Advanced Quantum Mechanics I

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Unit I: Abstract Formulation of Quantum Mechanics 1.1 Motivation for Linear Vector Space Formulation Quantum phenomena, such as superposition and entanglement, cannot be fully described by classical physics. A linear vector space (Hilbert space) provides the mathematical framework to represent quantum states as vectors and physical observables as operators acting on these vectors. This formalism allows for a consistent description of superposition, probability, and measurement. 1.2 Linear Vector Spaces & Dirac's Ket Notation Ket Vector: $|\psi\rangle$ denotes a state vector in a complex vector space $\mathcal{V}$. These vectors represent the quantum state of a system. Bra Vector: $\langle\phi|$ is an element of the dual space $\mathcal{V}^*$, which is the space of all linear functionals on $\mathcal{V}$. It is related to the ket vector $|\phi\rangle$ by Hermitian conjugation: $\langle\phi| = (|\phi\rangle)^\dagger$. Inner Product: The inner product of a bra $\langle\phi|$ and a ket $|\psi\rangle$ is a complex number, denoted $\langle\phi|\psi\rangle$. Properties: $\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^*$ (Hermitian symmetry) $\langle\phi|(a|\psi_1\rangle + b|\psi_2\rangle) = a\langle\phi|\psi_1\rangle + b\langle\phi|\psi_2\rangle$ (Linearity in ket) $(a\langle\phi_1| + b\langle\phi_2|)|\psi\rangle = a^*\langle\phi_1|\psi\rangle + b^*\langle\phi_2|\psi\rangle$ (Anti-linearity in bra) $\langle\psi|\psi\rangle \ge 0$, and $\langle\psi|\psi\rangle = 0 \iff |\psi\rangle = 0$ (Positive-definiteness) Norm: The norm (or length) of a state vector $|\psi\rangle$ is given by $|||\psi\rangle|| = \sqrt{\langle\psi|\psi\rangle}$. For a physically realizable state, it must be normalized, i.e., $\langle\psi|\psi\rangle = 1$. Schwarz Inequality: For any two vectors $|\phi\rangle$ and $|\psi\rangle$ in the Hilbert space, $$|\langle\phi|\psi\rangle|^2 \le \langle\phi|\phi\rangle\langle\psi|\psi\rangle$$ Equality holds if and only if $|\phi\rangle$ and $|\psi\rangle$ are linearly dependent. Orthonormal Basis: A set of vectors $\{|n\rangle\}$ in a vector space forms an orthonormal basis if they are mutually orthogonal and normalized: $\langle m|n\rangle = \delta_{mn}$. Completeness Relation (Resolution of Identity): For a complete orthonormal basis $\{|n\rangle\}$, the identity operator $I$ can be expressed as: $$\sum_n |n\rangle\langle n| = I$$ This means any state $|\psi\rangle$ can be expanded as $|\psi\rangle = \sum_n |n\rangle\langle n|\psi\rangle = \sum_n c_n|n\rangle$, where $c_n = \langle n|\psi\rangle$. Infinite Dimensional (Discrete) Vector Space: For systems with discrete infinite states (e.g., harmonic oscillator energy levels), the summation extends to infinity. Hilbert Space: A Hilbert space is a complete inner product space. "Complete" means that every Cauchy sequence in the space converges to a point in the space. Quantum mechanical state vectors reside in a Hilbert space. 1.3 Dynamical Observables as Linear Operators Linear Operator: An operator $A$ is linear if $A(a|\psi_1\rangle + b|\psi_2\rangle) = aA|\psi_1\rangle + bA|\psi_2\rangle$ for scalars $a, b$ and vectors $|\psi_1\rangle, |\psi_2\rangle$. Adjoint of a Linear Operator ($A^\dagger$): The adjoint of an operator $A$ is defined by the relation: $$\langle\phi|A\psi\rangle = \langle A^\dagger\phi|\psi\rangle$$ for all $|\phi\rangle, |\psi\rangle$ in the Hilbert space. Properties: $(A^\dagger)^\dagger = A$ $(A+B)^\dagger = A^\dagger + B^\dagger$ $(AB)^\dagger = B^\dagger A^\dagger$ $(cA)^\dagger = c^*A^\dagger$ for a complex scalar $c$. Hermitian (Self-Adjoint) Operators: An operator $A$ is Hermitian if $A = A^\dagger$. These operators represent physical observables. Properties of Hermitian Operators: Real Eigenvalues: If $A|\psi\rangle = a|\psi\rangle$, then $a$ is real. Derivation: $\langle\psi|A\psi\rangle = \langle\psi|a\psi\rangle = a\langle\psi|\psi\rangle$. Also, $\langle\psi|A\psi\rangle = \langle A\psi|\psi\rangle = \langle a\psi|\psi\rangle = a^*\langle\psi|\psi\rangle$. Thus, $a\langle\psi|\psi\rangle = a^*\langle\psi|\psi\rangle$. Since $\langle\psi|\psi\rangle \ne 0$, we have $a=a^*$, so $a$ is real. Orthogonal Eigenvectors: Eigenvectors corresponding to distinct eigenvalues are orthogonal. Derivation: Let $A|\psi_1\rangle = a_1|\psi_1\rangle$ and $A|\psi_2\rangle = a_2|\psi_2\rangle$ with $a_1 \ne a_2$. $\langle\psi_2|A\psi_1\rangle = a_1\langle\psi_2|\psi_1\rangle$. Also, $\langle\psi_2|A\psi_1\rangle = \langle A\psi_2|\psi_1\rangle = \langle a_2\psi_2|\psi_1\rangle = a_2^*\langle\psi_2|\psi_1\rangle = a_2\langle\psi_2|\psi_1\rangle$ (since $a_2$ is real). So, $a_1\langle\psi_2|\psi_1\rangle = a_2\langle\psi_2|\psi_1\rangle \implies (a_1-a_2)\langle\psi_2|\psi_1\rangle = 0$. Since $a_1 \ne a_2$, it must be that $\langle\psi_2|\psi_1\rangle = 0$, meaning they are orthogonal. Eigenvectors corresponding to degenerate eigenvalues can be orthogonalized (e.g., Gram-Schmidt process). Projection Operator: For an orthonormal basis vector $|n\rangle$, the projection operator onto $|n\rangle$ is $P_n = |n\rangle\langle n|$. Properties: $P_n^2 = (|n\rangle\langle n|)(|n\rangle\langle n|) = |n\rangle(\langle n|n\rangle)\langle n| = |n\rangle(1)\langle n| = P_n$ (Idempotent) $P_n^\dagger = (|n\rangle\langle n|)^\dagger = (|n\rangle)^\dagger(\langle n|)^\dagger = |n\rangle\langle n| = P_n$ (Hermitian) Complete Set of Basis: If $\{|n\rangle\}$ forms a complete orthonormal basis, then the sum of projection operators is the identity: $\sum_n P_n = \sum_n |n\rangle\langle n| = I$. 