Physics 505 Homework No. 9 Solutions S9-1

Transcription

1 Physics 505 Homewok No 9 s S9-1 1 As pomised, hee is the tick fo summing the matix elements fo the Stak effect fo the gound state of the hydogen atom Recall, we need to calculate the coection to the gound state enegy to second ode in the petubation due to an extenal field This coection is E 1 = e E m= m, 1, 0 z 1, 0, 0 E 1 E m To simplify the notation, lets call 1, 0, 0 = 0, the gound state with enegy E 0 and call m, 1, 0 = n with enegy E n and n 1 So, we want to compute E 0 = e E n=1 (a Suppose we had an opeato A such that n z 0 E 0 E n z 0 = (AH 0 H 0 A 0, (1 whee H 0 is the unpetubed Hamiltonian fo the hydogen atom, H 0 = p m e, whee m is the educed mass of the electon and poton Show that n z 0 = (E 0 E n n A 0 Also show that n z 0 = 0 za 0 0 z 0 0 A 0 = 0 za 0 E 0 E n n=1 n z 0 = n AH 0 0 n H 0 A 0 = (E 0 E n n A 0, Since H 0 is Hemitian Also, n z 0 = E n=1 0 E n n 1 0 z n (E 0 E n n A 0 E 0 E n = n 1 0 z n n A 0 = n 0 z n n A 0 0 z 0 0 A 0 = 0 za 0 0 z 0 0 A 0 = 0 za 0, Copyight c 01, Edwad J Goth

2 Physics 505 Homewok No 9 s S9- whee the second tem has been dopped, since 0 z 0 = 0 End (b So, if we knew A, we could get the answe just by calculating one matix element If we assume A is a function only of coodinates, then equation (1 is an inhomogeneous diffeential equation fo A If you e eally good at diffeential equations, you could solve it The esult is A = ma ( h + a z Show that this expession does, in fact, solve equation (1 (Note that the nomalization of 0 cancels out, so you can just take 0 = exp( /a We need to calculate ( ( h m + ( h 1 m sin θ θ sinθ θ ( ma h e ( cos + a θ e /a This is the second tem on the ight hand side of equation (1 Thee may be some ways to simplify this, but in the end, it appeas bute foce is equied, so, we ll just evaluate each tem Fist of all, the second deivative tem is cos θ a (( + a e /a = cos θ a (( ( + a 1 ( a + a e /a = cos θ a (1 a ( + a + 1a ( + a e /a ( a = cos θ + e /a 4a The fist deivative tem is cos θ a (( + a e /a = cos θ a The angula deivative tem is a ( 1 sin θ ( ( + a 1 ( a ( = cos θ a e /a θ sin θ ( cos θ θ + a e /a = a ( cos θ ( a = cos θ + a + a e /a + a e /a e /a Copyight c 01, Edwad J Goth

3 Physics 505 Homewok No 9 s S9-3 The potential tem is e ( ma h ( + a cos θ e /a = cos θ ( a e /a The fist tem on the ight of equation (1 is consideably easie to evaluate, AH 0 0 = A ( e a e /a ( = ( e ma a h ( = cos θ 4a + ( cos + a θ e /a e /a We now add up the esults of the last 5 calculations to find ( a (AH 0 H 0 A 0 = cos θ + 4a + a + a + a a + 4a + e /a = cos θ e /a = z 0, which is what was to be shown! End (c Calculate the Stak effect enegy shift fo the gound state of hydogen to second ode in the applied field We need to evaluate 0 za 0 Hee, the nomalization of 0 mattes, so we use 0 = (a 3/ (4π 1/ exp( /a Also, za is popotional to z As fa as evaluating the matix element, we can use symmety to eplace z by /3 So, ( e 0 za 0 = 0 + a 4 3 a 3e /a d = 9 4 a3 Finally, we need to include E in the enegy shift, so E 1 = 9 4 a3 E End Copyight c 01, Edwad J Goth

4 Physics 505 Homewok No 9 s S9-4 A paticle in a D box (Based on a poblem fom Mezbache A paticle is confined to a squae box, 0 x L and 0 y L We ae not inteested in the z-motion, so this is a D poblem (a Obtain the enegies and eigenfunctions What is the degeneacy of the few lowest levels? The wave function must vanish at the boundaies of the box This means nm = sin(nπx/l sin(mπy/l, L whee n and m ae integes geate than 0 The enegy is E nm = h π ml ( n + m The degeneacy has to do with how many ways one can choose n and m to give the same n + m The gound state is non-degeneate with n = m = 1 The fist excited state has a degeneacy of with n =, m = 1 o n = 1, m = The next excited state is non-degeneate with n = m = The thid excited state is doubly degeneate with n = 3, m = 1 o n = 1, m = 3 That s enough! End (b A small petubation V = Cxy, whee C is a constant, is applied Find the enegy change fo the gound state and the fist excited state in the lowest non-vanishing ode Constuct the appopiate eigenfunctions in the case of the fist excited state We calculate L 0 x L sin n 1πx L sin n πx L L π dx = sin n π 1 x sin n x x dx 0 = L π π 0 (cos((n 1 n x cos((n 1 + n x x dx The integals can be evaluated with an integation by pats If n 1 n, the esult is L 0 x L sin n 1πx L sin n πx L dx = L ( ( 1 n 1 n 1 1 π (n 1 n ( 1n1+n (n 1 + n If n 1 and n ae both odd o both even, the esult is 0 If one is odd and the othe is even, the esult is L x L sin n 1πx L sin n πx n 1 n dx = 8L L π (n 1 n 0 Copyight c 01, Edwad J Goth

