_1} \right\rangle} \otimes {\left\lvert {j_2 m_2} \right\rangle}$, a $(2 j_1 + 1)(2 j_2 + 1)$ dimensional space. How to find the eigenkets of $J^2$ and $J_z$?

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] Wikipedia. Spin.orbit interaction — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 2-November-2011]. http://en.wikipedia.org/w/index.php?title=Spin\%E2\%80\%93orbit_interaction&oldid=451606718.