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Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

by Pravegaa Education
Chapter 10

Spin

10.1  The Stern–Gerlach Experiment

  • 1921, Otto Stern and Walther Gerlach: a beam of neutral silver atoms from an oven is collimated and sent through an inhomogeneous magnetic field; a plate records the beam.

  • The magnetic moment of a silver atom comes from the spin of a single electron.

  • Uniform field \(\Rightarrow\) only a torque; inhomogeneous field \(\Rightarrow\) a net force that depends on the orientation of the moment.

  • Classical expectation: all orientations \(\Rightarrow\) a broad smear. Observed: the beam splits into two distinct parts — space quantization of spin.

Figure

Sequential Stern–Gerlach devices

  • Case 1: SG\(_z\) \(\to\) block \(z_-\) \(\to\) SG\(_z\): only \(z_+\) emerges — like classical polarization.

  • Case 2: SG\(_z\) \(\to\) block \(z_-\) \(\to\) SG\(_x\): both \(x_+\) and \(x_-\) emerge (50 : 50) — not classical.

  • Case 3: SG\(_z\) \(\to\) block \(z_-\) \(\to\) SG\(_x\) \(\to\) block \(x_-\) \(\to\) SG\(_z\): both \(z_+\) and \(z_-\) reappear. Measuring \(S_x\) destroys the earlier \(S_z\) information.

  • Intensities: \(I_0\xrightarrow{\hat z}I_0/2\xrightarrow{\hat y}I_0/4\) (\(+\)) and \(I_0/4\) (\(-\)) after a further \(\hat x\) stage, etc.

Figure

10.2  Spin Angular Momentum

  • By analogy with \(\hat L\): states \(\left|s,m_s\right\rangle\) with \(2s+1\) values of \(m_s\). SG shows two beams \(\Rightarrow2s+1=2\Rightarrow s=\tfrac12\), \(m_s=\pm\tfrac12\).

  • Notation: \(\left|\tfrac 12,\tfrac 12\right\rangle\equiv\left|\uparrow\right\rangle\equiv\chi_{+}\),   \(\left|\tfrac 12,-\tfrac 12\right\rangle\equiv\left|\downarrow\right\rangle\equiv\chi_{-}\).

  • Algebra identical to orbital angular momentum: \[\hat S^2\left|s,m_s\right\rangle=s(s+1)\hbar^2\left|s,m_s\right\rangle,\qquad \hat S_z\left|s,m_s\right\rangle=m_s\hbar\left|s,m_s\right\rangle,\] \[[\hat S_x,\hat S_y]=i\hbar\hat S_z\;(\text{cyclic}),\qquad [\hat S^2,\hat S_i]=0,\qquad [\hat S_z,\hat S_\pm]=\pm\hbar\hat S_\pm,\] \[\hat S_\pm\left|s,m_s\right\rangle=\sqrt{s(s+1)-m_s(m_s\pm1)}\,\hbar\left|s,m_s\pm 1\right\rangle.\]

  • Electron: \(\hat S^2\left|\tfrac 12,\pm\tfrac 12\right\rangle=\tfrac34\hbar^2\left|\tfrac 12,\pm\tfrac 12\right\rangle\),   \(\hat S_z\left|\tfrac 12,\pm\tfrac 12\right\rangle=\pm\tfrac\hbar2\left|\tfrac 12,\pm\tfrac 12\right\rangle\);

  • \(\hat S_+\left|\downarrow\right\rangle=\hbar\left|\uparrow\right\rangle\), \(\hat S_+\left|\uparrow\right\rangle=0\);   \(\hat S_-\left|\uparrow\right\rangle=\hbar\left|\downarrow\right\rangle\), \(\hat S_-\left|\downarrow\right\rangle=0\).

Matrix representation (basis \(\left|\uparrow\right\rangle,\left|\downarrow\right\rangle\))

\[\hat S^2=\frac34\hbar^2\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad \hat S_z=\frac\hbar2\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad \hat S_+=\hbar\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad \hat S_-=\hbar\begin{pmatrix}0&0\\1&0\end{pmatrix},\] \[\hat S_x=\frac{\hat S_++\hat S_-}{2}=\frac\hbar2\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \hat S_y=\frac{\hat S_+-\hat S_-}{2i}=\frac\hbar2\begin{pmatrix}0&-i\\i&0\end{pmatrix}.\]

Eigenvectors

Operator \(+\frac\hbar2\) \(-\frac\hbar2\)
\(\hat S_z\) \(\begin{pmatrix}1\\0\end{pmatrix}\) \(\begin{pmatrix}0\\1\end{pmatrix}\)
\(\hat S_x\) \(\frac1{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}\) \(\frac1{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}\)
\(\hat S_y\) \(\frac1{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}\) \(\frac1{\sqrt2}\begin{pmatrix}1\\-i\end{pmatrix}\)
  • \(\hat S^2\) has eigenvalue \(\tfrac34\hbar^2\), doubly degenerate; \(\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\) are the common eigenvectors of \(\hat S^2\) and \(\hat S_z\).

