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# Posts Tagged ‘spin angular momentum’

## 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.

## PHY356F: Quantum Mechanics I. Dec 7 2010 ; Lecture 11 notes. Rotations and Angular momentum.

Posted by peeterjoot on December 7, 2010

## This time. Rotations (chapter 26).

Why are we doing the math? Because it applies to physical systems. Slides of IBM’s SEM quantum coral and others shown and discussed.

PICTURE: Standard right handed coordinate system with point $(x,y,z)$. We’d like to discuss how to represent this point in other coordinate systems, such as one with the $x,y$ axes rotated to $x',y'$ through an angle $\phi$.

Our problem is to find in the rotated coordinate system from $(x,y,z)$ to $(x', y', z')$.

There’s clearly a relationship between the representations. That relationship between $x', y', z'$ and $x,y,z$ for a counter-clockwise rotation about the $z$ axis is

\begin{aligned}x' &= x \cos \phi - y \sin\phi \\ y' &= x \sin \phi + y \cos\phi \\ z' &= z\end{aligned} \hspace{\stretch{1}}(13.214)

Treat $(x,y,z)$ and $(x',y',z')$ like vectors and write

\begin{aligned}\begin{bmatrix}x' \\ y' \\ z' \end{bmatrix}=\begin{bmatrix}\cos \phi &- \sin\phi & 0 \\ \sin \phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(13.217)

Or

\begin{aligned}\begin{bmatrix}x' \\ y' \\ z' \end{bmatrix}=R_z(\phi)\begin{bmatrix}x \\ y \\ z \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(13.218)

\paragraph{Q: Is $R_z(\phi)$ a unitary operator?}

Definition $U$ is unitary if $U^\dagger U = \mathbf{1}$, where $\mathbf{1}$ is the identity operator. We take Hermitian conjugates, which in this case is just the transpose since all elements of the matrix are real, and multiply

\begin{aligned}(R_z(\phi))^\dagger R_z(\phi) &=\begin{bmatrix}\cos \phi & \sin\phi & 0 \\ -\sin \phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}\cos \phi &- \sin\phi & 0 \\ \sin \phi & \cos\phi & 0 \\ 0 & 0 & 1\end{bmatrix} \\ &=\begin{bmatrix}\cos^2 \phi + \sin^2\phi & -\sin\phi \cos\phi + \sin\phi \cos\phi & 0 \\ -\cos\phi \sin\phi + \cos\phi \sin\phi & \cos^2\phi + \sin^2 \phi & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ &=\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \\ &= \mathbf{1}\end{aligned}

Apply the above to a vector $\mathbf{v} = (v_x, v_y, v_z)$ and write $\mathbf{v}' = (v_x', v_y', v_z')$. These are related as

\begin{aligned}\mathbf{v} = R_z(\phi) \mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.219)

Now we want to consider the infinitesimal case where we allow the rotation angle to get arbitrarily small. Consider this specific $z$ axis rotation case, and assume that $\phi$ is very small. Let $\phi = \epsilon$ and write

\begin{aligned}\mathbf{v}' &=\begin{bmatrix}v_x' \\ v_y' \\ v_z' \end{bmatrix}=R_z(\phi)\begin{bmatrix}v_x \\ v_y \\ v_z \end{bmatrix}=\begin{bmatrix}\cos \epsilon &- \sin\epsilon & 0 \\ \sin \epsilon & \cos\epsilon & 0 \\ 0 & 0 & 1\end{bmatrix} \mathbf{v} \\ &\approx\begin{bmatrix}1 &- \epsilon & 0 \\ \epsilon & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \mathbf{v} =\left(\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} +\begin{bmatrix}0 &- \epsilon & 0 \\ \epsilon & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \right)\mathbf{v} \end{aligned} \hspace{\stretch{1}}(13.220)

Define

\begin{aligned}S_z = i \hbar\begin{bmatrix}0 &- 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(13.222)

which is the generator of infinitesimal rotations about the $z$ axis.

