Quantum Mechanics Notes

Probability Calculations and Eigenfunctions

  • The advantage of expressing the wave function in terms of eigenfunctions is to calculate probabilities.
    • For discrete eigenvalues, the square of the coefficients represents the probabilities.
    • For continuous eigenvalues, the square of the inner product provides the probability density.

Inner Product and Coefficient Determination

  • Inner product is used to find the components of a wavefunction.

  • The x component is obtained via the inner product:

    • Analogy: If r=ax^+by^\vec{r} = a \hat{x} + b \hat{y}, then a=rx^a = \vec{r} \cdot \hat{x}.
    • Only the term corresponding to the eigenfunction survives when the dot product is performed.
  • The inner product allows calculation of coefficients:

    • If ψ(x,t=0)\psi(x, t=0) is known, the coefficients can be determined.
    • The square of these coefficients gives the probability of measuring an observable.

Inner Product Details

  • Inner product is analogous to a dot product. It involves a complex conjugate, important for imaginary eigenfunctions.
  • Example problem: A particle is confined between x=0x = 0 and x=Lx = L.
    • Part a: Find b<em>kb<em>k, the coefficient on the momentum eigenfunction. b</em>k2b</em>k^2 corresponds to the probability density in k-space.
    • Part b: Find the coefficient cc for x=L/2x = L/2. c2c^2 corresponds to the probability density in real space (position eigenfunctions).

Delta Function and Probability Density

  • Using a delta function, with height 1/a1/\sqrt{a} and width aa, to calculate the inner product.
  • The probability of finding the particle between L/2L/2 and L/2+aL/2 + a is to be determined.
  • Both momentum and position operators have continuous eigenvalues leading to probability densities.

Integral Forms and Evaluation

  • For the momentum operator case, the answer can be left in integral form.
  • For the position operator case, the integral should be evaluated.
  • To find bkb_k, the complex conjugate of the momentum eigenfunction is needed:
    • eikxe^{ikx}
    • The function is ψ(x)=ex/L\psi(x) = e^{-x/L}.
    • b<em>k=</em>eikxψ(x)dx=eikxex/Ldxb<em>k = \int</em>{-\infty}^{\infty} e^{-ikx} \psi(x) dx = \int_{-\infty}^{\infty} e^{-ikx} e^{-x/L} dx
    • b<em>kb<em>k will be an imaginary number, but b</em>k2|b</em>k|^2 will be real, as required for probability density.

Fourier Transform Connection

  • The calculation of bkb_k resembles a Fourier transform of the wave function.

Probability Calculation with Delta Function

  • Integrating from -\infty to \infty, but the delta function is only non-zero at x=L/2x = L/2.

  • The integral becomes the wave function evaluated at L/2L/2.

  • δ(xL/2)ψ(x)dx=ψ(L/2)\int_{-\infty}^{\infty} \delta(x - L/2) \psi(x) dx = \psi(L/2)

  • With the delta function: height 1/a1/\sqrt{a}, width aa

  • The result of the integral is ψ(L/2)/a\psi(L/2) / \sqrt{a}.

  • Squaring this gives the probability density: ψ(L/2)a2\left| \frac{\psi(L/2)}{\sqrt{a}} \right|^2

  • This c2c^2 represents a probability from L/2L/2 to L/2+aL/2 + a, not a probability density because we're not working in the limit of an infinitesimally small dxdx.

Generalization to Other Operators

  • This procedure applies to any operator.
  • For finding the probability of energy between EE and E+δEE + \delta E, use the Hamiltonian eigenfunction.
  • Example: Kinetic energy eigenfunctions for energy range (e.g., 1 joule to 1.3×10171.3 \times 10^{-17} joules).

Practical Measurement and Wave Function Role

  • Practical measurement involves detectors. Detector size can correspond to 'a'.
  • The wave function is analogous to position and velocity in classical mechanics.
  • From the wave function, kinetic energy, momentum, and probabilities can be calculated.

Hamiltonian Operator

  • The Hamiltonian operator HH is the sum of kinetic energy operator TT and potential energy operator VV: H=T+VH = T + V.
  • HH operates on the spatial part of the wave function.

Potential Energy Operator

  • Potential energy VV is a function of position xx.
  • Example: V(x)=12kx2V(x) = \frac{1}{2} kx^2 (harmonic oscillator potential).
  • This could model an electron acted upon by a spring-like force.

Potential Energy and Systems

  • Potential energy always relates to a system (e.g., electron-proton system in the hydrogen atom).
  • For the hydrogen atom, V(r)=ke2rV(r) = -\frac{ke^2}{r}, where rr is the distance between proton and electron.
  • The potential energy is a function of position and converted into an operator.

