Model QM Problems with Exact Solutions

Lecture2A--Model QM Problems with Exact Solutions — (1-D)

(Ch 2.2-Levine, 3-3 Atkins, Ch. 2-R&S)

1. Free Particle -- If there is no potential then Schroedinger Equation becomes

T Y(x) = E Y(x) ==> -(h²/2m) d²/dx² Y(x) = E Y(x)

for this section: let underline "h" be "h-bar": h = h/2p

easiest solution to this has exponential form: (d/dx)e^x = e^x

but this has only one eigen value 1 and can’t well cope with eigen value

could use Y = e^-kx then k =(2mE/h²)^1/2 but would get wrong sign

so need e^-ikx recall: e^+ikx = cos kx + i sin kx

These solutions are plane waves INSERT GRAPH

starting point is due to phase (balance between sin & cos–arbitrary)

but don’t measure Y just |Y^*Y| –thus phase not significant (normally)

This describes motion of a free particle

no potential –> no force (F = -dV/dx)) — continuous motion

[Note these are eigen functions of momentum pY_± = (-ih) d/dx[e^±ikx] = ±hkY_±]

What is it doing?

e^±ikx --> motion to the pos x --> <p> = +kh

e^-ikx --> motion to the pos x --> <p> = -kh

2 solution: Y_± both solve the problem. These have different eigen values of momentum but different of energy depend on preparation–initial state

cos kx also an eigen function TY = Tcos(kx) = -k(-h²/2m)cos (kx) = EY

but p cos(kx) = -k(-ih)sin(kx) not eigen function

so real component, cos(kx), of e^±ikx has energy but ill defined momentum

— real wave function–represents motion left (-) and right (+)

--complex wave function–well defined linear momentum (-) or (+)

Note if energy is not well defined — then k varies - means wavelength varies

get wave packet -- super position of these with constructive interference -- if enough waves interfere see particle with some position, finite Dx

Time evolution:

Y (x,t) = Ae^-ikx e^-Et/h= Ae^i[kx ⁺ ^(k2h/2m)t]

evaluate at different times - point of constructive interference changes — phase shift

*try it with graphing calculator or program

2. Particle in a box with infinite sides–restrict motion - contain with V(x)

V = 0 à 0 < x < L

V = • à all else (x < 0, x > L)

INSERT GRAPH

Region I, III HY = EY --> -(h²/2m)(d²/dx²)Y = (E - •)Y

so particle of finite energy has no amplitude in this region Y (x)=0 (outside box)

special result due to V(x) = • outside (impenetrable) only need consider Region II

hence locally V = 0 — just as above for free particle Ae^-ikx

but V(x) provides restrictions on motion (in box) so that leads to quantized behavior

B.C. — boundary conditions, choose : Y = A cos kx + B sin kx

At wall Y(0) = Y(L) = 0 — must be a continuous, finite function

Y(0) = A sin kx (cos x ‚ 0 at 0 = x)

Y(L) = 0 ==> k = (np/L) where n = 1,2,3. . .integer, (n = 0 not allowed)

Energy is quantized by the B.C. (work it out)

k = (2mE/h²)^1/2 = np/L ===> E_n = n²p²h²/L²2m

Note: still need Aeⁱf, amplitude and phase--

Normalize: ÚY_n*Y_ndx = 1 ==> A = (2/L)^1/2===> Y_n (x) = (2/L)^1/2sin (npx/L)

Important to see what happens as modify B.C.

— as box enlarges Y toward free particle, i.e. the energy levels become continuous

— smaller box more energy goes as ~1/L² — note box constrains motion more,

Y has more curvature, second derivative is curvature, |T| inc. with curvature

— energy level separation expand with n^{2 --}property of steep sides

What are they? H-atom--example

more useful for smaller masses--eg. electron

If spectral transition: hn = DE = E_n - E_n' - change between levels

But–lower mass–electron~*2000 increase Energy over atom

---lower size--1Å (atomic) ~*100 inc Energy over 10Å (molecular)

conversely--> bigger box, heavier particle --> near continuous E --> classical corresp

