PHYSICS 721/821 - Spring Semester ODU. Graduate Quantum Mechanics II Midterm Exam - Solution

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1 PHYSICS 72/82 - Spring Semester 2 - ODU Graduate Quantum Mechanics II Midterm Exam - Solution Problem ) An electron (mass 5, ev/c 2 ) is in a one-dimensional potential well as sketched to the right (the potential becomes infinitely high below x = and above x = nm).. a) Write down the approximate ground state wave function ψ WKB in.5 the WKB approximation. What is the condition for the energy E of the. lowest bound state, using this WKB method? (Obviously, not enough x [nm] information is given for you to solve -. for E I just want to know how you would approach this task. Be as detailed as possible) The result equals the Bohr- Sommerfeld quantization condition. The general ansatz for the WKB solution is ψ(x) = C + with p(x) = p(x) ei/ x p(x')dx' + C p(x) e i/ x p(x')dx' 2m( E V(x) ). Since ψ(x = ) = ( potential wall), we have to require C C = C + and we can rewrite the solution as ψ(x) = p(x) sin x p(x')dx' ' with C = 2iC +. Since we must also require ψ(x = nm) =, we find that for any eigenstate nm ( ) dx' with energy E n we must have p(x')dx' = 2m E n V(x') = nπ (which equals the Bohr-Sommerfeld quantization condition). For the ground state, n = and we have to find a (analytical or numerical) solution E for the above equation where the lefthand side equals π. b) Continuing with your ground state wave function in the WKB approximation: There are two parameters: the ground state binding energy, which is E =. ev, and an overall normalization constant, which is C = 2.7 (ev/c) /2 nm -/2 (trust me ;-) With these values and within the WKB approximation, what is the probability to find the electron in the interval x = nm? Explain your reasoning V(x) [ev] nm

2 PHYSICS 72/82 - Spring Semester 2 - ODU Since the interval is very small, we can use the approximation d P(x nm) = ψ(x) 2 C 2 dx = p(x =.5 nm) sin2.5 nm p(x')dx' (. nm. ' From the graph, I read off that V(x =.5 nm) -.6 ev and therefore p(x) = 852 ev/c. Furthermore I observe that the potential (and therefore the function inside the integral) is symmetric around x =.5 nm, which means that the integral up to.5 nm should be ½ of the total integral up to nm, which we know is equal to π. Therefore, the 2 nd term on the r.h.s. simply becomes sin 2 (π/2) =. Plugging in the given constant, I get dp =.2 = 2.. Problem 2) In addition to quadrupole deformation, some nuclei even have octupole deformation. You don t need to know what that is, only that it is the ground state expectation value of one of a set of operators that together form a rank- spherical tensor T q. a) How many different components T q form this spherical tensor (i.e., what values can q assume)? Since the allowed values for q are -,-2,-,,,2,, there are 7 components of this (or any) spherical tensor of rank. b) What is the minimum total angular momentum J of a nucleus that has a non-zero octupole deformation in its ground state? Let s call its ground state wave functions α,j,m> with J M J. The octupole deformation is given by α, J, M T q α, J, M from which it follows that the total angular momentum of the tensor, k =, must be able to couple with J to yield J again: J J J +. This is only possible if J = /2 or larger, which is therefore the minimum ground state angular momentum of a nucleus with octupole deformation. c) How many non-zero matrix elements of the type α, J, M T q α, J, M ' can there be in principle? How many of them do I have to measure to be able to predict all others using only Clebsch-Gordan coefficients?

3 PHYSICS 72/82 - Spring Semester 2 - ODU There are four possible ground state wave functions if J = /2: M = -/2, -/2, /2 and /2. For any pair M, M, there is exactly one non-zero matrix element α, J, M T q α, J, M ', namely for q = M-M, for a total of 6. (Of course, many of them are trivially the same, since they are all real and therefore unchanged under exchange of M and M ). Only ONE of them has to be determined; that is sufficient to extract the reduced matrix element α, J T and the Wigner-Eckart theorem, all other 5. α, J and, with the help of Clebsch-Gordan coefficients Problem ) Consider particles of mass m in a three-dimensional box with sides a < b < c, i.e. the potential V(x,y,z) = if x a, y b, and z c; and infinite everywhere else. Eigenstates to this Hamiltonian can be characterized by three quantum numbers k, l, m = 8 kxπ lyπ mzπ,2, : ψ klm ( r ) = sin sin sin with abc a b c Hψ klm = E klm ψ klm ; E klm = 2 π 2 k 2 2m a + l 2 2 b + m2. Now assume that there are two identical 2 c 2 particles in this box. What are the two lowest possible energy levels of this two-particle system, their degeneracies, and the associated two-particle wave functions, if a) the two particles are Bosons (spin-)? For a single particle, the ground state wave function of course has the quantum numbers k,l,m =,,, and (apart from spin degrees of freedom) this state is non-degenerate. Since bosons can occupy the same state, the (also non-degenerate) ground state for two bosons is Φ ( r, r 2 ) = ψ ( r ) ψ ( r 2 ) (clearly symmetric under particle exchange) with an energy eigenvalue of E = 2 2 π 2 2m a + 2 b + ' = 2E 2 c 2. The lowest-energy excited state for a single particle must be ψ 2 with eigenvalue E 2 = 2 π 2 2m a + 2 b +. (Note that 2 c 2 increasing either k or l to 2 gives a LARGER energy increase over the ground state since a and b are smaller than c) Therefore, the st excited two-boson state must have one boson in the ground state and one in this first excited state. Since the wave function must be properly symmetrized, there is only one possible choice and the state is once again non-degenerate: Φ ( r, r 2 ) = ( 2 ψ ( r ) ψ 2 ( r 2 )+ψ ( r 2 ) ψ 2 ( r )) with E = E +E 2.

