Physics 115C Homework 2

Transcription

1 Physics 5C Homework Problem Our full Hamiltonian is H = p m + mω x +βx 4 = H +H where the unperturbed Hamiltonian is our usual and the perturbation is H = p m + mω x H = βx 4 Assuming β is small, the perturbation is small, and we can use perturbation theory to estimate the corrections to the energy levels. From perturbation theory, these first-order corrections are given by E n ) = ψ n ) H ψ n ) where the are the unperturbed eigenstates. For our H, this expression becomes ψ n ) E ) n = βn x 4 n where I m using the usual notation n ψ n ) for brevity. Now, we could turn the inner product into an integral and try to evaluate it explicitly; we ll have integrals over arbitrary Hermite polynomials, which will be pretty hideous although doable if we were to cleverly exploit generating functions). Instead, let s be less masochistic and use our usual ladder operators. Recall that we can express the position operator in terms of the ladder operators as x = a +a ) mω

2 We can therefore write E ) n = βn x 4 n ) = β n a +a ) 4 n mω We can save ourselves some time in evaluating this expression via the following observation: recall that the raising and lowering operators acting on energy eigenstates give a n = n n a n = n+ n Alsorecallthattheenergyeigenstatesareorthonormal:n m = δ nm. Now, inourexpression for E n ), we have a whole bunch of raising and lowering operators acting on n ; since the operators are being raised to the fourth power, the state a +a ) 4 n will be some linear combination of all the states from n 4 to n+4. But since we are then taking the inner product with n, the only contibuting terms in the expression for a +a ) 4 n will be those that are proportional to n. These terms will only appear if we have an equal number of raising and lowering operators acting on n, so that after raising and lowering, we end back up with something proportional to n. The conclusion is that the only terms we re interested in in the expansion of a +a ) 4 are those that contain two raising and two lowering operators; the rest we can discard. With this in mind, let s expand: a +a ) 4 n = a +a ) a +a ) a +a ) a +a ) n = [ a a aa+a aa a+a aaa +aa a a+aa aa +aaa a ] n +terms not containing n ) We evaluate the terms via the usual rules, e.g. a a aa n = na a a n Evaluating all the terms in this way, we get = n n a a n = nn )a n = nn ) n a +a ) 4 n = [ nn )+n +nn+)+nn+)+n+) +n+)n+) ] n +terms not containing n ) = 6n +6n+3) n +terms not containing n ) and therefore n a +a ) 4 n = 6n +6n+3

3 Our first-order energy shifts therefore become ) E n ) = β n a +a ) 4 n mω ) E n ) = β 6n +6n+3) mω Note that since the polynomial 6n +6n+3 is always positive, the first-order energy shifts are all positive: this is not surprising, since our perturbation is positive everywhere. 3

4 Problem a) The Hamiltonian for a particle with spin in a magnetic field is ignoring the spatial part) Our magnetic field is B = B ˆk +Bî, so H = γb S H = γb S z +BS x ) We can solve for the energy eigenvalues of this Hamiltonian in two different ways. The first is to use a brute-force approach: using the matrix representations of the spin matrices for spin-/, S z = ) S x = ) we can write the Hamiltonian as H = ) γ B B B B Finding the energy eigenvalues simply means finding the eigenvalues of the above matrix. This is a very simple and straightforward procedure, but I ll instead opt for a more physically intuitive approach. Note that the magnetic field is just a constant vector; in particular, since its x and z- components are B and B, respectively, its magnitude is B = B +B This field is pointing in a direction given by some unit normal ˆn which, if we wanted, we could express in terms of the unit vectors î and ˆk and B and B, but nevermind - we don t need to). Thus we can write B = B +B ˆn, so the Hamiltonian becomes H = γ B +B S n where S n is the component of the spin along the axis ˆn. Now, recall from last quarter or whenever you learned about spin) that we can diagolize any component of the spin that we want; in particular, we can diagonalize the component of the spin along any axis whatsoever, not just the x, y, z axes. By convention, we usually choose to diagonalize along the z axis i.e. we diagonalize S z ), but in this problem it seems natural to diagoanlize the component of the spin along the ˆn axis i.e. S n ). Recall that for 4