1.4 Matrix Representation of State Vectors and Operators Matrix Representation of a State Vector: Given an orthonormal basis $\{|n\rangle\}$, any state $|\psi\rangle$ can be expanded as $|\psi\rangle = \sum_n c_n|n\rangle$, where $c_n = \langle n|\psi\rangle$. The state vector is represented by a column matrix of its components: $$|\psi\rangle \leftrightarrow \begin{pmatrix} c_1 \\ c_2 \\ \vdots \end{pmatrix}$$ The bra vector $\langle\psi|$ is represented by a row matrix: $\langle\psi| \leftrightarrow \begin{pmatrix} c_1^* & c_2^* & \dots \end{pmatrix}$. Matrix Representation of an Operator: An operator $A$ is represented by a matrix whose elements are $A_{mn} = \langle m|A|n\rangle$. $$A \leftrightarrow \begin{pmatrix} A_{11} & A_{12} & \dots \\ A_{21} & A_{22} & \dots \\ \vdots & \vdots & \ddots \end{pmatrix}$$ The action of an operator on a state vector becomes matrix multiplication: $(A|\psi\rangle)_m = \sum_n A_{mn}c_n$. Properties in Matrix Form: $(A^\dagger)_{mn} = (A_{nm})^*$. For a Hermitian operator $A$, $A_{mn} = (A_{nm})^*$. The inner product $\langle\phi|\psi\rangle = \sum_n \phi_n^* \psi_n$. Unitary Operators and Change of Basis: A unitary operator $U$ preserves the inner product: $\langle U\phi|U\psi\rangle = \langle\phi|\psi\rangle$. This implies $U^\dagger U = UU^\dagger = I$. Unitary operators are crucial for transformations between different orthonormal bases. If $\{|n\rangle\}$ and $\{|n'\rangle\}$ are two orthonormal bases, then there exists a unitary operator $U$ such that $|n'\rangle = U|n\rangle$. Transformation of State Vectors: If $|\psi\rangle$ has components $c_n = \langle n|\psi\rangle$ in the old basis, its components $c_n' = \langle n'|\psi\rangle$ in the new basis are: $$c_n' = \sum_m \langle n'|m\rangle\langle m|\psi\rangle = \sum_m U_{nm}^* c_m$$ where $U_{nm} = \langle n'|m\rangle$. Transformation of Operators: An operator $A$ transforms as $A' = UAU^\dagger$. Derivation: $A'_{mn} = \langle m'|A|n'\rangle = \langle Um|AUn\rangle = \sum_{k,l} \langle Um|k\rangle\langle k|A|l\rangle\langle l|Un\rangle = \sum_{k,l} (\langle k|Um\rangle)^*\langle k|A|l\rangle\langle l|Un\rangle = \sum_{k,l} U_{mk}^* A_{kl} U_{ln}$. This is the matrix multiplication $(U^\dagger A U)_{mn}$. Unit II: Postulates, Measurement and Observables 2.1 Postulates of Quantum Mechanics These postulates form the foundation of quantum mechanics: State Space Postulate: The state of a physical system is completely described by a vector $|\psi\rangle$ in a complex, separable Hilbert space $\mathcal{H}$. This vector is called the state vector or wave function. It is normalized, $\langle\psi|\psi\rangle = 1$, and two state vectors that differ only by a phase factor ($|\psi'\rangle = e^{i\alpha}|\psi\rangle$) represent the same physical state. Observable Postulate: Every measurable physical quantity (observable) is represented by a unique Hermitian operator $A$ acting on the state vectors in $\mathcal{H}$. Measurement Postulate: The only possible results of a measurement of an observable $A$ are its eigenvalues $a_n$. If the system is in a state $|\psi\rangle$, the probability of obtaining the eigenvalue $a_n$ upon measurement is $P(a_n) = |\langle a_n|\psi\rangle|^2$, where $|a_n\rangle$ is the normalized eigenstate corresponding to $a_n$. If $a_n$ is degenerate, $P(a_n) = \sum_{k=1}^{g_n} |\langle a_n^{(k)}|\psi\rangle|^2$, where $g_n$ is the degeneracy and $|a_n^{(k)}\rangle$ are the orthonormal eigenstates for $a_n$. Immediately after the measurement, the state of the system collapses (or projects) to the eigenstate (or subspace) corresponding to the measured eigenvalue. For a non-degenerate eigenvalue $a_n$, the state becomes $|a_n\rangle$. The expectation value (average value) of an observable $A$ in a state $|\psi\rangle$ is given by $\langle A\rangle = \langle\psi|A\psi\rangle$. Time Evolution Postulate: The time evolution of the state vector $|\psi(t)\rangle$ of a system is governed by the time-dependent Schrödinger equation: $$i\hbar \frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle$$ where $H$ is the Hamiltonian operator, representing the total energy of the system. Composite Systems Postulate: The state space of a composite system formed by two subsystems $S_1$ and $S_2$ is the tensor product of their individual state spaces: $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$. If $|\psi_1\rangle \in \mathcal{H}_1$ and $|\psi_2\rangle \in \mathcal{H}_2$, then a state of the composite system is $|\psi\rangle = |\psi_1\rangle \otimes |\psi_2\rangle$. 