Space quantization of spin

  • \(|\vec S|=\dfrac{\sqrt3}{2}\hbar\) while \(S_z=\pm\dfrac\hbar2\): \[\cos\theta=\frac{\pm1/2}{\sqrt3/2}=\pm\frac{1}{\sqrt3}\]

  • \(\Rightarrow\theta\approx54.7^\circ\) or \(125.3^\circ\).

Figure

Example 10.1   Tilted Stern–Gerlach apparatus

Electrons with spin up along \(\hat z\) enter an SG apparatus whose field is at \(\theta=60^\circ\) from \(\hat z\) (in the \(xz\)-plane). Find the fractions emerging parallel and anti-parallel to the field.

Solution

  • Spin-up along \(\hat n(\theta,\phi)\): \(\left|+\right\rangle_{\hat n}=\begin{pmatrix}\cos\frac\theta2\\e^{i\phi}\sin\frac\theta2\end{pmatrix}\) — note the half angle.

  • Parallel: \(|\left\langle +_{\hat n}\middle|\uparrow\right\rangle |^2=\cos^2 30^\circ=\dfrac34\);   anti-parallel: \(\sin^230^\circ=\dfrac14\).

  • (Using \(\cos60^\circ\), \(\sin60^\circ\) as components — i.e. forgetting the half angle — wrongly gives \(\tfrac14\) and \(\tfrac34\).)

Example 10.2   Three spin-\(\frac12\) particles

\(\left|\psi\right\rangle=c_1\left|\uparrow\uparrow\uparrow\right\rangle+c_2\left|\uparrow\uparrow\downarrow\right\rangle+c_3\left|\uparrow\downarrow\uparrow\right\rangle+c_4\left|\uparrow\downarrow\downarrow\right\rangle+c_5\left|\downarrow\uparrow\uparrow\right\rangle+c_6\left|\downarrow\uparrow\downarrow\right\rangle+c_7\left|\downarrow\downarrow\uparrow\right\rangle+c_8\left|\downarrow\downarrow\downarrow\right\rangle\) for particles \(A,B,C\). A measurement finds \(A\) down. Find \(\left\langle S_z\right\rangle\) of \(C\).

Solution

  • Collapse: \(\left|\psi_{A\downarrow}\right\rangle\propto c_5\left|\downarrow\uparrow\uparrow\right\rangle+c_6\left|\downarrow\uparrow\downarrow\right\rangle+c_7\left|\downarrow\downarrow\uparrow\right\rangle+c_8\left|\downarrow\downarrow\downarrow\right\rangle\).

  • \(C\) up: \(|c_5|^2+|c_7|^2\);   \(C\) down: \(|c_6|^2+|c_8|^2\) (normalized by \(N=|c_5|^2+|c_6|^2+|c_7|^2+|c_8|^2\)): \[\left\langle S_z\right\rangle_C=\frac\hbar2\,\frac{|c_5|^2+|c_7|^2-|c_6|^2-|c_8|^2}{|c_5|^2+|c_6|^2+|c_7|^2+|c_8|^2}.\]

10.3  Pauli Spin Matrices

  • \(\hat{\vec S}=\dfrac\hbar2\vec\sigma\) with \[\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\qquad\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.\]

Properties

  • \(\sigma_x^2=\sigma_y^2=\sigma_z^2=I\); traceless; eigenvalues \(\pm1\); \(\det\sigma_i=-1\).

  • \(\sigma_x\sigma_y=i\sigma_z\), \(\sigma_y\sigma_z=i\sigma_x\), \(\sigma_z\sigma_x=i\sigma_y\) (reverse order gives \(-i\)): \(\sigma_i\sigma_j=\delta_{ij}I+i\epsilon_{ijk}\sigma_k\).

  • \([\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k\);   \(\{\sigma_i,\sigma_j\}=2\delta_{ij}I\).

  • \((\vec a\cdot\vec\sigma)(\vec b\cdot\vec\sigma)=(\vec a\cdot\vec b)I+i(\vec a\times\vec b)\cdot\vec\sigma\).

  • For a unit vector \(\hat n\): \((\hat n\cdot\vec\sigma)^{2p}=I\), \((\hat n\cdot\vec\sigma)^{2q+1}=\hat n\cdot\vec\sigma\), hence \[\exp\big[ia(\hat n\cdot\vec\sigma)\big]=I\cos a+i(\hat n\cdot\vec\sigma)\sin a.\]

  • Eigenvectors of \(\sigma_x,\sigma_y,\sigma_z\) are those of \(\hat S_x,\hat S_y,\hat S_z\) above, with eigenvalues \(\pm1\).