5 Physics 505 Homewok No 9 s S9-5 If n 1 = n, the esult is L/ So, the gound state changes enegy by E 11 = 11 Cxy 11 = CL /4, since the expectation value is the poduct of two of the integals just discussed with n 1 = n = m 1 = m = 1 The fist excited state is degeneate, so we need to choose a basis which diagonalizes the petubation We calculate the matix elements fo all the states: ( ( 1 V 1 1 V 1 = CL 1/4 56/81π 4 1 V 1 1 V 1 56/81π 4 1/4 The eigenvalues of this matix ae ( 1 E fist excited = CL 4 ± 56 81π 4, with eigenvectos fist excited ± = 1 ( 1 ± 1 End 3 Hypefine splitting of the hydogen gound state As you know, the spatial pat of the hydogen gound state is vey simple: ψ 100 (, θ, φ = exp( /a a ( 3/ π ( 1/ Since thee is no obital angula momentum, thee is no spin obit effect Howeve, the gound state has a degeneacy of 4 since both the poton and the electon have spin 1/ The spins can align, giving a tiplet state, o anti-align, giving a singlet state Since thee ae magnets associated with the spins, we expect that thee should be a diffeence in enegy between the tiplet and singlet states The nuclea spin is often denoted by I and poduces a magnetic moment µ p = eg p m p c I, whee µ p is the magnetic moment of the poton, g p is its g-facto and m p is the mass of the poton (Note: to conside othe nuclei, we would use the appopiate Z, g, and m We take the poton as fixed at the oigin and it poduces a magnetic field, B( = 3e (e µ p µ p 3 + 8π 3 µ pδ(, (Jackson, Classical Electodynamics, nd ed, p 184 The inteaction enegy of this field and the magnetic moment of the electon is H HF = µ e B = 3(e µ e (e µ p µ e µ p 3 8π 3 (µ e µ p δ( Copyight c 01, Edwad J Goth

6 Physics 505 Homewok No 9 s S9-6 Aside: If we wee to conside othe than s-states, the hypefine Hamiltonian would also include a spin obit tem due to the inteaction of the magnetic moment of the nucleus with magnetic field poduced by the moving electon(s Evaluate the hypefine Hamiltonian above fo the gound state of hydogen How does it depend on the poton and electon spins? O, what is the enegy diffeence between the singlet and tiplet states? Which is the actual gound state: tiplet o singlet? What ae the wavelength and fequency of the adiation emitted o absobed in the tansition between these states? Hint: can you show that the fist tem in H HF vanishes fo s-states? The fist tem in the hypefine Hamiltonian is zeo by the following agument The numeato times is 3x i µ ei x j µ pj µ e µ p When we take the expectation value with an s-state, the angula integal will eliminate coss tems x i x j with i j The squaed tems aveaged ove angles become /3 This leaves 3µ e µ p /3 µ e µ p = 0 The δ-function in the second tem is a thee dimensional delta function and it just picks out the value of the integand at the oigin So, The poduct of the magnetic moments is 100 8πµ e µ p /3 100 = 8µ e µ p /3a 3 µ e µ p = eg e m e c eg p m p c S I [ 1 = g eg p h µ Bµ N ( F S I ], whee µ B and µ N ae the Boh and nuclea magnetons, and F is the total spin S = I = 3 h /4 Fo the singlet state F = 0, F = 0 and the quantity in backets is 3 h /4 Fo a tiplet state, F = 1, F = h and the quantity in backets is + h /4 Putting eveything togethe, we wind up with H HFt = + g eg p µ B µ N 3a 3 H HFs = g eg p µ B µ N a 3 H HFt H HFs = + 8g eg p µ B µ N 3a 3 We see that the singlet is the gound state Copyight c 01, Edwad J Goth

