ECE 405 - Introduction to Quantum Mechanics

LaTeX Commands

• $\newcommand{\bra}[1]{\langle{#1}\rvert}$ $\bra{+}$: $\bra{+}$
• $\newcommand{\ket}[1]{\lvert{#1}\rangle}$ $\ket{+}$: $\ket{+}$
• $\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}$ $\braket{+}{+}$: $\braket{+}{+}$
• $\newcommand{\ketbra}[2]{\lvert{#1}\rangle\langle{#2}\rvert}$ $\ketbra{+}{+}$: $\ketbra{+}{+}$
• $\newcommand{\sandwich}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}$ $\sandwich{+}{A}{+}$: $\sandwich{+}{A}{+}$
• $\newcommand{\amp}[1]{{\lvert{#1}\rvert}^{2}}$ $\amp{\braket{\pm}{\psi}}$: $\amp{\braket{\pm}{\psi}}$
• $\newcommand{\avg}[1]{\langle{#1}\rangle}$ $\avg{S_{z}}$: $\avg{S_{z}}$
• $\newcommand{\det}[1]{\text{det}\lvert{#1}\rvert}$ $\det{A - \lambda I}$: $\det{A - \lambda I}$

Stern-Gerlach Experiments

Overview

• Produced by an oven, a beam of neutral silver atoms was split into two in its passage through an inhomogenous magnetic field.

Classical vs. Quantum Interpretation

1. The experiment suggests an interaction between a neutral particle and a magnetic field.
2. Accordingly, the particle possesses a magnetic moment $\mathbf{\mu}$ such that the potential energy of this interaction is $E = - \mathbf{\mu} \cdot \mathbf{B}$, which results in a force $\mathbf{F} = \nabla(\mathbf{\mu} \cdot \mathbf{B})$.
3. Because the magnetic field gradient is in the $z$-direction, the force is perpendicular to the direction of motion and deflects the beam in proportion to the component of the magnetic moment in the direction of the magnetic field gradient.
4. A charged particle in an atom has an orbital angular momentum $L$ and an intrinsic angular momentum, spin, $S$. $$ \begin{aligned} \mathbf{\mu} &= \frac{q}{2m} \mathbf{L} \\ \mathbf{\mu} &= g \frac{q}{2m} \mathbf{S} \end{aligned} $$
   • where $q$ is the charge.
   • where $m$ is the mass.
   • where $g$ is the gyroscopic ratio.
5. The magnetic moment of a neutral silver atom is the magnetic moment of its only outermost electron. $$ \begin{aligned} \mathbf{\mu} &= -g \frac{e}{2m_{e}} \mathbf{S} \\ F_{z} &= -g \frac{2}{2m_{e}} S_{z} \frac{\delta B_{z}}{\delta z} \\ S_{z} &= \lvert \mathbf{S} \rvert \cos \theta \end{aligned} $$
6. Thus, the deflection of the beam in the Stern-Gerlach experiment is a measure of the component $S_{z}$ of the spin along the $z$-axis, which is the orientation of the magnetic field gradient.
   • Classical: A continuous distribution of spin components from $-\lvert \mathbf{S} \rvert$ to $+\lvert \mathbf{S} \rvert$ was expected to be observed.
   • Quantum: A discrete distribution of spin components $S_{z} = \pm \frac{\hbar}{2}$ was actually observed.
     • where $\hbar = 6.5821 \times 10^{-16} eV \cdot s$ is Planck's constant $h$ divided by $2\pi$.
7. The result of the Stern-Gerlach experiment is evidence of the quantization of the electron's spin angular momentum component along an axis.

Stern-Gerlach Device

• Analyzer: A Stern-Gerlach device labeled with the axis along which the magnetic field is oriented.
• Observable: The physical quantity that is measured.
• The up and down arrows indicate the two possible measurement results for the device: spin up and spin down, respectively.
  • Upper Output Beam Ket: $\ket{+} \implies S_{z} = +\frac{\hbar}{2}$
  • Bottom Output Beam Ket: $\ket{-} \implies S_{z} = -\frac{\hbar}{2}$
  • Input Beam Ket: $\ket{\psi}$

Experiment 1

• Key Observation: The first analyzer (i.e., state preparation device) prepares the beam in a particular quantum state ($\ket{+}$) and the second analyzer measures the resultant beam. By preparing the state with the first analyzer, the details of the source of atoms can be ignored.

Experiment 2

• Key Observation: It is impossible to predict which of the second analyzer output ports any particular atom will come out.

