Quantum Mechanics Essentials

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1. Need for a Quantum Theory Classical Physics challenges: Black-body radiation, heat capacity of solids, discrete atomic spectra. Bohr model limitations: Electron radiating energy and collapsing. New Quantum Theory concepts: Particle-Wave Duality: Louis de Broglie's equation. $\lambda = \frac{h}{p}$ where $h$ is Planck's constant ($6.63 \times 10^{-34} \text{ J s}$). Heisenberg's Uncertainty Principle: For position $x$ and momentum $p$ in 1D. $\Delta x \Delta p \geq \frac{\hbar}{2}$ where $\hbar = h/2\pi$. 2. Postulates of Quantum Mechanics 2.1. Wavefunction $\psi(x)$ Describes the state of a particle in Quantum Mechanics (instead of fixed position/momentum). Related to probability density $P(x)$: $P(x) = \psi^*(x)\psi(x)$ where $\psi^*(x)$ is the complex conjugate. Normalization Condition (1D): The probability of finding the particle somewhere is 1. $\int_S \psi^*(x)\psi(x)dx = 1$ For all real values of $x$: $\int_{-\infty}^{\infty} \psi^*(x)\psi(x)dx = 1$ Wavefunction Conditions: Square integrable: $\int \psi^*(x)\psi(x)dx Single-valued, finite, continuous. First derivative must be continuous (can be relaxed in some cases). 2.2. Wavefunctions and Normalization (Summary) 1D Particle: State: $\psi(x)$ Normalization: $\int_S \psi^*(x)\psi(x)dx = 1$ 3D Particle (Cartesian): State: $\psi(x, y, z)$ or $\psi(\vec{r})$ Normalization: $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \psi^*(x, y, z)\psi(x, y, z)dxdydz = 1$ 3D Particle (Spherical Polar): State: $\psi(r, \theta, \phi)$ Normalization: $\int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} r^2 \sin \theta \psi^*(r, \theta, \phi)\psi(r, \theta, \phi) dr d\theta d\phi = 1$ Two Particles (1D): Probability $P(x_1, x_2)dx_1dx_2 = \psi^*(x_1, x_2)\psi(x_1, x_2)dx_1dx_2$ Identical Particles (3D): Indistinguishable. $|\psi(\vec{r}_1, \vec{r}_2)|^2 = |\psi(\vec{r}_2, \vec{r}_1)|^2 \implies \psi(\vec{r}_1, \vec{r}_2) = \pm \psi(\vec{r}_2, \vec{r}_1)$ Bosons: Symmetric ($+$ sign), integer spin. Fermions: Antisymmetric ($-$ sign), half-integer spin (Pauli's exclusion principle). 3. Operators corresponding to Physical Observables Every physical observable has a corresponding operator. An operator $\hat{O}$ acts on a function $\psi(x)$ to give a new function $\psi'(x)$: $\hat{O}\psi(x) = \psi'(x)$. Key Operators (1D): Position Operator: $\hat{x}\psi(x) = x\psi(x)$ Momentum Operator: $\hat{p}_x\psi(x) = -i\hbar \frac{d}{dx}\psi(x)$ Derived Operators: Kinetic Energy Operator: $\hat{T} = \frac{\hat{p}_x\hat{p}_x}{2m} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ Potential Energy Operator: $\hat{V} = V(x)$ Total Energy (Hamiltonian) Operator: $\hat{H} = \hat{T} + \hat{V} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$ Operator Properties: Linearity: $\hat{O}(c_1\psi_1 + c_2\psi_2) = c_1\hat{O}\psi_1 + c_2\hat{O}\psi_2$ Hermiticity: Quantum mechanical operators are Hermitian. Eigenvalues are real. Eigenvectors are orthogonal if eigenvalues are different: $\int_S \psi_1^*(x)\psi_2(x)dx = 0$ if $\lambda_1 \neq \lambda_2$. 