2.2 Continuous Basis: Position and Momentum Representations Position Basis: The position operator $X$ has a continuous spectrum of eigenvalues $x$. The corresponding eigenstates are $|x\rangle$, which are normalized to a Dirac delta function: $$\langle x|x'\rangle = \delta(x-x')$$ The completeness relation for the position basis is: $$\int_{-\infty}^{\infty} dx |x\rangle\langle x| = I$$ The wave function in position space is $\psi(x) = \langle x|\psi\rangle$. Position Operator in Position Representation: $X \to x$. Derivation: $X|x\rangle = x|x\rangle$. Apply $\langle x'|$ from left: $\langle x'|X|x\rangle = x\langle x'|x\rangle = x\delta(x'-x)$. For an arbitrary state $|\psi\rangle$, $\langle x'|X|\psi\rangle = \int dx \langle x'|X|x\rangle\langle x|\psi\rangle = \int dx \, x\delta(x'-x)\psi(x) = x'\psi(x')$. So $X$ acts as multiplication by $x$ on $\psi(x)$. Momentum Operator in Position Representation: $P \to -i\hbar \frac{\partial}{\partial x}$. Derivation: From the canonical commutation relation $[X, P] = i\hbar I$. In position representation, $[x, P_x] = i\hbar$. Let $P_x = f(x, \frac{\partial}{\partial x})$. $(x P_x - P_x x)\psi(x) = i\hbar\psi(x)$. $x(f\psi) - f(x\psi) = i\hbar\psi$. If $f = -i\hbar\frac{\partial}{\partial x}$, then $x(-i\hbar\frac{\partial\psi}{\partial x}) - (-i\hbar\frac{\partial}{\partial x}(x\psi)) = -i\hbar x\frac{\partial\psi}{\partial x} + i\hbar(\psi + x\frac{\partial\psi}{\partial x}) = i\hbar\psi$. This holds. Momentum Basis: The momentum operator $P$ has a continuous spectrum of eigenvalues $p$. The corresponding eigenstates are $|p\rangle$: $$\langle p|p'\rangle = \delta(p-p')$$ The completeness relation for the momentum basis is: $$\int_{-\infty}^{\infty} dp |p\rangle\langle p| = I$$ The wave function in momentum space is $\phi(p) = \langle p|\psi\rangle$. Momentum Operator in Momentum Representation: $P \to p$. Position Operator in Momentum Representation: $X \to i\hbar \frac{\partial}{\partial p}$. Relation between Position and Momentum Representations: The transformation between $\psi(x)$ and $\phi(p)$ is given by the Fourier transform: $$\psi(x) = \langle x|\psi\rangle = \int dp \langle x|p\rangle\langle p|\psi\rangle = \frac{1}{\sqrt{2\pi\hbar}}\int dp \, e^{ipx/\hbar}\phi(p)$$ $$\phi(p) = \langle p|\psi\rangle = \int dx \langle p|x\rangle\langle x|\psi\rangle = \frac{1}{\sqrt{2\pi\hbar}}\int dx \, e^{-ipx/\hbar}\psi(x)$$ where the transformation kernel is $\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}$. 2.3 Degenerate Eigenvalues and Complete Set of Commuting Observables (CSCO) Commutator: $[A, B] = AB - BA$. Compatibility of Observables: Two observables $A$ and $B$ are compatible if their corresponding operators commute, i.e., $[A, B] = 0$. If $[A, B] = 0$, then $A$ and $B$ can be measured simultaneously without affecting each other. They share a common set of eigenstates. Complete Set of Commuting Observables (CSCO): A set of commuting operators $\{A, B, C, \dots\}$ is a CSCO if their common eigenstates are uniquely specified by their eigenvalues. That is, for each unique set of eigenvalues $\{a, b, c, \dots\}$, there is only one linearly independent common eigenstate $|a,b,c,\dots\rangle$. 2.4 Generalized Uncertainty Principle For any two Hermitian observables $A$ and $B$, the product of their uncertainties ($\Delta A$ and $\Delta B$) is bounded by their commutator: $$\Delta A \Delta B \ge \frac{1}{2} |\langle[A, B]\rangle|$$ where $\Delta A = \sqrt{\langle A^2\rangle - \langle A\rangle^2}$ is the standard deviation of $A$. Derivation Sketch: Let $|\psi\rangle$ be the state. Define $A' = A - \langle A\rangle$ and $B' = B - \langle B\rangle$. Then $\Delta A^2 = \langle A'^2\rangle$ and $\Delta B^2 = \langle B'^2\rangle$. Consider an auxiliary state $|\chi\rangle = (A' + i\lambda B')|\psi\rangle$, where $\lambda$ is a real number. $\langle\chi|\chi\rangle = \langle (A' - i\lambda B')(A' + i\lambda B')\rangle = \langle A'^2 + \lambda^2 B'^2 + i\lambda(A'B' - B'A')\rangle = \Delta A^2 + \lambda^2 \Delta B^2 + i\lambda\langle[A', B']\rangle$. Since $[A', B'] = [A, B]$, we have $\langle\chi|\chi\rangle = \Delta A^2 + \lambda^2 \Delta B^2 + i\lambda\langle[A, B]\rangle \ge 0$. Since $\langle[A, B]\rangle$ is purely imaginary (for Hermitian $A, B$), let $\langle[A, B]\rangle = iC$ where $C$ is real. Then $\Delta A^2 + \lambda^2 \Delta B^2 - \lambda C \ge 0$. This is a quadratic in $\lambda$. For it to be non-negative for all real $\lambda$, its discriminant must be non-positive: $(-C)^2 - 4(\Delta B^2)(\Delta A^2) \le 0 \implies C^2 \le 4\Delta A^2 \Delta B^2$. $|C| \le 2\Delta A \Delta B \implies |\langle[A, B]\rangle| \le 2\Delta A \Delta B$. Rearranging gives the generalized uncertainty principle. Heisenberg Uncertainty Principle: For position and momentum, $[X, P] = i\hbar I$. $$\Delta X \Delta P \ge \frac{1}{2} |\langle i\hbar I\rangle| = \frac{\hbar}{2}$$ This is a fundamental limit on the precision with which position and momentum can be simultaneously known. Energy-Time Uncertainty Principle: $\Delta E \Delta t \ge \frac{\hbar}{2}$. This is not a direct consequence of $[H, T]$ (as $T$ is not a Hermitian operator in the same sense). It is derived from the generalized uncertainty principle using $A=H$ and $B$ being an observable whose expectation value changes with time. 2.5 Quantum Dynamics Unitary Time-Evolution: The Schrödinger equation $i\hbar \frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle$ can be formally integrated to give: $$|\psi(t)\rangle = U(t, t_0)|\psi(t_0)\rangle$$ where $U(t, t_0)$ is the time-evolution operator. For a time-independent Hamiltonian $H$, $U(t, t_0) = e^{-iH(t-t_0)/\hbar}$. Properties of $U$: $U(t, t) = I$. $U(t_2, t_1)U(t_1, t_0) = U(t_2, t_0)$. $U^\dagger(t, t_0) = U^{-1}(t, t_0) = U(t_0, t)$. This means $U$ is unitary, preserving the norm of the state vector over time. Derivation: $\frac{d}{dt}(U^\dagger U) = \frac{dU^\dagger}{dt}U + U^\dagger\frac{dU}{dt}$. From $i\hbar \frac{dU}{dt} = HU$, we get $\frac{dU}{dt} = -\frac{i}{\hbar}HU$. Taking adjoint: $-i\hbar \frac{dU^\dagger}{dt} = (HU)^\dagger = U^\dagger H^\dagger = U^\dagger H$ (since $H$ is Hermitian). So $\frac{dU^\dagger}{dt} = \frac{i}{\hbar}U^\dagger H$. Thus, $\frac{d}{dt}(U^\dagger U) = (\frac{i}{\hbar}U^\dagger H)U + U^\dagger(-\frac{i}{\hbar}HU) = \frac{i}{\hbar}U^\dagger HU - \frac{i}{\hbar}U^\dagger HU = 0$. Since $U^\dagger U$ is constant and $U^\dagger(t_0, t_0)U(t_0, t_0) = I$, then $U^\dagger(t, t_0)U(t, t_0) = I$. Schrödinger vs. Heisenberg Picture: Schrödinger Picture ($S$): States evolve in time: $i\hbar \frac{d}{dt}|\psi_S(t)\rangle = H_S|\psi_S(t)\rangle$. Operators are time-independent: $A_S(t) = A_S$. Expectation values: $\langle A\rangle(t) = \langle\psi_S(t)|A_S|\psi_S(t)\rangle$. Heisenberg Picture ($H$): States are time-independent: $|\psi_H\rangle = |\psi_S(t_0)\rangle$. Operators evolve in time: $A_H(t) = U^\dagger(t, t_0)A_S U(t, t_0)$. Derivation: $\frac{dA_H}{dt} = \frac{dU^\dagger}{dt}A_S U + U^\dagger A_S \frac{dU}{dt} + U^\dagger \frac{\partial A_S}{\partial t} U$. Using $\frac{dU}{dt} = -\frac{i}{\hbar}HU$ and $\frac{dU^\dagger}{dt} = \frac{i}{\hbar}U^\dagger H$: $\frac{dA_H}{dt} = \frac{i}{\hbar}U^\dagger H A_S U - \frac{i}{\hbar}U^\dagger A_S H U + U^\dagger \frac{\partial A_S}{\partial t} U$ $\frac{dA_H}{dt} = \frac{i}{\hbar}(U^\dagger H A_S U - U^\dagger A_S H U) + \left(\frac{\partial A_S}{\partial t}\right)_H$. Since $H_H = U^\dagger H U$, and $H_S = H$, we can write $H = U H_H U^\dagger$. $\frac{dA_H}{dt} = \frac{i}{\hbar}(H_H A_H - A_H H_H) + \left(\frac{\partial A_S}{\partial t}\right)_H$. This gives the Heisenberg Equation of Motion: $$\frac{dA_H}{dt} = \frac{1}{i\hbar}[A_H, H_H] + \left(\frac{\partial A_S}{\partial t}\right)_H$$ If $A_S$ has no explicit time dependence, the last term is zero. Expectation values are the same in both pictures: $\langle\psi_S(t)|A_S|\psi_S(t)\rangle = \langle\psi_S(t_0)|U^\dagger(t, t_0)A_S U(t, t_0)|\psi_S(t_0)\rangle = \langle\psi_H|A_H(t)|\psi_H\rangle$. Momentum as Generator of Translation in Space: The operator for spatial translation by $a$ is $T_a = e^{-iaP/\hbar}$. $$T_a|\psi(x)\rangle = |\psi(x-a)\rangle$$ Derivation: For infinitesimal translation $\delta a$: $T_{\delta a} = I - \frac{i}{\hbar}P\delta a$. $T_{\delta a}\psi(x) = (I - \frac{i}{\hbar}(-i\hbar\frac{\partial}{\partial x})\delta a)\psi(x) = (I - \delta a\frac{\partial}{\partial x})\psi(x) = \psi(x) - \delta a\frac{\partial\psi}{\partial x}$. This is the Taylor expansion of $\psi(x-\delta a)$. Repeated application or exponentiation leads to $T_a|\psi(x)\rangle = |\psi(x-a)\rangle$. Hamiltonian as Generator of Translation in Time: The time-evolution operator $U(t, t_0) = e^{-iH(t-t_0)/\hbar}$ generates translation in time. $$|\psi(t)\rangle = e^{-iH(t-t_0)/\hbar}|\psi(t_0)\rangle$$ Example: Simple Harmonic Oscillator in Heisenberg Picture Hamiltonian: $H = \frac{P^2}{2m} + \frac{1}{2}m\omega^2 X^2$. (No explicit time dependence) Heisenberg equations for $X_H(t)$ and $P_H(t)$: $\frac{dX_H}{dt} = \frac{1}{i\hbar}[X_H, H] = \frac{1}{i\hbar}[X_H, \frac{P_H^2}{2m}] = \frac{1}{i\hbar}\frac{1}{2m}(X_H P_H^2 - P_H^2 X_H)$. Using $[X, P^2] = XP^2 - P^2X = XPP - PPX = XPP - PXP + PXP - PPX = (XP-PX)P + P(XP-PX) = i\hbar P + P(i\hbar) = 2i\hbar P$. So, $\frac{dX_H}{dt} = \frac{1}{i\hbar}\frac{1}{2m}(2i\hbar P_H) = \frac{P_H}{m}$. $\frac{dP_H}{dt} = \frac{1}{i\hbar}[P_H, H] = \frac{1}{i\hbar}[P_H, \frac{1}{2}m\omega^2 X_H^2] = \frac{1}{i\hbar}\frac{1}{2}m\omega^2(P_H X_H^2 - X_H^2 P_H)$. Using $[P, X^2] = PX^2 - X^2P = PX^2 - XPX + XPX - X^2P = P(X^2-PX) + (PX-XP)X = -i\hbar X - i\hbar X = -2i\hbar X$. So, $\frac{dP_H}{dt} = \frac{1}{i\hbar}\frac{1}{2}m\omega^2(-2i\hbar X_H) = -m\omega^2 X_H$. These are the classical equations of motion: $$\frac{dX_H}{dt} = \frac{P_H}{m}$$ $$\frac{dP_H}{dt} = -m\omega^2 