5 spin-/, the eigenvalues of the components of the spin are ± /, and therefore S n has eigenvalues ± /. Since H is proportional to S n, we thus find that the eigenvalues of H are E = γ B +B ± ) E ± = ± γ B +B b) Let s write the Hamiltonian as H = H +H where H = γb S z is the unperturbed Hamiltonian and H = γbs x is the perturbation. Since the eigenvalues of S z are ± /, the eigenvalues of the unperturbed Hamiltonian are E ) ± = ± γb and the unperturbed spin states are the eigenstates of S z, χ z) with χ z) corresponding to the energy E ) + and vice versa). The first-order shifts in the energy are thus E ) ± = χ z) H χ z) = γb χ z) χ z) S x This expression can be evaluated by either expressing S x in terms of the ladder operators via S x = S + +S )/, or by using the matrix and column vector representations of S x and. I ll do the latter: we have χ z) and thus χ z) = S x = ) ) χ z) + = ) E ) + = γb ) ) ) = E ) = γb ) ) ) = Evidently, the first-order corrections are zero, and we need to move on to second-order perturbation theory to get the first non-vanishing corrections. The second-order energy 5

6 shifts are given by E ) n = m n ψ m ) ψ H ) n E n ) E m ) In our case, the Hilbert space is two-dimensional, so n and m only take on the values ±. Using the fact that our unperturbed states were =, we get ψ n ) E ) ± = χ z) χ m H z) ± m ± E ) ± E m ) χ z) χ H z) ± = = γ B E ) ± E ) χ z) χ z) S x ± E ) ± E ) From our expressions for the unperturbed energies, we have and thus E ) ± E ) = ± γb E ) ± = ± γb B χ z) χ z) S x ± χ z) ± Finally, the matrix elements we need are χ z) χ z) S x + = ) ) ) χ z) + χ z) S x = = ) ) ) = Using these expressions, our second-order energy shifts becomes ) E ) ± = ± γb B = ± γb 4B 6

7 E ) ± = ± ) B 4 γb B The energy levels up to second order in the perturbation B are therefore E ± = E ) ± + [ = ± γb + B 4 γb ± +E ) ± +E ) E ± = ± γb [ + B B B ) ) ] 3 B +O B ) ) ] 3 B +O B c) Let s go back to our exact energy expressions from part a) and rewrite them in terms of the small parameter B/B : E ± = ± ) B γ B + If we Taylor-expand the square root, we get [ E ± = ± γ B + B B B ) ) ] 4 B +O B which agrees precisely with the expression we obtained in part b) above using perturbation theory. 7

8 Problem 3 a) This is a standard freshman E&M problem. Because the charged sphere is spherically symmetric, we expect the electric field to only point in the radial direction. If we draw a spherical Gaussian surface of radius r concentric with the charged sphere and containing it, then the total enclosed charge is e remember that our charged sphere is the nucleus of a hydrogen atom, which just has charge e). Then by Gauss law, the total flux through the surface must be proportional to the enclosed charge: E da = e ǫ But the electric field at the surface is uniform and normal to the surface, so the flux through the surface is just E da = EA = 4πr E where A = 4πr is the area of the surface. Thus outside the sphere, we have E = e 4πǫ r r > R) If we instead consider a spherical Gaussian surface smaller than the sphere i.e. of radius r < R), then the enclosed charge will be the volume enclosed by the surface times the charge density ρ of the sphere: Q enclosed = 4 3 πr3 ρ But the charge density is just the total charge of the sphere divided by its volume, so ρ = e/4/3)πr 3, and thus Q enclosed = 4 ) 3e r 3 3 πr3 = e 4πR R) 3 Thus inside the sphere, Gauss law gives E da = Q enclosed ǫ 4πr E = e ǫ r R ) 3 In short, E = e 4πǫ r R 3 r < R) Er) = e 4πǫ { r r R 3 r R r < R 8