4. Eigenvalues and Eigenfunctions If $\hat{O}\psi_\lambda(x) = \lambda\psi_\lambda(x)$, then $\psi_\lambda(x)$ is an eigenfunction of $\hat{O}$ with eigenvalue $\lambda$. Example: For $\hat{p}_x$, $Ae^{ikx}$ is an eigenfunction with eigenvalue $\hbar k$. $\hat{p}_x(Ae^{ikx}) = -i\hbar \frac{d}{dx}(Ae^{ikx}) = -i\hbar (ikAe^{ikx}) = \hbar k Ae^{ikx}$ A constant multiplier of an eigenfunction still yields the same eigenvalue. Eigenvalues correspond to observable values upon measurement. Expectation Value (Average Value): For an observable $\hat{O}$. $\langle \hat{O} \rangle = \frac{\int_S \psi^*\hat{O}\psi dx}{\int_S \psi^*\psi dx}$ If $\psi = c_1\psi_1 + c_2\psi_2$, then $\langle \hat{O} \rangle = \frac{c_1^2\lambda_1 + c_2^2\lambda_2}{c_1^2 + c_2^2}$. 5. Equation of Motion (Time-Dependent Schrödinger Equation) General form: $i\hbar \frac{\partial\psi}{\partial t} = \hat{H}\psi$ Explicitly (1D): $i\hbar \frac{\partial\psi(x,t)}{\partial t} = \left( -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right)\psi(x,t)$ Stationary States: Eigenfunctions of the Hamiltonian $\hat{H}\psi_s(x,t) = E\psi_s(x,t)$ satisfy the time-independent Schrödinger equation. $i\hbar \frac{\partial\psi_s(x,t)}{\partial t} = E\psi_s(x,t)$ Solution: $\psi_s(x,t) = \psi_s(x,0)e^{-iEt/\hbar}$. Eigenvalues do not change with time. 6. Particle in a 1D Box Particle of mass $m$ confined to $0 Schrödinger Equation inside the box: $-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} = E\psi(x)$ With boundary conditions $\psi(0) = 0$ and $\psi(L) = 0$. This leads to: $\frac{d^2\psi(x)}{dx^2} = -k^2\psi(x)$, where $k^2 = \frac{2mE}{\hbar^2}$. General solution: $\psi(x) = A\sin(kx) + B\cos(kx)$. Applying boundary conditions: $\psi(0)=0 \implies B=0$. $\psi(L)=0 \implies A\sin(kL)=0$. Since $A \neq 0$, $kL = n\pi$, where $n = 1, 2, 3, \dots$ (not $0$ as it gives $\psi(x)=0$). Wavefunctions: $\psi_n(x) = A\sin\left(\frac{n\pi x}{L}\right)$, for $n = 1, 2, 3, \dots$ Normalized wavefunction (where $A = \sqrt{2/L}$): $\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$ Energies: $E_n = \frac{n^2\hbar^2\pi^2}{2mL^2} = \frac{n^2h^2}{8mL^2}$, for $n = 1, 2, 3, \dots$ (Energy is quantized). Ground state ($n=1$): $E_1 = \frac{h^2}{8mL^2}$ First excited state ($n=2$): $E_2 = \frac{4h^2}{8mL^2}$ Quantization arises from boundary conditions. 