Experiment 3

• Key Observation: The second analyzer has disturbed the system such that the spin component along the $z$-axis does not have a unique value, even though we measured it with the first analyzer.
• There is a fundamental incompatibility in trying to measure the spin component of the atom along two different directions; thus, $S_{x}$ and $S_{z}$ are incompatible observables.

Experiment 4

• Experiment 4a ~ Experiment 3.
• Experiment 4b ~ Experiment 3 because Experiment 2.
• Key Observation: By combining the two beams from the second analyzer, the quantum mechanical disturbance that was evident in Experiments 3, 4a, and 4b are avoided.

Stern-Gerlach Experiments and Quantum Mechanics

1. Spin measurements are quantized.
   • The possible results of a spin component measurement are quantized. Only these discrete values are measured.
2. Quantum mechanics is probabilistic.
   • We cannot predict the results of experiments precisely. We can predict only the probability that a certain result is obtained in a measurement.
   • Experiment 2.
3. Quantum measurements disturb the system.
   • Measuring one physical observable can "destroy" information about other observables.
   • Experiment 3 & 4.

Quantum State Vectors

• Kets $\implies$ Quantum State Vectors $\implies$ Hilbert Space.

Quantum Mechanical Basis Vectors

• Normalization: $\mathbf{\hat{i}} \cdot \mathbf{\hat{i}} = \mathbf{\hat{j}} \cdot \mathbf{\hat{j}} = \mathbf{\hat{k}} \cdot \mathbf{\hat{k}} = 1$
  • $\braket{+}{+} = 1$
  • $\braket{-}{-} = 1$
• Orthogonality: $\mathbf{\hat{i}} \cdot \mathbf{\hat{j}} = \mathbf{\hat{i}} \cdot \mathbf{\hat{k}} = \mathbf{\hat{j}} \cdot \mathbf{\hat{k}} = 0$
  • $\braket{+}{-} = 0$
  • $\braket{-}{+} = 0$
• Completeness: $\mathbf{A} = a_{x}\mathbf{\hat{i}} + a_{y}\mathbf{\hat{j}} + a_{z}\mathbf{\hat{k}}$
  • $\ket{\psi} = a\ket{+} + b\ket{-}$
• Thus, all possible keys can be expressed as a linear combination of these quantum mechanical basis vectors.

Kets (Quantum State Vector)(Column Vectors)

• Ket:$\ket{\psi} = a\ket{+} + b\ket{-}$
  • where $a$ and $b$ are complex scalar numbers.

Bras (Conjugate Quantum State Vector)(Row Vectors)

• Bra: $\bra{\psi} = a^{*}\bra{+} + b^{*}\bra{-}$
  • where $a$ and $b$ are complex scalar conjugates.

Inner Product

• Bracket: $\braket{bra}{ket} \implies \text{Dot Product}$
  • Note 1: A product of kets $\ket{+}\ket{+}$ is meaningless.
  • Note 2: A product of bras $\bra{+}\bra{+}$ is meaningless.
• Property 1: Distributive.
• Property 2: Scalars Move Freely Through Bras/Kets.
• Property 3: Normalization. $$ \begin{aligned} \braket{+}{\psi} &= a \\ \braket{-}{\psi} &= b \\ \braket{\psi}{+} &= a^{*} \\ \braket{\psi}{-} &= b^{*} \\ \braket{\phi}{\psi} &= \braket{\psi}{\phi}^{*} \\ \braket{\psi}{\psi} &= 1 \\ \end{aligned} $$

Cartesian Axes

$$ \begin{aligned} \ket{+} &= \ket{+}\\ \ket{-} &= \ket{-}\\ \ket{+}_{x} &= \frac{1}{\sqrt{2}}\left[ \ket{+} + \ket{-} \right]\\ \ket{-}_{x} &= \frac{1}{\sqrt{2}}\left[ \ket{+} - \ket{-} \right]\\ \ket{+}_{y} &= \frac{1}{\sqrt{2}}\left[ \ket{+} + i\ket{-} \right]\\ \ket{-}_{y} &= \frac{1}{\sqrt{2}}\left[ \ket{+} - i\ket{-} \right] \end{aligned} $$

Probability in Spin-1/2 System

• The probability of obtaining the value $\pm\frac{\hbar}{2}$ in a measurement of the observable $S_{z}$ on a system in the state $\ket{\psi}$ is the following. $$P_{\pm} = \amp{\braket{\pm}{\psi}}$$
  • where $\ket{\pm}$ is the basis ket of $S_{z}$ corresponding to the result $\pm\frac{\hbar}{2}$.
  • where $\amp{\braket{\pm}{\psi}}$ is the probability amplitude.
• $\braket{out}{in} = \braket{in}{out}^{*}$

Analysis of Experiment 1

$$ \begin{aligned} \mathcal{P}_{+} &= \amp{\braket{+}{+}} = 1\\ \mathcal{P}_{-} &= \amp{\braket{-}{+}} = 0 \end{aligned} $$

1. The first analyzer prepared the second analyzer an input state of $\ket{+}$.
2. The second analyzer measured this input state $\ket{+}$, so the probabilities can be calculated by evaluating the probability amplitudes for spin up and spin down.