Where is particle? -- Ú₀^LY*Ydx = 1 ==> it's in the box

Also do region probability, eg. Ú₀^L/2Y*Ydx = 1/2 ==> equal distribute by halves

But distribution is peaked (sin²) and this changes with different excited states

Alternate related Problems--Mess around with your box

2.a.. Translate: Box shift to run from -L/2 to L/2

INSERT GRAPH

Clearly energies must be the same but B.C. change so form of solution changes

Y₁ (x) = (2/L)^1/2cos (2px/L) --> sym

Y₂ (x) = (2/L)^1/2sin (4px/L) --> asym

Y₃(x) = (2/L)^1/2cos (6px/L) --> sym

Y₄ (x) = (2/L)^1/2sin (8px/L) --> asym

Phase shift, but states now have parity

2.b. Tie the ends together–particle on a ring: -- Looks like 2-D -- But really 1-D

because particle can’t move off the ring so only variable is q, R is fixed

need transform coordinates: x = R cos f and y = R sin f

d²/dx² + d²/dy² = d²/dR² + (1/R)d/dR + (1/R²)d²/df²

-h²/2m(d²/dx² + d²/dy²)Y = EY new variables: Y(R,f) = K_RF(f)

since R constant only the d²/df² term remains , K_R of the w/f is constant

-h²/2m(d²/df²)F(f) = EF(f) ===> F(f) = A e ^ibf

B.C. — correct F(0) = F(2p), ==> F(f) = (1/÷2p) e ^imf

2.c. Particle in 3-D box (Levine 3.5) if we go up to 3-D for one particle:

HY(xyz) = EY(xyz)

Now this is the same our simplest problem if we can do one coordinate at a time:

V=0 for 0<x<a, 0<y<b, 0<z<c and V= ƒ outside the box (rectangular)

INSERT DRAWING

Outside the box, it’s the same deal: Y(xyz)=0

Inside must solve

HY = -h²/2m(d²/dx² + d²/dy²+ d²/dz²)Y = -(h²/2m)—² Y(xyz) = EY (xyz)

Problem needs simplification-- Since —² = (d²/dx² + d²/dy²+ d²/dz²)

This is additive, no cross terms - can separate variables as for time dep. S.E.

-2mE/h²= (1/Y(xyz))—² Y(xyz) -- Choose Y(xyz)= X(x)Y(y)Z(z):

-2mE/h²= (1/X(x)) d²/dx² X(x) + (1/Y(y)) d²/dy² Y(y) + (1/Z(z)) d²/dz² Z(z)

Each of these is independent set (1/X(x)) d²/dx² X(x) = K_x,etc.

Then -2mE/h²= K_x + K_y+ K_z

Each one is simple 1-D particle in a box solution

X(x) = A sin (K_xx) = A sin (n_xpx/a) , E_nx = n_x²p²h²/a²2m

Put it together: E_n = (n_x²/ a² + n_y²/ b² + n_z²/ c²)p²h²/2m

Y_nxnynz (xyz) = (8/abc)^1/2sin(n_xpx/a) sin(n_ypy/b) sin(n_zpz/c)

These basically act independently but create a higher density of energy levels because we can excite modes (states) in 3 different coordinates

with proper (improper?) selection of a, b, c there could be states accidentally degenerate–e.g. let b = 2a, then E₄₁₁ = E₂₇₁

But it is more interesting to have symmetry, e.g. a=b=c now

E_n = (n_x² + n_y² + n_z²)p²h²/2m a² = (n_x² + n_y² + n_z²)E_{1 ,}where E₁ is part. box sol'n

Not important which coordinate is excited, just how many Quanta put in

E₁₁₁ = 3E₁ and E₂₁₁ = E₁₂₁ = E₁₁₂ = 6E₁ and E₁₂₁ = E₂₂₁ = E₂₁₂ = 9E₁ etc. Degenerate

INSERT GRAPH

2.d. Particle in a finite-well, what happens when V=0 0<x<L, but V=V outside

INSERT GRAPH

Expect the same form but need modify to account for

a) continuity at 0 = x, L = x but Y(0) ‚ 0 since V is finite, thus expect penetration as discussed based on curvature

b) also expect E_n to converge to continuous states as get higher E, higher n