7 Physics 505 Homewok No 9 s S9-7 Fo the numeical evaluation, we use g e =, g p = 559, µ B = eg G 1, µ N = eg G 1, a = cm We find, H HF = eg = ev ν HF = Hz λ HF = 1 cm This tansition is the famous 1 cm line of neutal Hydogen which is seen all ove the sky It s one of the pincipal ways to study ou galaxy and othe galaxies with adio telescopes! End 4 Zeeman splitting We conside an atom with a single valence electon, subject to a magnetic field B = Be z in the z-diection The Hamiltonian fo the electon is whee H = H 0 + H so + H B, H 0 = P m + V (, accounts fo the dominant electic inteaction of the electon (fo Hydogen, V ( = e /, fo alkali metals, V ( takes account of the filled shells in an appoximate way The spin obit inteaction is H so = 1 1 dv m c d L S = f( L S The inteaction with the applied magnetic field is H B = eb mc (L z + gs z = eb mc (L z + S z = eb mc (J z + S z, whee the tem popotional to B has been dopped Also, some othe small tems, fo example, the elativistic coection to the momentum, have been dopped since they don t give a splitting dependent on j, l and s In the calculations below, we ae inteested in H so and H B ; H 0 detemines the zeoth ode enegies and states which ae used in computing expectation values of, fo example, f(, but can othewise be ignoed (a Suppose the magnetic field is vey weak What ae the appopiate basis states and what ae the spin-obit and Zeeman splittings? If the magnetic field is completely tuned off, thee is only the spin-obit inteaction The appopiate states ae those of total angula momentum, z-component of angula momentum, and obital and spin angula momentum: njm j ls These states diagonalize the spin-obit inteaction which is assumed to be lage than the Zeeman inteaction h { l j = l + 1/ H so = f( nl (l + 1 j = l 1/ Copyight c 01, Edwad J Goth

8 Physics 505 Homewok No 9 s S9-8 Using these basis states we need to calculate the matix elements of the Zeeman tem, H B = njm j ls eb mc (J z + S z njm jls The matix element of J z is just m j h The matix element of S z equies moe wok In paticula, the states njm j ls must be witten in tems of nlm l sm s Recall the expansions given in lectue, l + mj + 1/ l mj + 1/ j = l + 1/, m j, l = + lm j 1/ + lm j + 1/ l + 1 l + 1 l mj + 1/ l + mj + 1/ j = l 1/, m j, l = lm j 1/ + lm j + 1/ l + 1 l + 1 With these expansions given in lectue, we find J z + S z = hm { j l + j = l + 1/ l + 1 l j = l 1/ Both cases ae coveed by J z + S z = hm j j + 1 l + 1 = hm jg, whee the g-facto vaies fom fo l = 0 to 1 fo l Then the Zeeman tem is H B = gm j µ B B Each spin-obit level is split into j + 1 equi-spaced levels by the Zeeman effect End (b Now suppose the magnetic field is vey stong so the Zeeman tem is lage than the spin-obit tem What ae the appopiate states and what ae the Zeeman and spin-obit splittings? The Zeeman tem is diagonal in the basis lm l m s and H B = eb mc L z + S z = eb mc h(m l + m s = (m l + m s µ B B In this basis, the spin-obit tem is easy to evaluate, L S = L x S x + L y S y + L z S z = L z S z = h m l m s Copyight c 01, Edwad J Goth

9 Physics 505 Homewok No 9 s S9-9 Then H so = m l m s h f( nl Note that in this limit thee ae some degeneacies Fo example, if l = 1, then states m l = 1, m s = +1/ and m l = +1, m s = 1/ ae degeneate End (c Suppose that neithe the spin-obit no the Zeeman effect is appeciably lage than the othe How would you detemine the level splittings in this case? (This is a shot essay question, no calculations ae equied! Pick a convenient basis fo the levels that ae degeneate in the absence of the spinobit and Zeeman inteactions Calculate the matix elements in this basis The eigenvalues of this matix ae the enegy shifts in the states that coespond to the eigenvalues End 5 Viial theoem fo a paticle in a fixed potential (See Schwabl, chapte 1 Conside x p and a Hamiltonian H = p /m + V (x (a Show that We conside p fist: ( p [H, x p] = i h m x V (x [p, x p] = p j p j x i p i x i p i p = p j (x i p j i hδ ij p i x i p i p = pjx i p j p i i hp x i p i p = (x i p j i hδ ij p j p i hp x i p i p = x i p i p i hp x i p i p = i hp Now V (x: So, [V (x, x i p i ] = V (xx i p i x i p i V (x ( h = V (xx i p i x i V (x + V (xp i i x i = +i hx V (x ( p [H, x p] = i h m x V (x Copyight c 01, Edwad J Goth

10 Physics 505 Homewok No 9 s S9-10 End (b If ψ is a stationay state of H, H ψ = E ψ, show that ψ p m ψ ψ x V (x ψ = 0, and theefoe, fo the Coulomb potential, ψ H ψ + ψ Ze ψ = 0 ψ [H, x p] ψ = ψ Hx p x ph ψ = ψ Ex p x pe ψ = E ψ x p x p ψ = 0 With the Coulomb potential, we have x ( Ze = x So, ( + Ze x = + Ze 3 0 = ψ p m ψ ψ x V (x ψ = ψ p m ψ ψ Ze ψ = ψ p m ψ ψ Ze ψ + = ψ H ψ + ψ Ze ψ End ψ Ze ψ (c Detemine 1/ nl fo the hydogen atom Fom the above, we have E n = 1 Ze, and we know, E n = Z e /an, so 1 nl = Z an End Copyight c 01, Edwad J Goth