X_H$$ Differentiating the first equation: $\frac{d^2X_H}{dt^2} = \frac{1}{m}\frac{dP_H}{dt} = \frac{1}{m}(-m\omega^2 X_H) = -\omega^2 X_H$. The solutions are $X_H(t) = X_H(0)\cos(\omega t) + \frac{P_H(0)}{m\omega}\sin(\omega t)$ and $P_H(t) = P_H(0)\cos(\omega t) - m\omega X_H(0)\sin(\omega t)$. Classical Limit (Correspondence Principle): Quantum mechanics must reproduce classical mechanics in the limit of large quantum numbers or when $\hbar \to 0$. Ehrenfest's Theorem: This theorem provides a bridge between quantum and classical mechanics. It states that the expectation values of position and momentum operators obey equations that resemble classical equations of motion: $$\frac{d}{dt}\langle X\rangle = \frac{1}{m}\langle P\rangle$$ $$\frac{d}{dt}\langle P\rangle = -\left\langle\frac{\partial V}{\partial X}\right\rangle$$ Derivation: Using the Heisenberg equation of motion for an operator $A$ without explicit time dependence: $\frac{d\langle A\rangle}{dt} = \frac{1}{i\hbar}\langle[A, H]\rangle$. For $A=X$: $\frac{d\langle X\rangle}{dt} = \frac{1}{i\hbar}\langle[X, \frac{P^2}{2m} + V(X)]\rangle = \frac{1}{i\hbar}\langle[X, \frac{P^2}{2m}]\rangle = \frac{1}{i\hbar}\frac{1}{2m}\langle 2i\hbar P\rangle = \frac{1}{m}\langle P\rangle$. For $A=P$: $\frac{d\langle P\rangle}{dt} = \frac{1}{i\hbar}\langle[P, \frac{P^2}{2m} + V(X)]\rangle = \frac{1}{i\hbar}\langle[P, V(X)]\rangle$. Using the relation $[P, V(X)] = -i\hbar \frac{\partial V}{\partial X}$ (from Taylor expansion of $V(X+\epsilon)$). So, $\frac{d\langle P\rangle}{dt} = \frac{1}{i\hbar}\langle -i\hbar \frac{\partial V}{\partial X}\rangle = -\langle \frac{\partial V}{\partial X}\rangle$. Unit III: Angular Momentum 3.1 Abstract Operator Approach to Angular Momentum Angular momentum is a fundamental observable in quantum mechanics, defined by its commutation relations rather than its classical definition ($\vec{L} = \vec{r} \times \vec{p}$). Commutation Relations: The components of angular momentum $J_x, J_y, J_z$ satisfy: $$[J_x, J_y] = i\hbar J_z$$ $$[J_y, J_z] = i\hbar J_x$$ $$[J_z, J_x] = i\hbar J_y$$ These can be compactly written as $[J_i, J_j] = i\hbar \epsilon_{ijk} J_k$. Total Angular Momentum Squared Operator: $J^2 = J_x^2 + J_y^2 + J_z^2$. Commutation with Components: $J^2$ commutes with each component of angular momentum: $$[J^2, J_x] = 0, \quad [J^2, J_y] = 0, \quad [J^2, J_z] = 0$$ Derivation for $[J^2, J_z]$: $[J^2, J_z] = [J_x^2 + J_y^2 + J_z^2, J_z] = [J_x^2, J_z] + [J_y^2, J_z]$. $[J_x^2, J_z] = J_x[J_x, J_z] + [J_x, J_z]J_x = J_x(-i\hbar J_y) + (-i\hbar J_y)J_x = -i\hbar(J_x J_y + J_y J_x)$. $[J_y^2, J_z] = J_y[J_y, J_z] + [J_y, J_z]J_y = J_y(i\hbar J_x) + (i\hbar J_x)J_y = i\hbar(J_y J_x + J_x J_y)$. Summing them: $[J^2, J_z] = -i\hbar(J_x J_y + J_y J_x) + i\hbar(J_y J_x + J_x J_y) = 0$. This implies that $J^2$ and any one component (conventionally $J_z$) can be simultaneously diagonalized, meaning they share a common set of eigenstates. Ladder Operators ($J_\pm$): Defined as $J_\pm = J_x \pm iJ_y$. Commutation Relations: $$[J_z, J_\pm] = \pm\hbar J_\pm$$ $$[J^2, J_\pm] = 0$$ Derivation for $[J_z, J_+]$: $[J_z, J_+] = [J_z, J_x + iJ_y] = [J_z, J_x] + i[J_z, J_y] = i\hbar J_y + i(-i\hbar J_x) = i\hbar J_y + \hbar J_x = \hbar(J_x + iJ_y) = \hbar J_+$. Product Relations: $$J_+ J_- = J^2 - J_z^2 + \hbar J_z$$ $$J_- J_+ = J^2 - J_z^2 - \hbar J_z$$ Derivation for $J_+ J_-$: $J_+ J_- = (J_x + iJ_y)(J_x - iJ_y) = J_x^2 - iJ_x J_y + iJ_y J_x + J_y^2 = J_x^2 + J_y^2 + i[J_y, J_x]$. Since $[J_y, J_x] = -[J_x, J_y] = -i\hbar J_z$. $J_+ J_- = J_x^2 + J_y^2 + i(-i\hbar J_z) = J_x^2 + J_y^2 + \hbar J_z$. Since $J^2 = J_x^2 + J_y^2 + J_z^2$, then $J_x^2 + J_y^2 = J^2 - J_z^2$. So, $J_+ J_- = J^2 - J_z^2 + \hbar J_z$. Eigenvalues and Eigenvectors: Let $|j, m\rangle$ be the common eigenstates of $J^2$ and $J_z$. $$J^2|j, m\rangle = \hbar^2 j(j+1)|j, m\rangle$$ $$J_z|j, m\rangle = \hbar m|j, m\rangle$$ $j$ can be a non-negative integer or half-integer ($0, 1/2, 1, 3/2, \dots$). $m$ (magnetic quantum number) ranges from $-j$ to $j$ in integer steps (i.e., $m \in \{-j, -j+1, \dots, j-1, j\}$). For a given $j$, there are $2j+1$ possible values of $m$. Action of Ladder Operators: $$J_+|j, m\rangle = \hbar\sqrt{j(j+1)-m(m+1)}|j, m+1\rangle$$ $$J_-|j, m\rangle = \hbar\sqrt{j(j+1)-m(m-1)}|j, m-1\rangle$$ These operators raise or lower the $m$ quantum number by one unit, without changing $j$. They produce zero when acting on the maximum ($m=j$) or minimum ($m=-j$) states, respectively. Derivation Sketch: Consider $J_+|j, m\rangle$. Applying $J_z$ to it: $J_z (J_+|j, m\rangle) = (J_+ J_z + [J_z, J_+])|j, m\rangle = (J_+ \hbar m + \hbar J_+)|j, m\rangle = \hbar(m+1)(J_+|j, m\rangle)$. This shows that $J_+|j, m\rangle$ is an eigenstate of $J_z$ with eigenvalue $\hbar(m+1)$. To find the normalization constant, let $J_+|j, m\rangle = C_{j,m}|j, m+1\rangle$. $|C_{j,m}|^2 = \langle j, m|J_- J_+|j, m\rangle = \langle j, m|(J^2 - J_z^2 - \hbar J_z)|j, m\rangle$. $= \langle j, m|(\hbar^2 j(j+1) - (\hbar m)^2 - \hbar (\hbar m))|j, m\rangle = \hbar^2(j(j+1) - m^2 - m) = \hbar^2(j(j+1) - m(m+1))$. Thus, $C_{j,m} = \hbar\sqrt{j(j+1)-m(m+1)}$. (Conventionally, $C_{j,m}$ is chosen to be real and positive). 