Spin rotation matrix

  • \(U[R(\theta)]=\exp\Big(-\dfrac{i\,\vec\theta\cdot\vec S}{\hbar}\Big)=\exp\Big(-\dfrac{i\theta}{2}\hat m\cdot\vec\sigma\Big)=I\cos\dfrac\theta2-i\sin\dfrac\theta2\,\hat m\cdot\vec\sigma\).

  • Rotation about \(x\): \(R_x(\theta)=\begin{pmatrix}\cos\frac\theta2&-i\sin\frac\theta2\\-i\sin\frac\theta2&\cos\frac\theta2\end{pmatrix}\).

  • To turn \(\hat z\) into \(\hat n(\theta,\phi)\), rotate by \(\theta\) about \(\hat m=(-\sin\phi,\cos\phi,0)\): \[U=\begin{pmatrix}\cos\frac\theta2&-\sin\frac\theta2\,e^{-i\phi}\\\sin\frac\theta2\,e^{i\phi}&\cos\frac\theta2\end{pmatrix}.\]

  • A \(2\pi\) rotation gives \(U=-I\): spinors change sign.

Example 10.3   The operator \(\hat n\cdot\vec\sigma\)

\(\hat A=\hat n\cdot\vec\sigma\). (a) Write its matrix in terms of \(\theta,\phi\). (b) For \(\hat n\) in the \(xz\)-plane find its eigenvalues.

Solution

  • (a)  \(\hat A=\sin\theta\cos\phi\,\sigma_x+\sin\theta\sin\phi\,\sigma_y+\cos\theta\,\sigma_z=\begin{pmatrix}\cos\theta&\sin\theta\,e^{-i\phi}\\\sin\theta\,e^{i\phi}&-\cos\theta\end{pmatrix}\).

  • (b)  \(\phi=0\): \(\begin{vmatrix}\cos\theta-\lambda&\sin\theta\\\sin\theta&-\cos\theta-\lambda\end{vmatrix}=\lambda^2-1=0\Rightarrow\lambda=\pm1\) (for any \(\hat n\)).

  • Eigenvectors: \(\left|+\right\rangle_{\hat n}=\begin{pmatrix}\cos\frac\theta2\\e^{i\phi}\sin\frac\theta2\end{pmatrix}\),   \(\left|-\right\rangle_{\hat n}=\begin{pmatrix}-e^{-i\phi}\sin\frac\theta2\\\cos\frac\theta2\end{pmatrix}\).

Example 10.4   Time-evolution operator in a field

\(H=\mu\vec\sigma\cdot\vec B\), \(\omega=\mu B/\hbar\). Show \(e^{iHt/\hbar}=I\cos\omega t+\dfrac{i\vec\sigma\cdot\vec B}{B}\sin\omega t\).

Solution

  • \(\dfrac{iHt}{\hbar}=i\,\dfrac{\mu Bt}{\hbar}\,(\hat n\cdot\vec\sigma)=i\omega t\,(\hat n\cdot\vec\sigma)\) with \(\hat n=\vec B/B\).

  • Use \(e^{ia\hat n\cdot\vec\sigma}=I\cos a+i\hat n\cdot\vec\sigma\sin a\) with \(a=\omega t\).

Example 10.5   Rotation about \(x\)

Spins in the \(\sigma_z=+1\) state are rotated by \(60^\circ\) about \(x\). Find the fractions with \(\sigma_z=+1\) and \(-1\).

Solution

  • \(R_x(\pi/3)\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}\cos30^\circ\\-i\sin30^\circ\end{pmatrix}=\begin{pmatrix}\sqrt3/2\\-i/2\end{pmatrix}\).

  • \(P(+1)=\tfrac34\),   \(P(-1)=\tfrac14\).

Example 10.6   Precession in an arbitrary field

An electron (\(g\)-factor \(g\), \(\vec S=\frac\hbar2\vec\sigma\)) is at rest in \(\vec B=(B_x,B_y,B_z)\), all components non-zero. At \(t=0\) it is in \(s_z=+\frac12\). Find the earliest time at which the spin state is orthogonal to the initial state.

Solution

  • \(H=-\vec\mu\cdot\vec B=\dfrac{ge}{2m}\vec S\cdot\vec B=E_0|B|\,\hat n\cdot\vec\sigma\),   \(E_0=\dfrac{ge\hbar}{4m}\); eigenvalues \(\pm E_0|B|\).

  • Eigenvectors \(\left|\phi_1\right\rangle=\left|+\right\rangle_{\hat n}\), \(\left|\phi_2\right\rangle=\left|-\right\rangle_{\hat n}\); initially \(\left|\uparrow\right\rangle=\cos\frac\theta2\,e^{-i\phi/2}\left|\phi_1\right\rangle-\sin\frac\theta2\,e^{i\phi/2}\left|\phi_2\right\rangle\) (up to phase).