Analysis of Experiment 2

1. The spin up and the spin down variants for the first analyzer yields the same experimental results. $$ \begin{aligned} \mathcal{P}_{1, +x} &= \amp{{}_{x}\braket{+}{+}} = \frac{1}{2}\\ \mathcal{P}_{1, -x} &= \amp{{}_{x}\braket{-}{+}} = \frac{1}{2}\\ \mathcal{P}_{2, +x} &= \amp{{}_{x}\braket{+}{-}} = \frac{1}{2}\\ \mathcal{P}_{2, -x} &= \amp{{}_{x}\braket{-}{-}} = \frac{1}{2} \end{aligned} $$
2. The kets $\ket{+}$ and $\ket{-}$ are known as they are the basis kets of a spin-1/2 system; the kets $\ket{+}_{x}$ and $\ket{-}_{x}$ are unknown.
3. Because of completeness, the kets $\ket{+}_{x}$ and $\ket{-}_{x}$ can be expressed by the following. $$ \begin{aligned} \ket{+}_{x} &= a\ket{+} + b\ket{-} \implies \bra{+}_{x} = a^{*}\bra{+} + b^{*}\bra{-}\\ \ket{-}_{x} &= c\ket{+} + d\ket{-} \implies \bra{+}_{x} = c^{*}\bra{+} + d^{*}\bra{-} \end{aligned} $$
   • where $a$, $b$, $c$, and $d$ are unknown complex coefficients.
4. Using the experimental results, the unknown complex coefficients can be determined. $$ \amp{a} = \amp{b} = \amp{c} = \amp{d} = \frac{1}{2} $$
5. Because each coefficient is complex, each has an amplitude and a phase. However, only the relative phase between different components of a quantum state vector is physically measurable. Thus, only one coefficient of each vector can be chosen to be real and positive without any loss of generality. $$ \begin{aligned} \ket{+}_{x} &= \frac{1}{\sqrt{2}}\left[ \ket{+} + e^{i\alpha}\ket{-} \right]\\ \ket{-}_{x} &= \frac{1}{\sqrt{2}}\left[ \ket{+} + e^{i\beta}\ket{-} \right] \end{aligned} $$
   • where $\alpha$ and $\beta$ are unknown relative phases.
6. If Experiment 1 was performed with both analyzers aligned along the $x$-axis, the $\ket{\pm}_{x}$ form the $S_{x}$ basis. $$ \begin{aligned} {}_{x}\braket{+}{+}_{x} &= 1\\ {}_{x}\braket{-}{+}_{x} &= 0\\ \end{aligned} $$
7. Using the observation that the kets $\ket{+}_{x}$ and $\ket{-}_{x}$ are orthogonal, the following can be derived. $$e^{i\alpha} = -e^{i\beta}$$
8. Accordingly, let $\alpha = 0 \implies e^{i\alpha} = 1 \land e^{i\beta} = -1$. $$ \begin{aligned} \ket{+}_{x} &= \frac{1}{\sqrt{2}}\left[ \ket{+} + \ket{-} \right]\\ \ket{-}_{x} &= \frac{1}{\sqrt{2}}\left[ \ket{+} - \ket{-} \right] \end{aligned} $$

Matrix Notation

$$ \begin{aligned} \ket{+} &= \begin{pmatrix}1 \\ 0\end{pmatrix}\\ \ket{-} &= \begin{pmatrix}0 \\ 1\end{pmatrix}\\ \ket{+}_{x} &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}\\ \ket{-}_{x} &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}\\ \ket{+}_{y} &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ i\end{pmatrix}\\ \ket{-}_{y} &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -i\end{pmatrix}\\ \ket{\psi} &= \begin{pmatrix}a \\ b\end{pmatrix}\\ \bra{\psi} &= \left(a^{*} \quad b^{*}\right) \end{aligned} $$

General Quantum Systems

$$ \begin{aligned} \braket{a_{i}}{a_{j}} &= \delta_{ij} \quad \text{orthonormality}\\ \ket{\psi} &= \sum_{i} \braket{a_{i}}{\psi}\ket{a_{i}} \quad \text{completeness}\\ \delta_{ij} &= \begin{cases}0 & i \ne j \\ 1 & i = j\end{cases}\\ \mathcal{P}_{a_{n}} &= \amp{\braket{a_{n}}{\psi_{in}}} \end{aligned} $$

