* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Quantum Mechanics Lecture 30 Dr. Mauro Ferreira
Perturbation theory wikipedia , lookup
Particle in a box wikipedia , lookup
Bohr–Einstein debates wikipedia , lookup
Second quantization wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Perturbation theory (quantum mechanics) wikipedia , lookup
Dirac equation wikipedia , lookup
Path integral formulation wikipedia , lookup
Schrödinger equation wikipedia , lookup
Probability amplitude wikipedia , lookup
Matter wave wikipedia , lookup
Hartree–Fock method wikipedia , lookup
Canonical quantization wikipedia , lookup
Copenhagen interpretation wikipedia , lookup
Renormalization group wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Molecular Hamiltonian wikipedia , lookup
Atomic theory wikipedia , lookup
Wave–particle duality wikipedia , lookup
Coupled cluster wikipedia , lookup
Tight binding wikipedia , lookup
Wave function wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Quantum Mechanics Lecture 30 Dr. Mauro Ferreira E-mail: [email protected] Room 2.49, Lloyd Institute Suppose we want to calculate the ground state energy Egs of a system described by the Hamiltonian H, but we’re unable to solve the Schroedinger equation. The variational method Pick any arbitrary normalized wave function the quantity |Ĥ| ⇥ Egs ⇥ ⇤ |Ĥ| ⌅ ⇤H⌅ and calculate Egs ⇥ ⇤ |Ĥ| ⌅ ⇤H⌅ Proof: We start by writing as a linear combination of the (unknown) eigenfunctions of H. = cn where n Ĥ n = En n n Because 1= is normalized: | ⇥= m| cm m cn n n⇥ = cm cn m n m| n⇥ = n |cn |2 Meanwhile Ĥ⇥ = m |Ĥ| cm m cn n n⇥ But by definition: Egs ⇥H⇤ Egs n |cn |2 = Egs = cm En cn m En Q.E.D. n m| n⇥ = n En |cn |2 By minimizing <H> we’re certain to approach the true ground state Suppose that we want to find out the ground state for the one-dimensional harmonic oscillator 2 2 d m⇥ 2 x2 + =E 2 2m dx 2 We pick as a trial wave function: (x) = A e A is determined by normalization: A = H⇥ = T ⇥ + V ⇥ ⇥T ⇤ = 2 2m 2 |A| m 2 2 V⇥= |A| 2 2 ⇤ +⇥ dx e bx 2 ⇥ +⇥ dx e ⇥ b m 2 H⇥ = + 2m 8b 2bx2 d2 e 2 dx bx 2 ⇥ 2b bx2 ⇥1/4 2 b = 2m m 2 x = 8b 2 Now all we need to do is to minimize <H> 2 b m 2 H⇥ = + 2m 8b 2 d ⇥H⇤ = db 2m m b= 2 H⇥min 1 = 2 m 2 =0 2 8b min = Ae m 2 x2 If a solution is known, the so-called ladder operators generate new solutions that are higher or lower in the energy spectrum. (â ) 0 =0 (Ground state) 1 d [ + m⇥x] 2 m⇥ dx d 0 = dx d 0 m⇥ = x 0 0 (x) = Ae =0 0 m⇥ m 2 0 x2 m⇥ 2 dx x ⇥ ln 0 = x +C 2 m⇥ 1 )4 Normalization: A = ( gs lecture 8 To obtain the energy E0 associated with the state ψ0, we write the Schroedinger Eq. in the form 1 ⇥(â+ â + ) 0 = E0 0 2 and make use of the