9 To find the potential, we need to take the line integral of the electric field: Vr) Vr ) = r r E dr where r is some arbitrary reference position. It is conventional to take the potential to be zero at infinity, so let s choose r = and Vr ) =. Since the electric field points radially outward, if we integrate radially inward from infinity, we have that E dr = Edr, and thus r Vr) = Er) dr Plugging in our expression for Er), we get Vr) = e 4πǫ Evaluating the integrals, we get Vr) = e 4πǫ { r R { r Vr) = e 4πǫ R dr r dr + r r R R + r R 3 R { R r 3 r R r R r dr r < R R 3 ) r R r < R r R r < R Note that in the limit R, we recover the usual potential for a point charge, as we should. b) Our first step is to find the perturbing Hamiltonian that corresponds to our correction of treating the nuclear as an extended object. Recall that the unperturbed point-charge nucleus) Hamiltonian was H = p m ev pointr) where V point r) is the Coulomb potential of a point charge. Our Hamiltonian now is H = p m ev spherer) where V sphere r) is the potential of our uniformly charged sphere. But the perturbing Hamiltonian is simply the difference between the unperturbed and exact Hamiltonians: H = H H = ev sphere V point ). Using the expression for V sphere that we found above, we have V sphere V point = = e 4πǫ R e 4πǫ R { ) R r R r 3 r r < R R { r R 3 r R R r 9 r < R ) e 4πǫ r

10 and thus the perturbing Hamiltonian is { H = α r R R r 3 + R r < R R r where α e /4πǫ. Now, finding the first-order energy shift of the ground state is straightforward: by the usual perturbation theory, we have E ) = H where is the usual unperturbed) ground state of the Hydrogen atom. Using the functional form ψ r) = πa 3 e r/a we have E ) = = 4π = 4π α R = 4α a 3 R = 4α a H r) ψ r) d 3 r R R R a H r) ψ r) r dr r R 3 + R ) r dr r πa 3e r/a r 4 R 3 ) r +Rr e r/a dr ) R r 4 R 3 r 4 R + r R ) r/a dr e R Now, actually computing this integral isn t too bad, but here s where our approximation comes in: the reason H is only a perturbation is because the radius of the nucleus R is very small; in particular, R/a. Thus in this case, lowest order in the perturbation means lowest order in the parameter R/a. Because r < R everywhere in the region of integration, inside the integral r/a is also very small; this prompts us to expand the exponential: e r/a = r a + But notice that the entire expression for E ) is already of order R/a) because of the overall prefactor in front of the integral), so in fact, to lowest order in R/a, we only need the zeroth order term in the expansion of the exponential; i.e. we can take e r/a

11 Changing variables to x = r/r, the integral is now simply E ) 4α a ) R a x4 3 ) x +x dx ) ) R 4α a α 5a a R a ) + Explicitly keeping track of the powers of R/a we neglected, and writing the unperturbed) hydrogen ground-state energy as E ) = α/a, we get E ) = 4 5 E) ) R +O a ) 3 R a c) Since the Bohr radius is a 5.3 m, we have R/a 5, and therefore the lowest-order correction to the energy due is E ) ev) 5 ) 4 9 ev This correction is really tiny: the hyperfine structure of hydrogen which comes from interactio between the spins of the electron and proton) gives a correction on the order of 8 ev, so our correction is one order of magnitude smaller than that.

12 Problem 4 a) The Hamiltonian of an electron in a uniform electric field E is H = er E In our case, E = E ˆk, so the perturbing Hamiltonian is H = ee z Since we re interested in evaluating the matrix elements nlm H n l m, it will be useful to convert the perturbing Hamiltonian to spherical coordinates. To that end, let s write z = rcosθ, so that H = ee o rcosθ Now, we re interested in calculating the matrix elements nlm H n l m for the n = states; thus all we re really interested in are the elements lm H l m. From now on, I ll drop the n = label in the kets, and denote the matrix elements as lm H l m. There are four states to consider:,,, and. These four states form a four-dimensional degenerate subspace, so we have sixteen matrix elements to compute i.e. the matrix representation of H in this basis is a 4 4 matrix). Since H must be Hermitian, we really only need to calculate ten matrix elements: the four diagonal ones nl H nl and six off-diagonal elements; the other elements are related to these by complex conjugation: nl H n l = n l H nl. To begin, let s write down the position-space representation of the states we re interested in which I got from Griffiths Tables 4.3 and 4.7): ψ = R Y = r ) e r/a 8πa 3 a ψ = R Y = 8 πa 3 r a e r/a sinθe iφ ψ = R Y = ψ = R Y = 3πa 3 r a e r/a cosθ 8 πa 3 r a e r/a sinθe iφ Let s begin calculating matrix elements. A ton of the matrix elements are zero; I ll