Postulates of Quantum Mechanics

1. The state of a quantum mechanical system, including all the information you can know about it, is represented mathematically by a normalized ket $\ket{\psi}$.
2. A physical observable is represented mathematically by an operator $A$ that acts on kets.
3. The only possible result of a measurement of an observable is one of the eigenvalues $a_{n}$ of the corresponding operator $A$.
4. The probability of obtaining the eigenvalue $a_{n}$ in a measurement of the observable $A$ on the system in the state $\ket{\psi}$ is: $$\mathcal{P}_{a_{n}} = \amp{\braket{a_{n}}{\psi}}$$
   • where $\ket{a_{n}}$ is the normalized eigenvector of $A$ corresponding to the eigenvalue $a_{n}$.
5. After a measurement of $A$ that yields the result $a_{n}$, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement: $$\ket{\psi'} = \frac{P_{n}\ket{\psi}}{\sqrt{\sandwich{\psi}{P_{n}}{\psi}}}$$
6. The time evolution of a quantum system is determined by the Hamiltonian or total energy operator $H(t)$ through a Schrodinger equation: $$i\hbar\frac{d}{dt}\ket{\psi(t)} = H(t)\ket{\psi(t)}$$

Operators and Measurement

Operators, Eigenvalues, and Eigenvectors

• An operator is a mathematical object that acts or operates on a ket and transforms it into a new ket: $A\ket{\psi} = \ket{\phi}$.
• An ket that is not changed by an operator, except for a possible multiplicative constant, is $A\ket{\psi} = a\ket{\psi}$.
  • where $A$ is the operator.
  • where $\ket{\psi}$ is the eigenvector.
  • where $a$ is the eigenvalue.
• In an eigenvalue equation, the observable is represented by an operator, the eigenvalue is one of the possible measurement results of the observable, and the eigenvector is the ket corresponding to the chosen eigenvalue of the operator.
• Important Note 1: An operator is always diagonal in its own basis.
• Important Note 2: Eigenvectors are unit vectors in their own basis.

Matrices

Matrix Addition

$$C_{ij} = A_{ij} + B_{ij}$$$$A + B = \begin{pmatrix}a & b\\ c & d\end{pmatrix} + \begin{pmatrix}e & f\\ g & h\end{pmatrix} = \begin{pmatrix}a + e & b + f\\ c + g & d + h\end{pmatrix}$$

Matrix Multiplication

$$C_{ij} = \sum^{n}_{k = 1} A_{ik} B_{kj}$$$$AB = \begin{pmatrix}a & b\\ c & d\end{pmatrix} \begin{pmatrix}e & f\\ g & h\end{pmatrix} = \begin{pmatrix}ae + bg & af + bh\\ ce + dg & cf + dh \end{pmatrix}$$

• To find the matrix element $C_{ij}$ in the $i$th row and $j$th column of $C$, that the scalar product of the $i$th row of $A$ and the $j$th column of $B$.

Determinant of a Matrix

$$\det{A} = \begin{vmatrix}a & b\\ c & d\end{vmatrix} = ad - bc$$

Matrix Representation of Operators

$$ \begin{aligned} A &= \begin{pmatrix}a & b\\ c & d\end{pmatrix}\\ A\ket{+} &= \begin{pmatrix}a & b\\ c & d\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}a \\ c\end{pmatrix}\\ \sandwich{+}{A}{+} &= \begin{pmatrix}1 & 0\end{pmatrix} \begin{pmatrix}a \\ c\end{pmatrix} = a\\ A &= \begin{pmatrix}\sandwich{+}{A}{+} & \sandwich{+}{A}{-}\\ \sandwich{-}{A}{+} & \sandwich{-}{A}{-}\end{pmatrix}\\ \sandwich{\text{bra}}{\text{OPERATOR}}{\text{ket}} &= \sandwich{\text{row}}{\text{OPERATOR}}{\text{column}} \end{aligned} $$

Diagonalization of Operators

1. Experimental Results $+$ Eigenvalue Equation $\implies$ Matrix Representation.
2. Matrix Represenation $\implies$ Theoretical Results.