7. Particle in a 2D Box Particle in $0 Schrödinger Equation inside: $-\frac{\hbar^2}{2m}\left(\frac{\partial^2\psi(x,y)}{\partial x^2} + \frac{\partial^2\psi(x,y)}{\partial y^2}\right) = E\psi(x,y)$ Method of separation of variables: $\psi(x,y) = X(x)Y(y)$. Leads to two independent 1D box problems: $E = E_x + E_y$ Wavefunctions: $\psi_{n_x,n_y}(x,y) = \sqrt{\frac{2}{L_x}}\sin\left(\frac{n_x\pi x}{L_x}\right)\sqrt{\frac{2}{L_y}}\sin\left(\frac{n_y\pi y}{L_y}\right)$ where $n_x, n_y = 1, 2, 3, \dots$ Energies: $E_{n_x,n_y} = \frac{n_x^2h^2}{8mL_x^2} + \frac{n_y^2h^2}{8mL_y^2}$ Degeneracy: Occurs when different $(n_x, n_y)$ pairs yield the same energy. For a square box ($L_x=L_y=L$): $E_{n_x,n_y} = \frac{h^2}{8mL^2}(n_x^2 + n_y^2)$. Ground state ($n_x=1, n_y=1$): $E_{1,1} = \frac{2h^2}{8mL^2}$ (non-degenerate). First excited state ($n_x=1, n_y=2$ or $n_x=2, n_y=1$): $E_{1,2}=E_{2,1} = \frac{5h^2}{8mL^2}$ (2-fold degenerate). For rectangular boxes ($L_x \neq L_y$), accidental degeneracies are less common. 8. 1D Harmonic Oscillator Particle of mass $m$ (or reduced mass $\mu$) with potential $V(x) = \frac{1}{2}kx^2$. Schrödinger Equation: $-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + \frac{1}{2}kx^2\psi(x) = E\psi(x)$ (Range of $x$ is $(-\infty, \infty)$). Allowed Energy Values: Quantized. $E_v = \left(v + \frac{1}{2}\right)h\nu$, where $v = 0, 1, 2, \dots$ $\nu$ is the oscillator frequency: $\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (or $\sqrt{\frac{k}{\mu}}$). Also, $h\nu = \hbar\omega$. Ground state ($v=0$): $E_0 = \frac{1}{2}h\nu$ (zero-point energy). First excited state ($v=1$): $E_1 = \frac{3}{2}h\nu$. Second excited state ($v=2$): $E_2 = \frac{5}{2}h\nu$. Spacing between energy levels is constant ($h\nu$). Wavefunctions: $\psi_v(x) = N_v H_v(y)e^{-y^2/2}$ where $y = \frac{x}{\alpha}$ and $\alpha = \left(\frac{\hbar^2}{m k}\right)^{1/4}$. $H_v(y)$ are Hermite polynomials. $H_0(y) = 1 \implies \psi_0(x) = N_0 e^{-x^2/(2\alpha^2)}$ $H_1(y) = 2y \implies \psi_1(x) = N_1 \frac{x}{\alpha} e^{-x^2/(2\alpha^2)}$ Normalization Constant: $N_v = \frac{1}{\sqrt{\alpha\sqrt{\pi}2^v v!}}$ Classical Turning Points ($x_{ctp}$): Where $E_v = V(x_{ctp})$. $x_{ctp} = \left(\frac{h\nu}{k}\right)^{1/2} = \alpha$ Tunneling: Wavefunction is non-zero in classically forbidden regions ($|x| > x_{ctp}$). Expectation Value of Position: $\langle x \rangle = \int_{-\infty}^{\infty} \psi^*(x)x\psi(x)dx = 0$ (for even/odd functions). Mean-Squared Displacement: For $v=0$ state: $\langle x^2 \rangle = \frac{\alpha^2}{2}$. Average Potential Energy: $\langle V(x) \rangle = \frac{1}{2}k\langle x^2 \rangle = \frac{1}{4}h\nu = \frac{1}{2}E_0$. 9. Rotational Motion: 1D Rigid Rotor (Particle on a Ring) Particle of mass $m$ (or reduced mass $\mu$) on a ring of radius $R$. Motion in XY plane, $\theta=\pi/2$. Wavefunction depends only on $\phi$. Moment of inertia $I = mR^2$ (or $\mu R^2$). Schrödinger Equation: $-\frac{\hbar^2}{2I}\frac{d^2\psi(\phi)}{d\phi^2} = E\psi(\phi)$ Boundary condition: $\psi(\phi) = \psi(\phi+2\pi)$ (periodic). Equation rewritten: $\frac{d^2\psi(\phi)}{d\phi^2} = -\frac{2IE}{\hbar^2}\psi(\phi) = -k^2\psi(\phi)$. Solution: $\psi(\phi) = Ae^{ik\phi}$. Applying