fact that (â ) E0 = 2 0 = 0 to conclude that Egs Once the ground state is known, we can obtain the excited states by repeatedly applying the ladder operator n (x) = An (â+ ) n 0 (x) In general, one can show that An = 1 En = (n + ) 2 1 n! lecture 8 2 b m 2 H⇥ = + 2m 8b 2 d ⇥H⇤ = db 2m m 2 =0 2 8b m b= 2 H⇥min min 1 = 2 H⇥min = Egs min = gs = Ae m 2 x2 Suppose that we want to find out the ground state for a delta-function potential 2 d2 ⇤ 2m dx2 ⇥(x) ⇤ = E⇤ We pick as a trial wave function: (x) = A e 2b A is determined by normalization: A = H⇥ = T ⇥ + V ⇥ ⇥T ⇤ = 2 2m 2 |A| ⇤ +⇥ dx e ⇥ +⇥ ⇥V ⇤ = 2 |A| 2 b ⇥H⇤ = 2m bx 2 dx e ⇥ 2b ⇥ 2bx2 d2 e 2 dx ⇥(x) = bx 2 ⇥ 2 ⇥1/4 b = 2m ⇥ 2b ⇤ bx2 same as before 2b ⇥ 2 b ⇥H⇤ = 2m 2 d ⇥H⇤ = db 2m ⇥H⇤min = m ⇥ ⌅ 2 2 2⇥b =0 2m2 2 b= ⇥ 4 The Hamiltonian contains a single bound state given by Ĥ (x) = E (x) where m ⇥(x) = and m 2 E= 1 e m |x|/ 2 2 Egs 2 (eigenfunction) (eigenvalue) (x) 0.75 0.5 0.25 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x 7 -0.25 lecture 10 2b ⇥ 2 b ⇥H⇤ = 2m 2 d ⇥H⇤ = db 2m ⇥H⇤min = m ⇥ H⇥min > Egs ⌅ 2⇥b =0 2m2 2 b= ⇥ 4 2 2 error ≈ 57% Let’s now use the variational method to estimate the ground state of the hydrogen atom We pick as a trial wave function: ⇤(r, , ⇥) = A e r/ 1 A is determined by normalization: A = ⇥ 3 H⇥ = T ⇥ + V ⇥ T⇥ = ⇥V ⇤ = ⇥H⇤ = 2 2m e2 2 2 2m e2 2 ⇥ ⇤H⌅min = m e4 2 2 d ⇥ ⇤H⌅ = d 2 m 3 + e2 2 =0 Hydrogen atom V (r) = scattering states e2 1 4⇥ 0 r Radial equation: r bound states 2 2 d u + 2 2m dr 1 r 2 e 4⇥ ⇥ 1 ⌅(⌅ + 1) + u = Eu 2 2m r 0 r 2 Vef f (x) Vef f = 2 e2 1 ⌅(⌅ + 1) + 4⇥ 0 r 2m r2 x =0 =1 =2 =3 lecture 21 cj+1 = 2(j + ⇥ + 1) 0 (j + 1)(j + 2⇥ + 2) ⇥ Recursive formula for the coefficients cj We can obtain all coefficients in terms of the initial one (co). The only way this expansion is compatible with the previous asymptotic analysis is if the series terminates at a finite value jmax. cjmax +1 = 0 ⇥ 2(jmax + ⇥ + 1) 0 =0 Defining the so-called principal quantum number n as n E= jmax + + 1 2 2 ⇥ = 2m ⇥ En = 0 me4 8⇤ 2 20 2 13.6 eV n2 ⌅20 Egs = 2n ⇥ En = ⇤ where n = 1, 2, 3, ... m 2 2 e2 4⇥ 0 ⇥2 ⌅ 1 n2 lecture 21 Let’s now use the variational method to estimate the ground state of the hydrogen atom We pick as a trial wave function: ⇤(r, , ⇥) = A e r/ 1 A is determined by normalization: A = ⇥ 3 H⇥ = T ⇥ + V ⇥ T⇥ = ⇥V ⇤ = ⇥H⇤ = 2 2m e2 2 2 2m e2 2 ⇥ ⇤H⌅min = m e4 2 2 d ⇥ ⇤H⌅ = d 2 m 3 + e2 2 =0 H⇥min = Egs Solution to the infinite well is 1.5 ⇥gs = ⇤ 1 2 sin L x⇥ L Egs = 0.5 0 0.5 2 2 2mL2 1 Trial wave function: = A x (L No variation parameter required x) A is determined by normalization: A = 30 L5 H⇥ = T ⇥ + V ⇥ H⇥ = T ⇥ = 2 5 2 A2 L3 = 6m mL2 error ≈ 1.3%