General Operator in Two-State System

$$ \begin{aligned} A\ket{a_{n}} &= a_{n}\ket{a_{n}}\\ A\ket{a_{n}} &= \begin{pmatrix}A_{11} & A_{12}\\ A_{21} & A_{22}\end{pmatrix} \begin{pmatrix}c_{n1} \\ c_{n2}\end{pmatrix} = a_{n}\begin{pmatrix}c_{n1} \\ c_{n2}\end{pmatrix}\\ \end{aligned} $$

• where $A$ is the operator.
• where $a_{n}$ is the corresponding eigenvalue.
• where $\ket{a_{n}}$ is the corresponding eigenvector.
• where $c_{n1}$ and $c_{n2}$ are the unknown coefficients of the eigenvector $\ket{a_{n}}$ for the eigenvalue $a_{n}$.

Solve for the eigenvalues and eigenvectors by solving $\det{A - \lambda I} = 0$.

Two-State System Operators/Eigenvectors

New Operators: Spin Component in a General Direction

1. Let $S_{n}$ be the operator for th spin component along a general direction $\mathbf{\hat{n}}$.
2. Let $\mathbf{\hat{n}}$ be a unit vector specified by the polar and azimuthal angles $\theta$ and $\phi$. $$\mathbf{\hat{n}} = \mathbf{\hat{i}}\sin{\theta}\cos{\phi} + \mathbf{\hat{j}}\sin{\theta}\sin{\phi} + \mathbf{\hat{k}}\cos{\theta}$$
3. The spin component along this direction is obtained by projecting the spin vector $\mathbf{S}$ onto this new unit vector. $$ \begin{aligned} S_{n} &= \mathbf{S \cdot \hat{n}} = S_{x}\sin{\theta}\cos{\phi} + S_{y}\sin{\theta}\sin{\phi} + S_{z}\cos{\theta}\\ S_{n} &= \frac{\hbar}{2}\begin{pmatrix}\cos{\theta} & \sin{\theta}e^{-i\phi}\\ \sin{\theta}e^{i\phi} & -\cos{\theta} \end{pmatrix} \end{aligned} $$
4. Eigenvalues: $\pm\frac{\hbar}{2}$
5. Eigenvectors: $$ \begin{aligned} \ket{+}_{n} &= \cos{\frac{\theta}{2}}\ket{+} + \sin{\frac{\theta}{2}}e^{i\phi}\ket{-}\\ \ket{-}_{n} &= \sin{\frac{\theta}{2}}\ket{+} - \cos{\frac{\theta}{2}}e^{i\phi}\ket{-} \end{aligned} $$
6. General State: $$\ket{\psi} = \lvert{a}\rvert\ket{+} + \sqrt{1 - {\lvert{a}\rvert}^{2}}e^{i\phi}\ket{-}$$

Hermitian Operators

• Hermitian Adjoint of $A$: The operator $A^{\dagger}$ is the transposition-and-complex-conjugation of $A$. $$ \begin{aligned} \ket{\phi} &= A\ket{\psi}\\ \braket{\phi}{\beta} &= \braket{\beta}{\phi}^{*}\\ \sandwich{\psi}{A^{+}}{\beta} &= \sandwich{\beta}{A}{\psi}^{*} \end{aligned} $$
• If an operator $A \iff A^{\dagger}$, then $\bra{\phi} = \bra{\psi}A \iff \ket{\phi} = A\ket{\psi}$.

Projection Operators

• Identity Operator: $\ketbra{+}{+} + \ketbra{-}{-} = I$
• Projection Operator,+: $P_{+} = \ketbra{+}{+} = \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$
• Projection Operator,-: $P_{-} = \ketbra{-}{-} = \begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}$
• When a projection operator for a particular eigenstate acts on a state $\ket{\psi}$, it produces a new ket that is aligned along the eigenstate and has a magnitude equal to the amplitude (including the phase) for the state $\ket{\psi}$ to be in that eigenstate.

Analysis of Experiments 3 and 4

• Important Note 1: The total probability is the product of the individual probabilities of each measurement: second analyzer and third analyzer.
• Important Note 2: The product of individual probabilities is read right to left.

Figure 2.7(a)

