< E >= 9/10E 1 +1/10E 3 = 9/10E 1 +9/10E 1 = 1.8E 1. no measurement can obtain this as its not an eigenvalue [U:1 mark]

Transcription

1 QM1) (a) normalisation ψ ψdx = 1 A (3ψ 1 +ψ 3 ) dx = A (9ψ 1 +ψ 3 +6ψ 1 ψ 3 )dx = 1 wavefunctions are orthonormal so = A (9 + 1) = 1 so A = 1/ 1 prob E 1 is 9/1. so prob < E 3 > is 1/1 < E >= 9/1E 1 +1/1E 3 = 9/1E 1 +9/1E 1 = 1.8E 1. no measurement can obtain this as its not an eigenvalue (b) ψmψdx = ψm nc n ψ n dx ψ m ψdx = nc n ψ m ψ n dx = nc n δ mn ψ m ψdx = c m so c n = ψ n ψdx c 1 = 1 ψ 1 ψdx = 3 1 x(1 x)sin(πx)dx = 6( 1 xsinπxdx 1 x sinπxdx) = 6[1/π (1/π 4/π 3 )] = 6.4/π 3 c = 1 ψ ψ(x)dx = 3 1 x(1 x)sinπxdx = 6( 1 xsinπxdx 1 x sinπxdx) = 6[1/π (1/π)] = [U:1mark] as expected as ψ(x) is symmetric about x = 1/ so will have a lot of ψ as this is also symmetric about this point but none of ψ as this is antisymmetric about x = 1/. (c) p = i hd/dx [x,p]ψ = (xpψ pxψ) [x,p]ψ = i h ( xdψ/dx d(xψ)/dx ) = i h ( xdψ/dx (xdψ/dx+ψ) ) = i hψ [S: marks] (d) < x >= 1 xsin (nπx)dx =.1/4 = 1/ < p >= i h 1 sin(nπx)d(sin(nπx)/dxdx = i h 1 sin(nπx)nπcos(nπx)dx = (could also get these from symmetry) 1

2 < x >= 1 x sin (nπx)dx = ( 1 1 ) = π n 3 π n < p >= h 1 sin(nπx)d (sin(nπx)/dx dx = h 1 n π sin (nπx)n π = h n π x p = (1/3 1/(n π ) 1/4)( h n π ) = h n π /1 1/ so minimize for n = 1 (where its.56 h so still above minimum limit from heisenburg) (e) dv(r)/dr = e /(4πǫ )d(1/r)/dr = e /(4πǫ )1/r = V(r)/r hence rdv/dr = V so < T >=< rdv/dr >= < V > But < E >=< T > + < V >=< T > < T >= < T > hence for any eigenstate ψ nlm where < E >= E n we have < T >= E n. this is a lot easier than calculating < T > as kinetic energy is a second order differential operator (f) (g) D(r)dr = = 1 64πa 3 π π π π ψ 11ψ 11 r sinθdrdθdφ r 4 a e r/a sin 3 θdθdφ = r4 e r/a π 64πa π sin 3 θdθ 5 = r4 e r/a 4 3a 5 3 = r4 e r/a 4a 5 max when dd/dr = so d(r 4 e r/a )/dr = 4r 3 e r/a + 1/ar 4 e r/a = r 3 e r/a (4 r/a) = so at r = 4a ( E) + = g eg + m p m e a 3 H ( E) H m + m e a 3 + g e g p [S: marks] = g + g p m p m + a 3 H a 3 +

3 a H a + = µ + µ H = m em + m e +m + m e +m p m e m p = m e m e m e +m p m p = 1 as m e << m p so ratio is / (1/) 3 = 8 (h) E 1 1 = 1/+ / 1/ / (α/ )sin πxdx. E 1 1 = α/(4π )[π +sinπ(1/ /) sinπ(1/+ /)]. [U:1mark] = α/(π )[π +sin(π π) sin(π+ π)] = α/(π )[π + sin π] α/(π )[π + π] = α. E 1 = 1/+ / 1/ / (α/ )sin πxdx. = α/(8π )[4π +sin4π(1/ /) sin4π(1/+ /)] = α/(4π )[4π sin(π )] α/(4π )[4π 4π ] = 3