Our rotated coordinate vector becomes

\begin{aligned}\mathbf{v}' &= \left(\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} +\frac{i \hbar \epsilon}{i\hbar}\begin{bmatrix}0 &- 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix} \right)\mathbf{v} \\ &=\left(\mathbf{1} + \frac{\epsilon}{i \hbar} S_z\right)\mathbf{v}\end{aligned}

Or

\begin{aligned}\mathbf{v}'=\left(\mathbf{1} - \frac{i \epsilon}{\hbar} S_z\right)\mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.223)

Many infinitesimal rotations can be combined to create a finite rotation via

\begin{aligned}\lim_{N \rightarrow \infty} \left( 1 + \frac{\alpha}{N} \right)^N = e^\alpha\end{aligned} \hspace{\stretch{1}}(13.224)

\begin{aligned}\alpha = -i \phi S_z/\hbar\end{aligned} \hspace{\stretch{1}}(13.225)

For a finite rotation

\begin{aligned}\mathbf{v}'=e^{ -i \frac{\phi S_z}{\hbar} }\mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.226)

Now think about transforming $g(x,y,z)$, an arbitrary function. Take $\epsilon$ is very small so that

\begin{aligned}x' &= x \cos \phi - y \sin\phi = x \cos \epsilon - y \sin\epsilon \approx x - y \epsilon \\ y' &= x \sin \phi + y \cos\phi = x \sin \epsilon + y \cos\epsilon \approx x \epsilon + y \\ z' &= z\end{aligned} \hspace{\stretch{1}}(13.227)

\paragraph{Question: Why can we assume that $\epsilon$ is small.}
\paragraph{Answer: We declare it to be small because it is simpler, and eventually build up to the general case where it is larger. We want to master the easy task before moving on to the more difficult ones.}

Our function is now transformed

\begin{aligned}g(x', y', z') \approx g( x - y \epsilon, y + x \epsilon, z) \\ &= g( x , y , z) - \epsilon y \frac{\partial {g}}{\partial {x}} + \epsilon x \frac{\partial {g}}{\partial {y}} + \cdots \\ &=\left( \mathbf{1} - \epsilon y \frac{\partial {}}{\partial {x}}+ \epsilon x \frac{\partial {}}{\partial {y}}\right)g( x, y ,z )\end{aligned}

Recall that the coordinate definition of the angular momentum operator is

\begin{aligned}L_z = -i \hbar \left( x \frac{\partial {}}{\partial {y}} - y \frac{\partial {}}{\partial {x}} \right) = x p_y - y p_x\end{aligned} \hspace{\stretch{1}}(13.230)

We can now write

\begin{aligned}g(x', y', z') &=\left( \mathbf{1} +\frac{-i \hbar \epsilon}{-i\hbar} \left(x \frac{\partial {}}{\partial {y}}- y \frac{\partial {}}{\partial {x}}\right)\right)g( x, y ,z ) \\ &=\left( \mathbf{1} +\frac{i \epsilon}{\hbar} L_z\right)g( x, y ,z )\end{aligned}

For a finite rotation with angle $\phi$ we have

\begin{aligned}g(x', y', z') =e^{i \frac{\phi L_z}{\hbar}}g( x, y ,z )\end{aligned} \hspace{\stretch{1}}(13.231)

\paragraph{Question: somebody says that the rotation is clockwise not counterclockwise.}

I didn’t follow the reasoning briefly mentioned on the board since it looks right to me. Perhaps this is the age old mixup between rotating the coordinates and the basis vectors. Review what’s in the text carefully. Can also check by

If you rotate a ket, and examine how the state representation of that ket changes under rotation, we have

\begin{aligned}{\lvert {x', y', z'} \rangle} = {\lvert {x - \epsilon y, y + \epsilon x, z} \rangle}\end{aligned} \hspace{\stretch{1}}(13.232)

Or

\begin{aligned}\left\langle{{\Psi}} \vert {{x', y', z'}}\right\rangle &= \Psi^{*}(x', y', z') \\ &=\Psi^{*}(x - \epsilon y, y + \epsilon x, z) \\ &=\Psi^{*}(x , y , z) - \epsilon \frac{\partial {\Psi^{*}}}{\partial {y}}+ \epsilon \frac{\partial {\Psi^{*}}}{\partial {x}} \\ &=\left( \mathbf{1} + \frac{i \epsilon} {\hbar} L_z \right) \Psi^{*}(x , y , z) \end{aligned}