1. Upper Positive Probability: $\mathcal{P}_{\text{upper},+} = \amp{\braket{+}{+}_{x}} \amp{{}_{x}\braket{+}{+}}$
   1. Second Analyzer: $S_{x} = +\hbar/2 \implies \amp{{}_{x}\braket{+}{+}}$
   2. Third Analyzer: $S_{z} = +\hbar/2 \implies \amp{\braket{+}{+}_{x}}$
2. Upper Negative Probability: $\mathcal{P}_{\text{upper},-} = \amp{\braket{-}{+}_{x}} \amp{{}_{x}\braket{+}{+}}$
   1. Second Analyzer: $S_{x} = +\hbar/2 \implies \amp{{}_{x}\braket{+}{+}}$
   2. Third Analyzer: $S_{z} = -\hbar/2 \implies \amp{\braket{-}{+}_{x}}$
3. Lower Positive Probability: $\mathcal{P}_{\text{lower},+} = \amp{\braket{+}{-}_{x}} \amp{{}_{x}\braket{-}{+}}$
   1. Second Analyzer: $S_{x} = -\hbar/2 \implies \amp{{}_{x}\braket{-}{+}}$
   2. Third Analyzer: $S_{z} = +\hbar/2 \implies \amp{\braket{+}{-}_{x}}$
4. Lower Negative Probability: $\mathcal{P}_{\text{lower},-} = \amp{\braket{-}{-}_{x}} \amp{{}_{x}\braket{-}{+}}$
   1. Second Analyzer: $S_{x} = -\hbar/2 \implies \amp{{}_{x}\braket{-}{+}}$
   2. Third Analyzer: $S_{z} = -\hbar/2 \implies \amp{\braket{-}{-}_{x}}$

Figure 2.7(b)

Input State of Third Analyzer* $\iff$ Output State of Second Analyzer

1. The output state of the second analyzer is aligned along the sum of the two projection operators for each port. $$\ket{\psi_{2}} = \frac{(P_{+x} + P_{-x})\ket{+}}{\sqrt{\sandwich{+}{(P_{+x} + P_{-x})}{+}}}$$
2. The sum of the two projection operators, $P_{+x} + P_{-x}$, is the identity operator, so the denominator is one. $$\ket{\psi_{2}} = \ket{+}_{x}{}_{x}\braket{+}{+} + \ket{-}_{x}{}_{x}\braket{-}{+}$$
3. Thus, the input state of the third analyzer is the coherent superposition of the eigenstates of the second analyzer. $$ \begin{aligned} \mathcal{P}_{+} &= \amp{\braket{+}{\psi_{2}}}\\ &= \amp{\braket{+}{+}_{x}{}_{x}\braket{+}{+} + \braket{+}{-}_{x}{}_{x}\braket{-}{+}}\\ &= \mathcal{P}_{\text{upper},+} + \mathcal{P}_{\text{lower},+} + \text{Interference}\\ \mathcal{P}_{-} &= \amp{\braket{-}{\psi_{2}}}\\ &= \amp{\braket{-}{+}_{x}{}_{x}\braket{+}{+} + \braket{-}{-}_{x}{}_{x}\braket{-}{+}}\\ &= \mathcal{P}_{\text{upper},-} + \mathcal{P}_{\text{lower},-} + \text{Interference} \end{aligned} $$
4. The interference arises from the nature of the superposition of states that enter the third analyzer.
5. If no measurement is made on the intermediate state, then we add amplitudes and then square to find the probability, while if an intermediate measurement is performed (i.e., watching), then we square the amplitudes first and then add to find the probability. One is the square of a sum and the other is the sum of squares, and only the former exhibits interference.

Measurement

• The expected value of a measurement is the sum of the products of each possible result and its probability.
  • Spin-1/2 Measurement: $\avg{S_{z}} = \left(+\cfrac{\hbar}{2}\right)\mathcal{P}_{+} + \left(-\cfrac{\hbar}{2}\right)\mathcal{P}_{-} = \sandwich{\psi}{S_{z}}{\psi}$
• The expectation value for a general quantum mechanical observable is the following. $$\avg{A} = \sandwich{\psi}{A}{\psi} = \sum_{n} a_{n} \mathcal{P}_{a_{n}}$$
• The root-mean-square deviation for a general quantum mechanical observable is the following. $$\Delta{A} = \sqrt{\avg{A^{2}} - \avg{A}^{2}}$$
• The standard deviation $\Delta{A}$ represents the minimum uncertainty in the results of an experiment.

Commuting Observables

• A commutator of two operators is the difference between the products of the two operators taken in alternate orders. $$[A,B] = AB - BA$$
  • If $[A,B] = 0$, then the operators commute / compatible.
• If operators commute, then they share common eigenstates; these observables can be measured simultaneously.

Uncertainty Principle

• The uncertainty principle of quantum mechanics states the uncertainties that would arise in any experiment is at least the following. $$\Delta{A}\Delta{B} \ge \frac{1}{2}\lvert\avg{[A,B]}\rvert$$

$\mathbf{S}^{2}$ Operator

• The $\mathbf{S}^{2}$ operator represents the magnitude of the spin vector without any information about the direction. $$\mathbf{S}^{2} = S_{x}^{2} + S_{y}^{2} + S_{z}^{2} = \frac{3}{4}\hbar^{2}\mathbf{I}$$
  • The $\mathbf{S}^{2}$ operator is proportional to the identity operator.
  • The $\mathbf{S}^{2}$ operator commutes with $S_{x}$, $S_{y}$, and $S_{z}$.
  • All states are eigenstates of the $\mathbf{S}^{2}$ operator: $\mathbf{S}^{2}\ket{\psi} = \frac{3}{4}\hbar^{2}\ket{\psi}$.
• As the length of the spin vector is $\lvert{\mathbf{S}}\rvert = \sqrt{\avg{\mathbf{S}^{2}}} = \sqrt{3}\frac{\hbar}{2}$, which is longer than the measured component of $\hbar/2$, the spin vector can never be fully aligned along any axis.