boundary condition: $Ae^{ik(\phi+2\pi)} = Ae^{ik\phi}e^{i2\pi k} = Ae^{ik\phi} \implies e^{i2\pi k} = 1$. This means $2\pi k = 2\pi m_l$, so $k = m_l$, where $m_l = 0, \pm 1, \pm 2, \dots$. Wavefunctions: $\psi_{m_l}(\phi) = Ae^{im_l\phi}$ Normalized wavefunction (where $A = \frac{1}{\sqrt{2\pi}}$): $\psi_{m_l}(\phi) = \frac{1}{\sqrt{2\pi}}e^{im_l\phi}$ Energies: $E_{m_l} = \frac{\hbar^2 m_l^2}{2I}$ Ground state ($m_l=0$): $E_0 = 0$ (non-degenerate). First excited level ($m_l=\pm 1$): $E_1 = E_{-1} = \frac{\hbar^2}{2I}$ (doubly degenerate). All states except ground state are doubly degenerate. 10. Rotational Motion: 3D Rigid Rotor (Particle on a Sphere) Particle of mass $m$ (or reduced mass $\mu$) on a sphere of radius $R$. Moment of inertia $I = \mu R^2$. Schrödinger Equation involves angular momentum operators: $\hat{L}^2\psi(\theta, \phi) = E\psi(\theta, \phi)$ where $\hat{L}^2 = -\hbar^2 \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right]$ Solutions are Spherical Harmonics $Y_{l,m_l}(\theta, \phi)$. Quantum Numbers: Angular momentum quantum number $l = 0, 1, 2, \dots$ Magnetic quantum number $m_l = 0, \pm 1, \dots, \pm l$ Energies: Quantized, depends only on $l$. $E_l = \frac{\hbar^2}{2I}l(l+1)$ Ground state ($l=0$): $E_0 = 0$. First excited state ($l=1$): $E_1 = \frac{2\hbar^2}{2I}$. Second excited state ($l=2$): $E_2 = \frac{6\hbar^2}{2I}$. Degeneracy: For a given $l$, there are $(2l+1)$ possible values of $m_l$. Degeneracy = $2l+1$ $l=0$: Degeneracy = 1 (non-degenerate). $l=1$: Degeneracy = 3 (triply degenerate). $l=2$: Degeneracy = 5. Wavefunctions: $Y_{l,m_l}(\theta, \phi) = N P_{l, |m_l|}(\cos\theta)e^{im_l\phi}$ where $P_{l, |m_l|}(\cos\theta)$ are associated Legendre polynomials. $Y_{0,0}(\theta, \phi) = \frac{1}{\sqrt{4\pi}}$ $Y_{1,0}(\theta, \phi) = \sqrt{\frac{3}{4\pi}}\cos\theta$ $Y_{1,1}(\theta, \phi) = \sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi}$ $Y_{1,-1}(\theta, \phi) = \sqrt{\frac{3}{8\pi}}\sin\theta e^{-i\phi}$ Angular Momentum: $\hat{L}^2 Y_{l,m_l}(\theta, \phi) = \hbar^2 l(l+1) Y_{l,m_l}(\theta, \phi)$ (eigenvalue of total angular momentum squared) $\hat{L}_z Y_{l,m_l}(\theta, \phi) = m_l \hbar Y_{l,m_l}(\theta, \phi)$ (eigenvalue of z-component of angular momentum) Vector Quantization: Angular momentum vector makes discrete angles with the Z-axis. $\cos\theta = \frac{m_l}{\sqrt{l(l+1)}}$ Example for $l=1$: $m_l=0 \implies \cos\theta = 0 \implies \theta = \pi/2$ $m_l=1 \implies \cos\theta = 1/\sqrt{2} \implies \theta = \pi/4$ $m_l=-1 \implies \cos\theta = -1/\sqrt{2} \implies \theta = 3\pi/4$ Commutators: $[\hat{L}_x, \hat{L}_y] = i\hbar \hat{L}_z$. This means $\hat{L}_x, \hat{L}_y, \hat{L}_z$ cannot be simultaneously diagonalized. $[\hat{L}^2, \hat{L}_x] = [\hat{L}^2, \hat{L}_y] = [\hat{L}^2, \hat{L}_z] = 0$. $\hat{L}^2$ and one component (conventionally $\hat{L}_z$) can be simultaneously diagonalized.