3.2 Pauli Matrices and Spin Angular Momentum Spin is an intrinsic form of angular momentum unique to quantum particles. For spin-1/2 particles (like electrons), the spin operators are proportional to the Pauli matrices. Pauli Matrices: $$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ Properties: $\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = I$ (Identity matrix). Anticommutation relation: $\{\sigma_i, \sigma_j\} = \sigma_i\sigma_j + \sigma_j\sigma_i = 2\delta_{ij}I$. Commutation relation: $[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k$. (This implies the angular momentum commutation relations). $\sigma_x\sigma_y = i\sigma_z$, $\sigma_y\sigma_z = i\sigma_x$, $\sigma_z\sigma_x = i\sigma_y$. Determinant: $\det(\sigma_i) = -1$. Trace: $\text{Tr}(\sigma_i) = 0$. Spin Angular Momentum Operators: For a spin-1/2 particle, the spin operators are given by $S_i = \frac{\hbar}{2}\sigma_i$. $S_x = \frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, $S_y = \frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$, $S_z = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. These operators satisfy the angular momentum commutation relations: $[S_x, S_y] = i\hbar S_z$, etc. Eigenvalues and Eigenvectors for Spin-1/2: $S^2 = S_x^2 + S_y^2 + S_z^2 = \frac{\hbar^2}{4}(\sigma_x^2 + \sigma_y^2 + \sigma_z^2) = \frac{\hbar^2}{4}(I+I+I) = \frac{3}{4}\hbar^2 I$. This corresponds to $s(s+1)\hbar^2$ with $s=1/2$. So, $s=1/2$. $S_z$ eigenvalues are $\pm\frac{\hbar}{2}$. The corresponding eigenvectors are: $|\uparrow\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (spin up, $m_s=+1/2$) $|\downarrow\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ (spin down, $m_s=-1/2$) Spin Ladder Operators: $S_\pm = S_x \pm iS_y = \frac{\hbar}{2}(\sigma_x \pm i\sigma_y)$. $S_+ = \hbar\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, $S_- = \hbar\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. $S_+|\downarrow\rangle = \hbar|\uparrow\rangle$, $S_-|\uparrow\rangle = \hbar|\downarrow\rangle$. $S_+|\uparrow\rangle = 0$, $S_-|\downarrow\rangle = 0$. General Spin State for Spin-1/2: A general spin state is a superposition: $$|\chi\rangle = c_1|\uparrow\rangle + c_2|\downarrow\rangle = c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$$ where $|c_1|^2 + |c_2|^2 = 1$. Spin 1 Systems: For a spin-1 particle, $j=1$. The possible $m$ values are $m \in \{-1, 0, 1\}$. The dimension of the spin space is $2j+1 = 3$. Matrix representation of $S_z$: $$S_z = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$ The eigenstates are $|1,1\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $|1,0\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, $|1,-1\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$. Matrix representation of $S_x, S_y$: $$S_x = \frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad S_y = \frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{pmatrix}$$ These can be derived from the ladder operators $S_\pm$. 3.3 Addition of Angular Momentum When two angular momenta $\vec{J_1}$ and $\vec{J_2}$ combine, the total angular momentum $\vec{J} = \vec{J_1} + \vec{J_2}$ is also a valid angular momentum. Uncoupled Basis: The eigenstates of $J_1^2, J_{1z}, J_2^2, J_{2z}$ are denoted by $|j_1 m_1; j_2 m_2\rangle$. These are product states. $J_1^2|j_1 m_1; j_2 m_2\rangle = \hbar^2 j_1(j_1+1)|j_1 m_1; j_2 m_2\rangle$ $J_{1z}|j_1 m_1; j_2 m_2\rangle = \hbar m_1|j_1 m_1; j_2 m_2\rangle$ Similarly for $J_2^2$ and $J_{2z}$. Coupled Basis: The eigenstates of $J^2, J_z, J_1^2, J_2^2$ are denoted by $|j_1 j_2; J M\rangle$. $J^2|j_1 j_2; J M\rangle = \hbar^2 J(J+1)|j_1 j_2; J M\rangle$ $J_z|j_1 j_2; J M\rangle = \hbar M|j_1 j_2; J M\rangle$ The possible values for $J$ range from $|j_1-j_2|$ to $j_1+j_2$ in integer steps. For each $J$, $M$ ranges from $-J$ to $J$ in integer steps. The total $z$-component quantum number is additive: $M = m_1 + m_2$. Clebsch-Gordan Coefficients (CGCs): These are the expansion coefficients that relate the coupled basis states to the uncoupled basis states: $$|j_1 j_2; J M\rangle = \sum_{m_1, m_2} \langle j_1 m_1; j_2 m_2 | j_1 j_2; J M \rangle |j_1 m_1; j_2 m_2\rangle$$ The sum is over all $m_1, m_2$ such that $m_1+m_2=M$. The CGCs are zero if $M \ne m_1+m_2$. CGCs of Addition for specific systems: (i) $j_1 = 1/2, j_2 = 1/2$: Total $J$ can be $0$ or $1$. $J=1$ (Triplet states, symmetric): $$|1/2, 1/2; 1, 1\rangle = |1/2, 1/2; 1/2, 1/2\rangle \equiv |\uparrow\uparrow\rangle$$ $$|1/2, 1/2; 1, 0\rangle = \frac{1}{\sqrt{2}}(|1/2, 1/2; 1/2, -1/2\rangle + |1/2, -1/2; 1/2, 1/2\rangle) \equiv \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)$$ $$|1/2, 1/2; 1, -1\rangle = |1/2, -1/2; 1/2, -1/2\rangle \equiv |\downarrow\downarrow\rangle$$ $J=0$ (Singlet