Taking the complex conjugate we have

\begin{aligned}\Psi(x', y', z') \left( \mathbf{1} - \frac{i \epsilon} {\hbar} L_z \right) \Psi(x , y , z) \end{aligned} \hspace{\stretch{1}}(13.233)

For infinitesimal rotations about the $z$ axis we have for functions

\begin{aligned}\Psi(x', y', z') =e^{ - \frac{i \epsilon} {\hbar} L_z } \Psi(x , y , z) \end{aligned} \hspace{\stretch{1}}(13.234)

For finite rotations of a vector about the $z$ axis we have

\begin{aligned}\mathbf{v}'=e^{ - \frac{i \phi S_z} {\hbar} } \Psi(x , y , z) \mathbf{v}\end{aligned} \hspace{\stretch{1}}(13.235)

and for functions

\begin{aligned}\Psi(x', y', z') =e^{ - \frac{i \phi L_z} {\hbar} } \Psi(x , y , z) \end{aligned} \hspace{\stretch{1}}(13.236)

Vatche has mentioned some devices being researched right now where there is an attempt to isolate the spin orientation so that, say, only spin up or spin down electrons are allowed to flow. There are some possible interesting applications here to Quantum computation. Can we actually make a quantum computing device that is actually usable? We can make NAND devices as mentioned in the article above. Can this be scaled? We don’t know how to do this yet.

Recall that one description of a “particle” that has both a position and spin representation is

\begin{aligned}{\lvert {\Psi} \rangle} = {\lvert {u} \rangle} \otimes {\lvert {s m} \rangle}\end{aligned} \hspace{\stretch{1}}(13.237)

where we have a tensor product of kets. One usually just writes the simpler

\begin{aligned}{\lvert {u} \rangle} \otimes {\lvert {s m} \rangle} \equiv {\lvert {u} \rangle} {\lvert {s m} \rangle} \end{aligned} \hspace{\stretch{1}}(13.238)

An example of the above is

\begin{aligned}\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \\ \end{bmatrix}= \Bigl( {\langle {\mathbf{r}} \rvert} {\langle { s m} \rvert} \Bigr) {\lvert {\Psi} \rangle}\end{aligned} \hspace{\stretch{1}}(13.239)

where $u_1$ is spin component one. For $s=1$ this would be $m=-1, 0, 1$.

Here we have also used

\begin{aligned}{\lvert {\mathbf{r}} \rangle}= {\lvert {x} \rangle}\otimes{\lvert {y} \rangle}\otimes{\lvert {z} \rangle} \\ &={\lvert {x} \rangle}{\lvert {y} \rangle}{\lvert {z} \rangle} \\ &={\lvert {x y z} \rangle}\end{aligned}

We can now ask the question of how this thing transforms. We transform each component of this as a vector. The transformation of

\begin{aligned}\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \end{bmatrix}\end{aligned}

results in

\begin{aligned}{\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \end{bmatrix}}'=e^{ -i \phi (S_z + L_z)/\hbar }\begin{bmatrix}u_1(\mathbf{r}) \\ u_2(\mathbf{r}) \\ u_3(\mathbf{r}) \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(13.240)

Or with $J_z = S_z + L_z$

\begin{aligned}{\lvert {\Psi'} \rangle} = e^{-i \phi J_z/\hbar } {\lvert {\Psi} \rangle} \end{aligned} \hspace{\stretch{1}}(13.241)

Observe that this separates out nicely with the $S_z$ operation acting on the vector parts, and the $L_z$ operator acting on the functional dependence.