Schrödinger Time Evolution

Schrödinger Equation

• The time evolution of a quantim system is determined by the Hamiltonian or total energy operator $H(t)$ through the Schrödinger equation. $$i\hbar\frac{d}{dt}\ket{\psi(t)} = H(t)\ket{\psi(t)}$$

Hamiltonian Operator ($H$)

1. $H$ is an observable.
2. $H$ is a Hermitian operator.
3. $\ket{E_{n}}$ is the energy eigenvectors of $H$.
4. $H\ket{E_{n}} = E_{n}\ket{E_{n}}$ is the energy eigenvalue equation of $H$.
5. The energy eigenvectors of $H$ form a complete basis, energy basis. $$ \begin{aligned} \ket{\psi(t)} &= \sum_{n} c_{n}(t) \ket{E_{n}}\\ \braket{E_{k}}{E_{n}} &= \delta_{kn} \end{aligned} $$
   • Where $c_{n}(t)$ are the expansion coefficients that contain the time dependence of a general state $\ket{\psi}$.
6. The energy eigenstates of $H$ are stationary states, if a system begins and remains in that state.

Time-Independent Hamiltonian ($H$)

• The time evolution of an original state vector under the action of the time-independent Hamiltonian $H$ is the following. $$ \begin{aligned} \ket{\psi(0)} &= \sum_{n} c_{n} \ket{E_{n}}\\ \ket{\psi(t)} &= \sum_{n} c_{n} e^{-i E_{n} t / \hbar} \ket{E_{n}} \end{aligned} $$
  • Where $e^{-i E_{n} t / \hbar}$ is the complex exponential, phase factor for the eigenstate $E_{n}$.

No Input Superposition / No Ouput Superposition

1. Let $\ket{\psi(0)} = \ket{E_1}$ be the initial input state.
2. Let $\ket{\psi(t)} = e^{-i E_{1} t / \hbar} \ket{E_{1}}$ be the time evolution of the initial input state.
3. Let $A$ be the output observable whose eigenstate correspond to the eigenvalue $a_{1}$.

$$ \begin{aligned} \mathcal{P}_{a_{1}} &= \amp{\braket{a_{1}}{\psi(t)}}\\ &= \amp{\sandwich{a_{1}}{e^{-i E_{1} t / \hbar}}{E_{1}}}\\ &= \amp{\braket{a_{1}}{E_{1}}} \end{aligned} $$

• Observation: The probability is time independent and is equal to the probability at the initial time.

Input Superposition / No Output Superposition

1. Let $\ket{\psi(0)} = c_{1}\ket{E_1} + c_{2}\ket{E_2}$ be the initial input state.
2. Let $\ket{\psi(t)} = c_{1}e^{-i E_{1} t / \hbar} \ket{E_{1}} + c_{2}e^{-i E_{2} t / \hbar} \ket{E_{2}}$ be the time evolution of the initial input state.
3. Let $A$ be the output observable whose eigenstate correspond to the eigenvalue $a_{1}$ such that $\ket{a_{1}} = \alpha_{1}\ket{E_{1}}$.

$$ \begin{aligned} \mathcal{P}_{a_{1}} &= \amp{\braket{a_{1}}{\psi(t)}}\\ &= \amp{\alpha_{1}^{*}\bra{E_{1}}\left[ c_{1}e^{-i E_{1} t / \hbar} \ket{E_{1}} + c_{2}e^{-i E_{2} t / \hbar} \ket{E_{2}} \right]}\\ &= \amp{\alpha_{1}}\amp{c_{1}} \end{aligned} $$

• Observation: The probability is time independent and is equal to the probability at the initial time.