  • \(\left\langle \psi(0)\middle|\psi(t)\right\rangle =\cos^2\frac\theta2\,e^{-iE_0|B|t/\hbar}+\sin^2\frac\theta2\,e^{iE_0|B|t/\hbar}\).

  • This vanishes only if \(\cos^2\frac\theta2=\sin^2\frac\theta2\), i.e. \(\theta=\pi/2\) (\(B_z=0\)). With \(B_z\ne0\) the spin precesses on a cone about \(\vec B\) and never becomes orthogonal.

  • If \(B_z=0\): \(\cos(E_0|B|t/\hbar)=0\Rightarrow t=\dfrac{\pi\hbar}{2E_0|B|}=\dfrac{2\pi m}{ge|B|}\).

Example 10.7   Eigenvalues of \(\hat S_x+\hat S_y\)

Solution

  • \(\hat S_x+\hat S_y=\dfrac\hbar2\begin{pmatrix}0&1-i\\1+i&0\end{pmatrix}\);   \(\lambda^2-\dfrac{\hbar^2}{4}\cdot2=0\Rightarrow\lambda=\pm\dfrac{\hbar}{\sqrt2}\).

Example 10.8   Measurements on \(\binom{3+4i}{3}\)

For \(\left|\psi\right\rangle=\binom{3+4i}{3}\) find the outcomes and probabilities of (a) \(\hat S^2\), (b) \(\hat S_z\), (c) \(\hat S_x\).

Solution

  • \(\left\langle \psi\middle|\psi\right\rangle =25+9=34\).

  • (a)  \(\tfrac34\hbar^2\) with probability 1.

  • (b)  \(P(+\tfrac\hbar2)=\dfrac{25}{34}\),   \(P(-\tfrac\hbar2)=\dfrac{9}{34}\).

  • (c)  \(\left\langle x_+\middle|\psi\right\rangle =\dfrac{6+4i}{\sqrt2}\), \(\left\langle x_-\middle|\psi\right\rangle =\dfrac{4i}{\sqrt2}\) \(\Rightarrow P(+\tfrac\hbar2)=\dfrac{26}{34}=\dfrac{13}{17}\),   \(P(-\tfrac\hbar2)=\dfrac{8}{34}=\dfrac{4}{17}\).

Example 10.9   Change of basis and time evolution

\(\left|\psi\right\rangle=\sqrt{\tfrac23}\left|\uparrow\right\rangle+\sqrt{\tfrac13}\left|\downarrow\right\rangle\). (a) Write \(\left|\psi\right\rangle\) in the \(\hat S_x\) eigenbasis \(\left|\phi_\pm\right\rangle\). (b) Find \(\left\langle \hat S_x\right\rangle\). (c) With \(H=-\mu_0B\hat S_x\) find \(\left|\psi(t)\right\rangle\).

Solution

  • (a)  Insert \(I=\left|\phi_+\right\rangle\left\langle\phi_+\right|+\left|\phi_-\right\rangle\left\langle\phi_-\right|\): \[\left|\psi\right\rangle=\Big(\frac1{\sqrt3}+\frac1{\sqrt6}\Big)\left|\phi_+\right\rangle+\Big(\frac1{\sqrt3}-\frac1{\sqrt6}\Big)\left|\phi_-\right\rangle=\frac{\sqrt2+1}{\sqrt6}\left|\phi_+\right\rangle+\frac{\sqrt2-1}{\sqrt6}\left|\phi_-\right\rangle.\]

  • (b)  \(\left\langle \hat S_x\right\rangle=\dfrac\hbar2\big(P_+-P_-\big)=\dfrac\hbar2\cdot\dfrac{4}{\sqrt{18}}=\dfrac{\sqrt2}{3}\hbar\).

  • (c)  \(E_\pm=\mp\dfrac{\mu_0B\hbar}{2}\):   \(\left|\psi(t)\right\rangle=\dfrac{\sqrt2+1}{\sqrt6}e^{i\mu_0Bt/2}\left|\phi_+\right\rangle+\dfrac{\sqrt2-1}{\sqrt6}e^{-i\mu_0Bt/2}\left|\phi_-\right\rangle\).

Example 10.10   General spinor

\(\left|\psi\right\rangle=\cos\alpha\,\chi_{+}+\sin\alpha\,e^{i\beta}\chi_{-}\) (\(\alpha,\beta\) real). Find \(P(S_x=+\frac\hbar2)\).

Solution

  • \(\left|\phi_+\right\rangle=\frac{1}{\sqrt2}(\chi_++\chi_-)\):   \(P=\tfrac12\big|\cos\alpha+\sin\alpha\,e^{i\beta}\big|^2=\tfrac12\big(1+\sin2\alpha\cos\beta\big)\)

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