4 PHYSICS 72/82 - Spring Semester 2 - ODU b) the two particles are Fermions (spin-/2)? For the Fermions, we have to also take spin degrees into account. The two spin-/2 fermions can couple to total spin S = (symmetric) or S = (antisymmetric). Correspondingly, the spatial wave function must have the opposite symmetry to give an overall antisymmetric result. Therefore, the same symmetric spatial wave function Φ given above must be combined with spin S = to give the non-degenerate ground state with total energy E = 2E. For the first excited state, there are now more options: we can choose the symmetric wave function Φ from above and combine again with spin S =. Or we can replace the + sign in Φ with a minus sign to get the antisymmetric spatial wave function which must be combined with one of the three spin wave functions with total spin S = (m s = -,, and ). All of these states will have energy E = E +E 2, which means the st excited state is fourfold degenerate. Problem ) Consider the -dimensional Hamiltonian x H = x P2 +V(x) x where V(x) is some 2m arbitrary potential with two requirements: V(x) < for all x, and V(x) for x ±. Use the ideas underlying the variational method to show that there must be at least one bound eigenstate ψ E to this Hamiltonian with eigenvalue E <. You may use the fact that there is a well-known solution of the stationary Schrödinger equation H = E with E <, for the case were V(x) is replaced by V the potential V(x), V (x) =, x < L else parameters >. ', with V and L being some arbitrary ( The figure below gives one example of how V(x) and V(x) might look like

5 PHYSICS 72/82 - Spring Semester 2 - ODU Following the hint, we are going to show that there is a trial wave function for which the expectation value of the energy of the original Hamiltonian is negative, and therefore that the lowest (ground state) energy eigenvalue for this Hamiltonian negative as well, as it MUST be less or equal to the expectation value for any trial wave function. Since the potential goes to at large absolute values of x, any state with E < is by definition bound because beyond a certain range in x, E V(x) will become negative and therefore the wave function will fall off exponentially. Since V(x) is negative for all x, we can always find parameters L and V for the potential V(x) such that V(x) V(x) for all x. For instance, we can choose L pretty much arbitrarily and then simply define V = max(v(x), x = L... + L) which must be positive. We use as our trial wave function the ground state eigenstate solution of the modified Hamiltonian with THIS potential V(x), which we may assume has E <. (It is straightforward if someone tedious to derive the exact form of this solution and the value of E ). We know that H (x) = 2 2 2m x (x)+v (x)ϕ 2 (x) = E (x) for all x. Now all we have to do is calculate the expectation value <E> for the energy of the original Hamiltonian with this wave function: ( ) E = H = H + H H = E + ϕ * (x) V(x) V (x) (x)dx ( ) = E + V(x) V (x) (x) 2 dx

6 PHYSICS 72/82 - Spring Semester 2 - ODU Since (x) 2 is always positive and V(x) V (x) ( ) is always negative or equal to zero (by construction of V(x)), the integral is less or equal to zero. Therefore, the expectation value of H for is negative, and therefore its lowest energy eigenvalue must be negative, as well, q.e.d. Problem 5) The density matrix for an ensemble of N spin-/2 particles is given by ρ =. Calculate the polarization vector P. What is your interpretation in terms of the spins of the individual particles? According to the known properties of the density matrix, the expectation value (averaged over the ensemble) of any operator O is given by O (Oρ). The components of the polarization vector are given by the expectation values of the three Pauli matrices: P x = σ x = 2 P x = σ y i i i i i i = P z = σ z = 2 So there is some net polarization in x and z direction, but not in y-direction. Furthermore, the total length of the polarization vector is P = / 2 which means the ensemble is

7 PHYSICS 72/82 - Spring Semester 2 - ODU NOT totally polarized (e.g. along the 5 degree line in the x-z plane). Some possible interpretation is that ½ of all particles are polarized along the x-axis and the other half along the z-axis (meaning they are in eigenstates of the corresponding spin components). Or, it could also be that / 2 of the particles (around 7) are polarized along the 5 degree line and the remainder are completely randomly oriented. Or many other possibilities the density matrix cannot distinguish between them (and, therefore, neither can any conceivable experiment).