state, antisymmetric): $$|1/2, 1/2; 0, 0\rangle = \frac{1}{\sqrt{2}}(|1/2, 1/2; 1/2, -1/2\rangle - |1/2, -1/2; 1/2, 1/2\rangle) \equiv \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$$ (ii) $j_1 = 1/2, j_2 = 1$: Total $J$ can be $1/2$ or $3/2$. $J=3/2$: (4 states) $$|1/2, 1; 3/2, 3/2\rangle = |1/2, 1/2; 1, 1\rangle$$ $$|1/2, 1; 3/2, 1/2\rangle = \sqrt{\frac{2}{3}}|1/2, 1/2; 1, 0\rangle + \sqrt{\frac{1}{3}}|1/2, -1/2; 1, 1\rangle$$ $$|1/2, 1; 3/2, -1/2\rangle = \sqrt{\frac{1}{3}}|1/2, 1/2; 1, -1\rangle + \sqrt{\frac{2}{3}}|1/2, -1/2; 1, 0\rangle$$ $$|1/2, 1; 3/2, -3/2\rangle = |1/2, -1/2; 1, -1\rangle$$ $J=1/2$: (2 states) $$|1/2, 1; 1/2, 1/2\rangle = \sqrt{\frac{1}{3}}|1/2, 1/2; 1, 0\rangle - \sqrt{\frac{2}{3}}|1/2, -1/2; 1, 1\rangle$$ $$|1/2, 1; 1/2, -1/2\rangle = \sqrt{\frac{2}{3}}|1/2, 1/2; 1, -1\rangle - \sqrt{\frac{1}{3}}|1/2, -1/2; 1, 0\rangle$$ (iii) $j_1 = 1, j_2 = 1$: Total $J$ can be $0, 1, 2$. Unit IV: Density Matrix Formalism & Identical Particles 4.1 Density Operator and Matrix The density operator (or density matrix) provides a more general description of a quantum system, especially useful for mixed states (statistical ensembles) or subsystems of entangled systems. Pure State: A system is in a pure state if its state can be described by a single, normalized state vector $|\psi\rangle$. The density operator for a pure state is $\rho = |\psi\rangle\langle\psi|$. Properties: $\text{Tr}(\rho) = \text{Tr}(|\psi\rangle\langle\psi|) = \langle\psi|\psi\rangle = 1$. $\rho^2 = (|\psi\rangle\langle\psi|)(|\psi\rangle\langle\psi|) = |\psi\rangle(\langle\psi|\psi\rangle)\langle\psi| = |\psi\rangle(1)\langle\psi| = \rho$. (Idempotent) Mixed State: A system is in a mixed state if it is in an ensemble of pure states $\{|\psi_i\rangle\}$ with classical probabilities $p_i$, where $\sum_i p_i = 1$. The system is *not* in any single $|\psi_i\rangle$. The density operator for a mixed state is $\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|$. Properties: $\text{Tr}(\rho) = \sum_i p_i \text{Tr}(|\psi_i\rangle\langle\psi_i|) = \sum_i p_i \langle\psi_i|\psi_i\rangle = \sum_i p_i = 1$. $\rho^2 = (\sum_i p_i |\psi_i\rangle\langle\psi_i|)(\sum_j p_j |\psi_j\rangle\langle\psi_j|) = \sum_{i,j} p_i p_j |\psi_i\rangle\langle\psi_i|\psi_j\rangle\langle\psi_j|$. In general, $\rho^2 \ne \rho$. Specifically, $\text{Tr}(\rho^2) \le 1$. For a pure state, $\text{Tr}(\rho^2) = 1$. For a mixed state, $\text{Tr}(\rho^2) < 1$. $\rho$ is Hermitian: $\rho^\dagger = \rho$. $\rho$ is positive semi-definite (eigenvalues are non-negative). Expectation Value of an Observable: The expectation value of an observable $A$ for a system described by $\rho$ is: $$\langle A\rangle = \text{Tr}(\rho A)$$ Derivation: For a mixed state, $\langle A\rangle = \sum_i p_i \langle\psi_i|A|\psi_i\rangle$. Expanding in a basis $\{|n\rangle\}$: $\langle\psi_i|A|\psi_i\rangle = \sum_n \langle\psi_i|A|n\rangle\langle n|\psi_i\rangle = \sum_n \langle n|\psi_i\rangle\langle\psi_i|A|n\rangle$. So $\langle A\rangle = \sum_i p_i \sum_n \langle n|\psi_i\rangle\langle\psi_i|A|n\rangle = \sum_n \sum_i p_i \langle n|\psi_i\rangle\langle\psi_i|A|n\rangle$. Since $\rho = \sum_i p_i |\psi_i\rangle\langle\psi_i|$, we have $\langle n|\rho A|n\rangle = \sum_k \langle n|\rho|k\rangle\langle k|A|n\rangle$. The expression $\sum_n \langle n|\rho A|n\rangle$ is $\text{Tr}(\rho A)$. And $\sum_n \sum_i p_i \langle n|\psi_i\rangle\langle\psi_i|A|n\rangle = \sum_n \langle n|(\sum_i p_i |\psi_i\rangle\langle\psi_i|)A|n\rangle = \sum_n \langle n|\rho A|n\rangle = \text{Tr}(\rho A)$. Time Evolution of Density Matrix: The time evolution of the density operator is given by the Liouville-von Neumann equation: $$i\hbar \frac{d\rho}{dt} = [H, \rho]$$ Derivation: If $|\psi(t)\rangle = U(t,t_0)|\psi(t_0)\rangle$, then $\rho(t) = |\psi(t)\rangle\langle\psi(t)| = U(t,t_0)\rho(t_0)U^\dagger(t,t_0)$. $\frac{d\rho}{dt} = \frac{dU}{dt}\rho(t_0)U^\dagger + U\rho(t_0)\frac{dU^\dagger}{dt}$. Using $\frac{dU}{dt} = -\frac{i}{\hbar}HU$ and $\frac{dU^\dagger}{dt} = \frac{i}{\hbar}U^\dagger H$: $\frac{d\rho}{dt} = -\frac{i}{\hbar}HU\rho(t_0)U^\dagger + U\rho(t_0)\frac{i}{\hbar}U^\dagger H = -\frac{i}{\hbar}H\rho + \frac{i}{\hbar}\rho H = \frac{1}{i\hbar}(H\rho - \rho H) = \frac{1}{i\hbar}[H, \rho]$. This holds for both pure and mixed states. Reduced Density Matrix for Subsystems: For a composite system $C = A \otimes B$ described by $\rho_{AB}$, the state of subsystem $A$ is described by the reduced density matrix $\rho_A$, obtained by tracing out the degrees of freedom of subsystem $B$: $$\rho_A = \text{Tr}_B(\rho_{AB})$$ If $\rho_{AB} = \sum_{ij,kl} C_{ijkl} |i_A\rangle\langle j_A| \otimes |k_B\rangle\langle l_B|$, then $\rho_A = \sum_{i,j,k} C_{ijkk} |i_A\rangle\langle j_A|$. Example: Entangled Spin-1/2 Pair: Consider two spin-1/2 particles in the entangled singlet state: $$|\Psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$$ The density operator for the composite system is $\rho_{12} = |\Psi\rangle\langle\Psi|$. $\rho_{12} = \frac{1}{2}(|\uparrow\downarrow\rangle\langle\uparrow\downarrow| - |\uparrow\downarrow\rangle\langle\downarrow\uparrow| - |\downarrow\uparrow\rangle\langle\uparrow\downarrow| + |\downarrow\uparrow\rangle\langle\downarrow\uparrow|)$. To find the reduced density matrix for particle 1, $\rho_1 = \text{Tr}_2(\rho_{12})$. Let the basis for particle 2 be $\{|\uparrow_2\rangle, |\downarrow_2\rangle\}$. $\rho_1 = \langle\uparrow_2|\rho_{12}|\uparrow_2\rangle + \langle\downarrow_2|\rho_{12}|\downarrow_2\rangle$. $\langle\uparrow_2|\rho_{12}|\uparrow_2\rangle = \frac{1}{2}(|\uparrow_1\rangle\langle\uparrow_1|\langle\downarrow_2|\uparrow_2\rangle\langle\uparrow_2|\downarrow_2\rangle - |\uparrow_1\rangle\langle\downarrow_1|\langle\downarrow_2|\uparrow_2\rangle\langle\uparrow_2|\uparrow_2\rangle - |\downarrow_1\rangle\langle\uparrow_1|\langle\uparrow_2|\uparrow_2\rangle\langle\uparrow_2|\downarrow_2\rangle + |\downarrow_1\rangle\langle\downarrow_1|\langle\uparrow_2|\uparrow_2\rangle\langle\uparrow_2|\downarrow_2\rangle)$. Since $\langle\downarrow_2|\uparrow_2\rangle=0$ and $\langle\uparrow_2|\uparrow_2\rangle=1$, this term is 0. $\langle\downarrow_2|\rho_{12}|\downarrow_2\rangle = \frac{1}{2}(|\uparrow_1\rangle\langle\uparrow_1|\langle\downarrow_2|\downarrow_2\rangle\langle\downarrow_2|\downarrow_2\rangle - |\uparrow_1\rangle\langle\downarrow_1|\langle\downarrow_2|\downarrow_2\rangle\langle\downarrow_2|\uparrow_2\rangle - |\downarrow_1\rangle\langle\uparrow_1|\langle\uparrow_2|\downarrow_2\rangle\langle\downarrow_2|\downarrow_2\rangle + |\downarrow_1\rangle\langle\downarrow_1|\langle\uparrow_2|\downarrow_2\rangle\langle\downarrow_2|\uparrow_2\rangle)$. Since $\langle\downarrow_2|\downarrow_2\rangle=1$ and $\langle\downarrow_2|\uparrow_2\rangle=0$, this simplifies to $\frac{1}{2}|\uparrow_1\rangle\langle\uparrow_1|$. Similarly, $\langle\uparrow_2|\rho_{12}|\uparrow_2\rangle = \frac{1}{2}|\downarrow_1\rangle\langle\downarrow_1|$. So, $\rho_1 = \frac{1}{2}(|\uparrow_1\rangle\langle\uparrow_1| + |\downarrow_1\rangle\langle\downarrow_1|) = \frac{1}{2}I_1$. This shows that particle 1 (and similarly particle 2) is in a completely mixed state, even though the total system is in a pure, entangled state. This is a hallmark of entanglement. 4.2 Identical Particles Many-Particle Systems: A system of $N$ identical particles. The state space is the tensor product of single-particle Hilbert spaces. Exchange Degeneracy: If particles are identical, permuting them should not lead to a new physical state. For example, for two identical particles in states $\psi_a(\vec{r}_1)$ and $\psi_b(\vec{r}_2)$, the state $\psi_a(\vec{r}_2)\psi_b(\vec{r}_1)$ has the same physical properties (like energy) as $\psi_a(\vec{r}_1)\psi_b(\vec{r}_2)$. Indistinguishability Postulate: For identical particles, all physical observables (like the Hamiltonian) are symmetric under the exchange of any two particles. This means the Hamiltonian $H$ commutes with the permutation operator $P_{ij}$ for any pair $(i,j)$ of identical particles: $[H, P_{ij}] = 0$. Permutation Operator ($P_{ij}$): Swaps the coordinates (position, spin, etc.) of particle $i$ and particle $j$. $$P_{ij} \psi(\dots, \vec{r}_i, \dots, \vec{r}_j, \dots) = \psi(\dots, \vec{r}_j, \dots, \vec{r}_i, \dots)$$ Since $P_{ij}^2 = I$, the eigenvalues of $P_{ij}$ are $\pm 1$. Symmetric and Anti-symmetric Wavefunctions: Symmetric Wavefunctions ($P_{ij}\Psi_S = +\Psi_S$): Particles described by symmetric wavefunctions are called bosons (e.g., photons, alpha particles). For two particles, $\Psi_S(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}}[\psi_a(\vec{r}_1)\psi_b(\vec{r}_2) + \psi_a(\vec{r}_2)\psi_b(\vec{r}_1)]$. Anti-symmetric Wavefunctions ($P_{ij}\Psi_A = -\Psi_A$): Particles described by anti-symmetric wavefunctions are called fermions (e.g., electrons, protons, neutrons). For two particles, $\Psi_A(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}}[\psi_a(\vec{r}_1)\psi_b(\vec{r}_2) - \psi_a(\vec{r}_2)\psi_b(\vec{r}_1)]$. Pauli Exclusion Principle: "No two identical fermions can occupy the same quantum state." This is a direct consequence of the anti-symmetry requirement. Derivation: If two fermions were to occupy the same state, meaning $\psi_a = \psi_b$, then the anti-symmetric wavefunction would become: $$\Psi_A(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}}[\psi_a(\vec{r}_1)\psi_a(\vec{r}_2) - \psi_a(\vec{r}_2)\psi_a(\vec{r}_1)] = 0$$ A zero wavefunction means the state is impossible. Thus, two identical fermions cannot be in the same quantum state. Concept of Parity: The parity operator $P$ inverts the spatial coordinates: $P\psi(\vec{r}) = \psi(-\vec{r})$. $P^2 = I$, so eigenvalues are $\pm 1$. Eigenvalue $+1$: Even parity, $\psi(-\vec{r}) = +\psi(\vec{r})$. Eigenvalue $-1$: Odd parity, $\psi(-\vec{r}) = -\psi(\vec{r})$. If the Hamiltonian is invariant under spatial inversion (i.e., $V(\vec{r}) = V(-\vec{r})$), then $[P, H] = 0$, and parity is a conserved quantity.