Forming the Potential Energy Operator

  • Replace position variables with operators.
  • For V(x)=12kx2V(x) = \frac{1}{2} kx^2, the operator is V^=12kx^x^\hat{V} = \frac{1}{2} k \hat{x} \hat{x}.
  • Applying operator twice, not squaring it. Not x^2\hat{x}^2 but x^(x^f(x))\hat{x}(\hat{x} f(x)).

Uniqueness of Quantum Mechanical Problems

  • Each quantum mechanics problem is unique due to its potential function V(x)V(x).
  • Different V(x)V(x) changes the Hamiltonian, thus changing the energy eigenstates.
  • Kinetic energy operator is the same for different interactions between the particle and surroundings.

Classical vs. Quantum Mechanics

  • Classical Mechanics: Problems differ due to forces acting on particle.
  • Quantum Mechanics: Problems differ due to potential energy function.
  • Newton's second law: F=maF = ma.
  • In quantum mechanics, knowing the potential energy function is essential. No potential energy means the problem is a free particle.

Free Particle

  • If the potential energy V=0V = 0, it's called a free particle problem.
  • The Hamiltonian is only the kinetic energy operator.
  • Potential energies can be derived from forces: ΔV=<em>x</em>1x2Fdx\Delta V = -\int<em>{x</em>1}^{x_2} F dx

Connecting Classical and Quantum Mechanics

  • Classical force can be used to find potential, which then becomes an operator for the Hamiltonian.
  • This applies at the atomic level where quantum mechanics is used.
  • Semiclassical models like Bohr's use classical mechanics to derive quantum models.

Energy Eigenfunctions and Potentials

  • Energy eigenfunctions depend on the potential energy.
  • To find eigenfunctions, solve the eigenvalue problem for the Hamiltonian.

Particle in a Box Problem

  • Simplest quantum mechanics problem.

  • Free particle eigenfunctions are also Hamiltonian eigenfunctions since H=TH = T.

  • Box potential: Potential energy is written as VV instead of PEPE (common for electron problems, where VV stands for voltage).

  • Potential: V(x)=0V(x) = 0 for 0 < x < L, V(x)=V0V(x) = V_0 elsewhere.

Classical Understanding of the Box

  • Classically, the particle is trapped in the box because the force is large at the boundaries.
  • Take the limit as V0 approaches infinity for trapping.

Quantum Mechanical Treatment

  • Time-independent Schrodinger equation Hψ=EψH \psi = E \psi.
  • In regions where VV approaches infinity, ψ\psi must approach zero.
  • ψ(x)=0\psi(x) = 0 for x<0x < 0 and x>Lx > L.
  • Both classical and quantum mechanics agree on this point.

Solving Inside the Box

  • Inside the box (0 < x < L), V=0V = 0, so H=TH = T.
  • Schrodinger equation becomes: 22md2ψdx2=Eψ-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi.
  • The function must be continuous and differentiable, so ψ(0)=0\psi(0) = 0 and ψ(L)=0\psi(L) = 0.

Boundary Conditions

  • Since ψ\psi has to be zero at x=0x=0 and x=Lx=L.
  • General solution is ψ(x)=Asin(kx)\psi(x) = A \sin(kx), where k=2mE2k = \sqrt{\frac{2mE}{\hbar^2}}.
  • To satisfy the condition ψ(L)=0\psi(L) = 0, kL=nπkL = n\pi, where nn is an integer.

Quantization of Energy

  • Energy is quantized: En=n2π222mL2E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}, where n is an integer.
  • The cosine function is not suitable because cos(0)=10\cos(0) = 1 \neq 0.
  • Sine function and linear combinations of exponential functions work because they vanish at x=0x = 0.

Discrete Energy Values and Eigenfunctions

  • Energy eigenfunctions are discrete and can be denoted by ψn(x)\psi_n(x).
  • The solutions are of the form ψn(x)=Asin(nπxL)\psi_n(x) = A \sin(\frac{n \pi x}{L}) where n=1,2,3n = 1, 2, 3…
  • Examples of Wavefunctions:
    • n = 1; A single hump
    • n = 2; two humps

Probability Density

  • Squaring the wave function gives probability density ψn(x)2|\psi_n(x)|^2.
  • For n=2n = 2, the probability density is zero at the middle of the box ψ(x=L/2)\psi(x=L/2).
  • In Classical Mechanics the probability is uniform across the box.
  • QM is different, there are positions where you will not measure the particle.

Space Quantization

  • Quantum mechanics results in space quantization, where particles are more likely to be found in certain locations.
  • This is analogous to Bohr's model where electrons can only be in certain orbits.

Large n Values and Correspondence Principle

  • For large n, the quantum mechanical probability density tends towards the classical prediction of uniform distribution.
  • This aligns with Bohr's correspondence principle.