## My submission for PHY356 (Quantum Mechanics I) Problem Set 3.

Posted by peeterjoot on November 30, 2010

# Problem 1.

## Statement

A particle of mass $m$ is free to move along the x-direction such that $V(X)=0$. The state of the system is represented by the wavefunction Eq. (4.74)

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k x} e^{- i \omega t} f(k)\end{aligned} \hspace{\stretch{1}}(1.1)

with $f(k)$ given by Eq. (4.59).

\begin{aligned}f(k) &= N e^{-\alpha k^2}\end{aligned} \hspace{\stretch{1}}(1.2)

Note that I’ve inserted a $1/\sqrt{2\pi}$ factor above that isn’t in the text, because otherwise $\psi(x,t)$ will not be unit normalized (assuming $f(k)$ is normalized in wavenumber space).

\begin{itemize}
\item
(a) What is the group velocity associated with this state?
\item
(b) What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(c) What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(d) Explain the physical meaning of the above results.
\end{itemize}

## Solution

### (a). group velocity.

To calculate the group velocity we need to know the dependence of $\omega$ on $k$.

Let’s step back and consider the time evolution action on $\psi(x,0)$. For the free particle case we have

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} = -\frac{\hbar^2}{2m} \partial_{xx}.\end{aligned} \hspace{\stretch{1}}(1.3)

Writing $N' = N/\sqrt{2\pi}$ we have

\begin{aligned}-\frac{i t}{\hbar} H \psi(x,0) &= \frac{i t \hbar }{2m} N' \int_{-\infty}^\infty dk (i k)^2 e^{i k x - \alpha k^2} \\ &= N' \int_{-\infty}^\infty dk \frac{-i t \hbar k^2}{2m} e^{i k x - \alpha k^2}\end{aligned}

Each successive application of $-iHt/\hbar$ will introduce another power of $-it\hbar k^2/2 m$, so once we sum all the terms of the exponential series $U(t) = e^{-iHt/\hbar}$ we have

\begin{aligned}\psi(x,t) =N' \int_{-\infty}^\infty dk \exp\left( \frac{-i t \hbar k^2}{2m} + i k x - \alpha k^2 \right).\end{aligned} \hspace{\stretch{1}}(1.4)

Comparing with 1.1 we find

\begin{aligned}\omega(k) = \frac{\hbar k^2}{2m}.\end{aligned} \hspace{\stretch{1}}(1.5)

This completes this section of the problem since we are now able to calculate the group velocity

\begin{aligned}v_g = \frac{\partial {\omega(k)}}{\partial {k}} = \frac{\hbar k}{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

## (b). What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

In order to evaluate the probability, it looks desirable to evaluate the wave function integral 1.4.
Writing $2 \beta = i/(\alpha + i t \hbar/2m )$, the exponent of that integral is

\begin{aligned}-k^2 \left( \alpha + \frac{i t \hbar }{2m} \right) + i k x&=-\left( \alpha + \frac{i t \hbar }{2m} \right) \left( k^2 - \frac{i k x }{\alpha + \frac{i t \hbar }{2m} } \right) \\ &=-\frac{i}{2\beta} \left( (k - x \beta )^2 - x^2 \beta^2 \right)\end{aligned}

The $x^2$ portion of the exponential

\begin{aligned}\frac{i x^2 \beta^2}{2\beta} = \frac{i x^2 \beta}{2} = - \frac{x^2 }{4 (\alpha + i t \hbar /2m)}\end{aligned}

then comes out of the integral. We can also make a change of variables $q = k - x \beta$ to evaluate the remainder of the Gaussian and are left with

\begin{aligned}\psi(x,t) =N' \sqrt{ \frac{\pi}{\alpha + i t \hbar/2m} } \exp\left( - \frac{x^2 }{4 (\alpha + i t \hbar /2m)} \right).\end{aligned} \hspace{\stretch{1}}(1.7)

Observe that from 1.2 we can compute $N = (2 \alpha/\pi)^{1/4}$, which could be substituted back into 1.7 if desired.