13 highlight these by extracting the particular angular integral that integrates to zero: H = ee ψ rcosθr sinθdrdθdφ π H = ee π H = ee cosθsinθdθ = ψ ψ rcosθr sinθdrdθdφ e iφ dφ = = ee 8πa 3 = ee a 8 = ee a 8 = ee a = ee a = 3eE a H = ee π H = ee π H = ee π H = ee ψ ψ rcosθr sinθdrdθdφ 3πa 3 r a) 4 r a r a r 3r a ) r/a dr e a x 4 )e x x dx x 4 )e x x dx 4! 5! ) ) π e r/a dr π [ 3 cos3 θ ψ ψ rcosθr sinθdrdθdφ e iφ dφ = ψ rcosθr sinθdrdθdφ sin 3 θcosθdθ = ψ ψ rcosθr sinθdrdθdφ e iφ dφ = ψ ψ rcosθr sinθdrdθdφ ] π cos θdcosθ) π cos θsinθdθ ) dφ π e iφ dφ = 3

14 H = ee ψ rcosθr sinθdrdθdφ π H = ee π H = ee cos 3 θsinθdθ = ψ ψ rcosθr sinθdrdθdφ e iφ dφ = ψ rcosθr sinθdrdθdφ π sin 3 θcosθdθ = Whew! Those are all the independent matrix elements; the rest follow from complexconjugating the off-diagonal ones. In the basis {,,, }, the matrix for H is then lm H l m = 3eE a b) Recall that in degenerate perturbation theory, the first-order energy shifts are simply the eigenvalues of the perturbing Hamiltonian H ; that is, we need to diagonalize H the good states to use in perturbation theory are then the eigenstates of H ). Our first-order energy shifts are therefore simply the eigenvalues of the matrix lm H l m that we found above. We find these as usual: defining β 3eE a, the characteristic equation is = deth λi) λ β = λ β λ λ = λ λ β ) = λ λ+β)λ β) Thus the eigenvalues of H in our degenerate basis) are λ = and λ = ±3E ae with zero a double eigenvalue). These eigenvalues are just the first-order energy shifts, so the shifts are 3E ae E ) n= = 3E ae 4

15 Thus the applied electric field has broken the four-fold degeneracy into three distinct energies, leaving just a two-fold degeneracy at this order, anyway). c) Remember that our goal was to find the states that diagonalize the perturbation H ; these states are the good states that we use in perturbation theory. The good states can therefore be represented by the eigenvectors of our matrix lm H l m. From our expression for this matrix above, we see right away that two of the eigenvectors are = = Both of these eigenvectors correspond to the double) eigenvalue λ = that we found. The other two eigenvectors are slightly more complicated: because of the mixing between the and states as indicated by the fact that the matrix elements H and H are nonzero), the other two eigenvectors will be some nontrivial linear combination of and ; their column vector representation will therefore be of the form a b We could find the coefficients a and b by brute force computation, but the form of our matrix for H is so simple that we can simply read off what these remaining two eigenvectors must be: they must be proportional to the vectors with a proportionality constant of / for normalization. Thus our remaining two eigenvectors are 3 = 4 = The state 3 corresponds to the eigenvalue 3E ae, and the state 4 corresponds to the eigenvalue +3E ae. Remembering that these column vectors are representations in 5

16 the basis {,,, }, we can write the good states as = = E ) = ) 3 = + ) E ) = 3E ae) 4 = ) E ) = +3E ae) In particular, we notice that the applied electric field breaks the degeneracy between the states with quantum number m =, but doesn t affect the states with m = ±. 6