Input Superposition / Output Superposition

1. Let $\ket{\psi(0)} = c_{1}\ket{E_1} + c_{2}\ket{E_2}$ be the initial input state.
2. Let $\ket{\psi(t)} = c_{1}e^{-i E_{1} t / \hbar} \ket{E_{1}} + c_{2}e^{-i E_{2} t / \hbar} \ket{E_{2}}$ be the time evolution of the initial input state.
3. Let $A$ be the output observable whose eigenstate correspond to the eigenvalue $a_{1}$ such that $\ket{a_{1}} = \alpha_{1}\ket{E_{1}} + \alpha_{2}\ket{E_{2}}$.

$$ \begin{aligned} \mathcal{P}_{a_{1}} &= \amp{\braket{a_{1}}{\psi(t)}}\\ &= \amp{\left[ \alpha_{1}^{*}\bra{E_{1}} + \alpha_{2}^{*}\bra{E_{2}} \right] \left[ c_{1}e^{-i E_{1} t / \hbar} \ket{E_{1}} + c_{2}e^{-i E_{2} t / \hbar} \ket{E_{2}} \right]}\\ &= \amp{\alpha_{1}^{*}c_{1}e^{-i E_{1} t / \hbar} + \alpha_{2}^{*}c_{2}e^{-i E_{2} t / \hbar}}\\ &= \amp{e^{-i E_{1} t / \hbar}}\amp{\alpha_{1}^{*}c_{1} + \alpha_{2}^{*}c_{2}e^{-i (E_{2} - E_{1}) t / \hbar}}\\ &= \amp{\alpha_{1}}\amp{c_{1}} + \amp{\alpha_{2}}\amp{c_{2}} + 2\text{Re}\left( \alpha_{1}c_{1}^{*}\alpha_{2}c_{2}^{*}e^{-i (E_{2} - E_{1}) t / \hbar} \right) \end{aligned} $$

• Observation: The probability is time dependent because of the different time-evolution phases of the two components of $\ket{\psi(t)}$.

Bohr Frequency

• The time dependence is determined by the difference of the energies of the two states involved in the superposition. $$\omega_{21} = \frac{E_{2} - E_{1}}{\hbar}$$

Solving Problems with Time-Independent Hamiltonian ($H$)

Given a Hamiltonian $H$ and an initial state $\ket{\psi(0)}$, what is the probability that the eigenvalue $a_{j}$ of the observable $A$ is measured at time $t$?

1. Diagonalize $H$ by finding the eigenvalues $E_{n}$ and eigenvectors $\ket{E_{n}}$.
2. Write $\ket{\psi(0)}$ in terms of the energy eigenstates $\ket{E_{n}}$.
3. Multiply each eigenstate coefficient by $e^{-i E_{n} t / \hbar}$ to get $\ket{\psi(t)}$.
4. Calculate the probability $\mathcal{P}_{a_{j}} = \amp{\braket{a_{j}}{\psi(t)}}$.

Spin Precession

• The Hamiltonian of a spin-1/2 system is the following. $$H = \frac{e}{m_{e}} \mathbf{S} \cdot \mathbf{B}$$
• Larmor Precession: The precession of a spin vector around a magnetic field.
• Larmor Frequency: The frequency of precession of a spin vector around a magnetic field.

Uniform Magnetic Field in $z$-Direction

Operator

$$ \begin{aligned} H &= \omega_{0} S_{z}\\ H &= \frac{\hbar\omega_{0}}{2} \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix} \end{aligned} $$

• Where $\omega_{0} = \frac{e B_{0}}{m_{e}}$.

Eigenvalue Equations

$$ \begin{aligned} H\ket{+} &= \omega_{0}S_{z}\ket{+} = \frac{\hbar\omega_{0}}{2}\ket{+} = E_{+}\ket{+}\\ H\ket{-} &= \omega_{0}S_{z}\ket{-} = -\frac{\hbar\omega_{0}}{2}\ket{+} = E_{-}\ket{-}\\ E_{+} &= \frac{\hbar\omega_{0}}{2}\\ E_{-} &= -\frac{\hbar\omega_{0}}{2}\\ \ket{E_{+}} &= \ket{+}\\ \ket{E_{-}} &= \ket{-} \end{aligned} $$

Spin Components Expectations

$$ \ket{\psi(0)} = \ket{+}_{n} = \cos{\frac{\theta}{2}}\ket{+} + \sin{\frac{\theta}{2}}e^{i\phi}\ket{-} \implies\\ \begin{aligned} \avg{S_{z}} &= \sandwich{\psi(t)}{S_{z}}{\psi(t)} = \frac{\hbar}{2}\cos\theta\\ \avg{S_{y}} &= \sandwich{\psi(t)}{S_{y}}{\psi(t)} = \frac{\hbar}{2}\sin\theta\sin(\phi + \omega_{0}t)\\ \avg{S_{x}} &= \sandwich{\psi(t)}{S_{x}}{\psi(t)} = \frac{\hbar}{2}\sin\theta\cos(\phi + \omega_{0}t) \end{aligned} $$