Our probability density is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 &=\frac{1}{{2 \pi}} N^2 {\left\lvert{ \frac{\pi}{\alpha + i t \hbar/2m} }\right\rvert} \exp\left( - \frac{x^2}{4} \left( \frac{1}{{(\alpha + i t \hbar /2m)}} + \frac{1}{{(\alpha - i t \hbar /2m)}} \right) \right) \\ &=\frac{1}{{2 \pi}} \sqrt{\frac{2 \alpha}{\pi} } \frac{\pi}{\sqrt{\alpha^2 + (t \hbar/2m)^2 }} \exp\left( - \frac{x^2}{4} \frac{1}{{\alpha^2 + (t \hbar/2m)^2 }} \left( \alpha - i t \hbar /2m + \alpha + i t \hbar /2m \right)\right) \\ &=\end{aligned}

With a final regrouping of terms, this is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 =\sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right).\end{aligned} \hspace{\stretch{1}}(1.8)

As a sanity check we observe that this integrates to unity for all $t$ as desired. The probability that we find the particle at position $x > x_0$ is then

\begin{aligned}P_{x>x_0}(t) = \sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\int_{x=x_0}^\infty dx \exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right)\end{aligned} \hspace{\stretch{1}}(1.9)

The only simplification we can make is to rewrite this in terms of the complementary error function

\begin{aligned}\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^\infty e^{-t^2} dt.\end{aligned} \hspace{\stretch{1}}(1.10)

Writing

\begin{aligned}\beta(t) = \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 },\end{aligned} \hspace{\stretch{1}}(1.11)

we have

\begin{aligned}P_{x>x_0}(t_0) = \frac{1}{{2}} \text{erfc} \left( \sqrt{\beta(t_0)/2} x_0 \right)\end{aligned} \hspace{\stretch{1}}(1.12)

Sanity checking this result, we note that since $\text{erfc}(0) = 1$ the probability for finding the particle in the $x>0$ range is $1/2$ as expected.

## (c). What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

This unit length probability is thus

\begin{aligned}P_{x>x_0+1/2}(t_0) - P_{x>x_0-1/2}(t_0) &=\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0+\frac{1}{{2}} \right) \right) -\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0-\frac{1}{{2}} \right) \right) \end{aligned} \hspace{\stretch{1}}(1.13)

## (d). Explain the physical meaning of the above results.

To get an idea what the group velocity means, observe that we can write our wavefunction 1.1 as

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k (x - v_g t)} f(k)\end{aligned} \hspace{\stretch{1}}(1.14)

We see that the phase coefficient of the Gaussian $f(k)$ “moves” at the rate of the group velocity $v_g$. Also recall that in the text it is noted that the time dependent term 1.11 can be expressed in terms of position and momentum uncertainties $(\Delta x)^2$, and $(\Delta p)^2 = \hbar^2 (\Delta k)^2$. That is

\begin{aligned}\frac{1}{{\beta(t)}} = (\Delta x)^2 + \frac{(\Delta p)^2}{m^2} t^2 \equiv (\Delta x(t))^2\end{aligned} \hspace{\stretch{1}}(1.15)

This makes it evident that the probability density flattens and spreads over time with the rate equal to the uncertainty of the group velocity $\Delta p/m = \Delta v_g$ (since $v_g = \hbar k/m$). It is interesting that something as simple as this phase change results in a physically measurable phenomena. We see that a direct result of this linear with time phase change, we are less able to find the particle localized around it’s original time $x = 0$ position as more time elapses.

# Problem 2.

## Statement

A particle with intrinsic angular momentum or spin $s=1/2$ is prepared in the spin-up with respect to the z-direction state ${\lvert {f} \rangle}={\lvert {z+} \rangle}$. Determine

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.16)

and

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.17)

and explain what these relations say about the system.

## Solution: Uncertainty of $S_z$ with respect to ${\lvert {z+} \rangle}$

Noting that $S_z {\lvert {f} \rangle} = S_z {\lvert {z+} \rangle} = \hbar/2 {\lvert {z+} \rangle}$ we have

\begin{aligned}{\langle {f} \rvert} S_z {\lvert {f} \rangle} = \frac{\hbar}{2} \end{aligned} \hspace{\stretch{1}}(2.18)

The average outcome for many measurements of the physical quantity associated with the operator $S_z$ when the system has been prepared in the state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$ is $\hbar/2$.

\begin{aligned}\Bigl(S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \Bigr) {\lvert {f} \rangle}&= \frac{\hbar}{2} {\lvert {f} \rangle} -\frac{\hbar}{2} {\lvert {f} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.19)

We could also compute this from the matrix representations, but it is slightly more work.