4 QM) (a) so eigenenergies are 1 l(l +1) h Y m e a L lm = Y m e a lm E lm = E l = l(l+1) h m e a l = means m = so degeneracy 1 l = 1 can have m = 1,,1 degeneracy 3 general formula is l + 1 degeneracy (b) dp = Y Y sinθdθdφ prob within π/3 θ /π/ is = π π/ dφ YY sinθdθdφ π/3 = ( 15 ) π π/ sin θe iφ sin θe iφ sinθdθdφ 3π π/3 = ( 15 ) π (1 µ ) ( dµ) 3π 1/ = ( 15 ) 1/ (1 µ ) dµ = ( 15) [µ µ 3 /3+µ 5 /5] 1/ = ( 15) [1/ (1/) 3 /3+(1/) 5 /5] = ( 15) (1/) 5 [ 4 3 /3+1/5] = ( 15 ) 3 [ 8/3+1/5] = =.396 1

5 (c) L (Y ) = N. he iφ( θ icotθ ) sin θe iφ φ = N. he iφ( e iφ (sin θ) θ icotθsin θ (eiφ )) φ = hn e iφ( e iφ sinθcosθ icosθsinθie iφ) = hn e iφ( sinθcosθ +cosθsinθ ) = 4 hn sinθcosθe iφ = 4 hn Y 1 /N 1 = AY 1 (d) [L z,l ] = [L z,l x il y ] = [L z,l x ] i[l z,l y ] = i hl y i( i hl x ) = hl x +i hl y = hl L z (L Y lm ) = (L z L L L z +L L z )Y lm = hl Y lm +L L z Y lm = ( h+m h)l Y lm = (m 1) h(l Y lm ) (e) L + L = (L x +il y )(L x il y ) = L x +il yl x il x L y +L y = L x +L y i(l xl y L y L x ) = L x +L y i[l x,l y ] = L x +L y ii hl z = L L z + hl z so eigenvalues l(l + 1) h m h + hm h = h [l(l + 1) m(m 1)]

6 (a) ψ nx,ny,nz = X nx (x)y ny (y)z nz (z) and V = V x (x)+v y (y)+v z (z) so ( ) h YZ X Y m e x +XZ y +XY XYZ +(V z x +V y +V z )XYZ = EXYZ divide by XYZ to get ( h 1 ) X m e X x +V x + ( h 1 ) Y m e Y y +V y + ( h 1 ) Z m e Z z +V z = E these 3groupsdependonlyonx,y andz respectively (sowe canreplace / x with d/dx) so NONE of them can depend on x,y or z - they must be constants which we ll call E x,e y,e z. Then we get the three equations h 1 d X m e X dx +V x = E x h 1 d Y m e Y dy +V y = E y h 1 d Z m e Z dz +V z = E z where each is the same as the 1D schroedinger equation for the harmonic oscillator. and E x +E y +E z = E so E = (n x +1/) hω +(n y +1/) hω +(n z + 1/) hω = (n x +n y +n z +3/) hω (b) E 1,, = Ne ax / e ay / e az / λx yzne ax / e ay / e az / dxdydz = N λ x yze ax e ay e az dxdydz = as ye ay dy = (and same in z) 1

7 (c) W 11 = A z e ax e ay e az λx yzdxdydz = W = A y e ax e ay e az λx yzdxdydz = W 33 = A x e ax e ay e az λx yzdxdydz = W 1 = A yze ax e ay e az λx yzdxdydz = A λ x e ax dx y e ay dx z e az dz ( ) 1/ ( ) 1/ ( ) 1/ = A λ 1 π 1 π 1 π a 3 a 3 a 3 = a ( ) a 3/λ ( ) 3/ 1 π π 8 a = λ 3 4a W 13 = A xze ax e ay e az λx yzdxdydz = W 3 = A yxe ax e ay e az λx yzdxdydz = W ij = W ji and all terms are real so W ij = W ji (d) let b = λ/(4a ) then the matrix is E 1 b α b E 1 β E 1 γ = non trivial solution when determinant is zero so E 1 (E 1 ) b( be 1 ) = so E 1 [b (E 1 ) ] = i.e. E 1 = and E 1 = ±b. so the degeneracy is completely lifted. for E 1 = b then the matrix becomes b b α b b β = b γ so bα+bβ = so α = β and γ = so χ 1 = 1 (ψ 1 +ψ ) for E 1 = b we have bα + bβ = so α = β and γ = so χ = 1 (ψ 1 ψ ) fo E 1 = the matrix gives all α,β,γ = so χ 3 = ψ 3 as this is the only one which works for all values of E 1.