Operating once more with $S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ on the zero ket vector still gives us zero, so we have zero in the root for 2.16

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = 0\end{aligned} \hspace{\stretch{1}}(2.20)

What does 2.20 say about the state of the system? Given many measurements of the physical quantity associated with the operator $V = (S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1})^2$, where the initial state of the system is always ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, then the average of the measurements of the physical quantity associated with $V$ is zero. We can think of the operator $V^{1/2} = S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ as a representation of the observable, “how different is the measured result from the average ${\langle {f} \rvert} S_z {\lvert {f} \rangle}$”.

So, given a system prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, and performance of repeated measurements capable of only examining spin-up, we find that the system is never any different than its initial spin-up state. We have no uncertainty that we will measure any difference from spin-up on average, when the system is prepared in the spin-up state.

## Solution: Uncertainty of $S_x$ with respect to ${\lvert {z+} \rangle}$

For this second part of the problem, we note that we can write

\begin{aligned}{\lvert {f} \rangle} = {\lvert {z+} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ).\end{aligned} \hspace{\stretch{1}}(2.21)

So the expectation value of $S_x$ with respect to this state is

\begin{aligned}{\langle {f} \rvert} S_x {\lvert {f} \rangle}&=\frac{1}{{2}}( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) S_x ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) \\ &=\hbar ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( {\lvert {x+} \rangle} - {\lvert {x-} \rangle} ) \\ &=\hbar ( 1 + 0 + 0 -1 ) \\ &= 0\end{aligned}

After repeated preparation of the system in state ${\lvert {f} \rangle}$, the average measurement of the physical quantity associated with operator $S_x$ is zero. In terms of the eigenstates for that operator ${\lvert {x+} \rangle}$ and ${\lvert {x-} \rangle}$ we have equal probability of measuring either given this particular initial system state.

For the variance calculation, this reduces our problem to the calculation of ${\langle {f} \rvert} S_x^2 {\lvert {f} \rangle}$, which is

\begin{aligned}{\langle {f} \rvert} S_x^2 {\lvert {f} \rangle} &=\frac{1}{{2}} \left( \frac{\hbar}{2} \right)^2 ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( (+1)^2 {\lvert {x+} \rangle} + (-1)^2 {\lvert {x-} \rangle} ) \\ &=\left( \frac{\hbar}{2} \right)^2,\end{aligned}

so for 2.22 we have

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = \frac{\hbar}{2}\end{aligned} \hspace{\stretch{1}}(2.22)

The average of the absolute magnitude of the physical quantity associated with operator $S_x$ is found to be $\hbar/2$ when repeated measurements are performed given a system initially prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$. We saw that the average value for the measurement of that physical quantity itself was zero, showing that we have equal probabilities of measuring either $\pm \hbar/2$ for this experiment. A measurement that would show the system was in the x-direction spin-up or spin-down states would find that these states are equi-probable.

I lost one mark on the group velocity response. Instead of 3.23 he wanted

\begin{aligned}v_g = {\left. \frac{\partial {\omega(k)}}{\partial {k}} \right\vert}_{k = k_0}= \frac{\hbar k_0}{m} = 0\end{aligned} \hspace{\stretch{1}}(3.23)

since $f(k)$ peaks at $k=0$.

I’ll have to go back and think about that a bit, because I’m unsure of the last bits of the reasoning there.

I also lost 0.5 and 0.25 (twice) because I didn’t explicitly state that the probability that the particle is at $x_0$, a specific single point, is zero. I thought that was obvious and didn’t have to be stated, but it appears expressing this explicitly is what he was looking for.

Curiously, one thing that I didn’t loose marks on was, the wrong answer for the probability per unit length. What he was actually asking for was the following

\begin{aligned}\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 - \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x_0, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.24)

That’s a whole lot more sensible seeming quantity to calculate than what I did, but I don’t think that I can be faulted too much since the phrase was never used in the text nor in the lectures.