Double Well Quantum Mechanics

Transcription

1 Quantum Mechanics The double well potential is a rich toy example for understanding numerous quantum mechanical systems. At the time I made these notes, I was actually motivated by thinking about the decoupling limit and the AdS CFT correspondence. Here, I outline a method for modeling the double well dynamics that I have not seen elsewhere in the literature. The standard reference I know (which uses instantons) is in Coleman s Aspects of Symmetry, the chapter titled The double well done doubly well. We want to think of perturbatively relaxing the limit in which the barrier is infinitely high, and there is no tunneling. To do this we wish to add interaction terms which annihlate a particle in the left well and create a particle in the right well (and vice-versa). The Nonrelativistic Toy Problem We want to consider non-interacting non-relativistic particles in the potential V For convenience we define x x < x x x < x x V (x) =. () V x < x x 3 otherwise x = L x x = a x 3 x = L. ()

2 Our ultimate goal is to approximate this problem as two isolated square wells of widths L and L with some small interaction. This description should be valid for energies,, such that < V <. (3) For simplicity, we set L = L = L. The interaction should annihlate a particle in the left well and create a particle in the right well (and vice-versa). Thus, this problem naturally requires the use of a multiparticle formulation of nonrelativistic quantum mechanics. We first review how to second quantize the schrodinger equation to find such a description. Then, we argue that one can drop the time-derivative kinetic term of the action in the barrier region and integrate out those degrees of freedom. This removes the barrier region of the action, but leaves behind boundary terms. The boundary terms define nonvanishing boundary conditions for the interior edges of the two decouopled wells, and provide a coupling term. We treat the coupling term as a perturbation. To zeroeth order, then, one can neglect it and canonically quantize the two wells separately. Finally, one sees with what strength the coupling term couples modes in the left mode to modes in the right well. Second Quantized Schrodinger quation Note that the Lagrangian L = i has the Schrodinger quation, ( ) Ψ Ψ Ψ Ψ Ψ m Ψ V (x)ψ Ψ (4) i Ψ = m Ψ + V (x)ψ, (5) and its Hermitian conjugate as its equation of motion. We wish to canonically quantize this theory to describe multiparticle nonrelativistic quantum mechanics. We begin by finding the canonical momenta π = L Ψ = i Ψ which give us two primary constraints π = L Ψ = i Ψ, (6) The naive Hamiltonian has no kinetic term, φ = π i Ψ φ = π + i Ψ. (7) H N = π Ψ + π Ψ L = m Ψ Ψ + V (x)ψ Ψ, (8) double-well.tex January,, 7:5pm

3 a key hint that we need to use Dirac s generalization of Hamiltonian mechanics. The two primary constraints are second-class and generate no secondary constraints. The total Hamiltonian is given by H = H N + u φ + u φ. (9) Demanding that φ = {φ, H } P.B. = (weakly), () fixes one of the arbitrary functions u = iv (x)ψ, () and similarly one finds from the other consistency condition that u = iv (x)ψ. () Since we are interested in quantizing the system, we find the canonical Dirac brackets and then we can use the naive Hamiltonian. The Dirac brackets are defined by where {f, g} D.B. = {f, g} P.B. {f, φ i } P.B. ( M ) ij {φ j, g} P.B., (3) M ij = {φ i, φ j } P.B. = (σ ) ij = iɛ ij One finds that the nonvanishing canonical Dirac brackets are = ( M ) ij. (4) {Ψ(x), Ψ (y)} D.B. = iδ(x y) {Ψ(x), π(y)} D.B. = δ(x y) (5) {Ψ (x), π (y)} D.B. = δ(x y). pon canonically quantizing, then, one can describe the system as H ˆ = m ˆΨ ˆΨ + V (x) ˆΨ ˆΨ [ ˆΨ(x), ˆΨ (y)] = δ(x y). (6) Suppose that there is some complete set of orthonormal solutions to the classical energy-eigenvalue wave equation: We can expand ˆΨ in these modes as { n, f n } m f n + V (x)f n = n f n. (7) ˆΨ(x) = n â n f n (x) [â m, â n] = δ mn. (8) double-well.tex 3 January,, 7:5pm

4 The Hamiltonian becomes Ĥ = m,n = m,n [ â mâ n dx ] m f n + V (x)f n fm n â mâ n dx f n fm = n n â nâ n. (9) 3 Integrating out Barrier Degrees of Freedom We can formally break the action into three parts S tot. = S + S barrier + S = dt dxl + 3. Integrating out the Barrier dt +a L dxl + dt +a L+a dxl () We now want to use our assumption that the energies under consideration are much smaller than the potential energy in the barrier region Consider that and therefore typ.. () Ψ e it, L B = i ( ) Ψ Ψ ΨΨ Ψ m Ψ V (x)ψ Ψ = ( )Ψ Ψ Ψ m Ψ Ψ m Ψ V (x)ψ Ψ. () In this approximation, we have replaced, and thereby dropped the kinetic term. By dropping the kinetic term, we have made the barrier degrees of freedom nondynamical. We can integrate them out as auxillary variables by solving the equation of motion and then plugging it back into the action. For this + -dimensional example, the equation for the barrier degrees of freedom is which has a solutions of the form Ψ (x) = mψ(x) L < x < L + a, (3) Ψ(x) = A cosh ( m(x L) ) + B sinh ( m(x L) ). (4) double-well.tex 4 January,, 7:5pm

5 We can eliminate A and B in terms of the boundary degrees of freedom by using A = Ψ(L) B = Ψ(L + a) csch( ma) Ψ(L) coth( ma). (5) We then plug this into the barrier Lagrangian, to find L B = +a L L B [Ψ], (6) [ ] L B = m coth( ma) Ψ (L)Ψ(L) + Ψ (L + a)ψ(l + a) [ ] + m sinh( Ψ (L)Ψ(L + a) + Ψ (L + a)ψ(l). ma) (7) The first term provides the boundary conditions of the solution in the two regions independently, whereas the second term provides the coupling between the two wells. 3. Solving the two regions with boundary terms We wish to work in a the limit where the second term is a small effect and can be treated perturbatively. In which case, the unperturbed action for the left region is given by { i S = dt dx (Ψ Ψ Ψ Ψ) } Ψ m Ψ m coth( ma) dt Ψ (L)Ψ(L). (8) To determine the equations of motion we should demand that the linear variation of the action under arbitrary variations of Ψ and Ψ vanishes. The variations δψ and δψ must be subject to appropriate boundary conditions. In this case, we should leave the right edge unconstrained while the left edge should vanish. Therefore, we must be careful to not drop all of the boundary terms when integrating by parts. One finds { δs = dt dx i Ψ + } [ ] m Ψ δψ dt m Ψ (L) + m coth( ma)ψ(l) δψ (L), which implies that m Ψ (x, t) = i Ψ(x, t) Ψ (L, t) = m coth( ma)ψ(l, t) Ψ(, t) =. (3) The boundary term gave us a nonvanishing boundary condition. For convenience we define µ = m coth( ma). (3) (9) double-well.tex 5 January,, 7:5pm

6 The eigenvalue problem, then, is m g (x) = g (x) g() = g (L) = µg(l). (3) We take the ansatz g(x) = sin( mx), (33) which satisfies the D and the first boundary condition. The second boundary condition quantizes the energy: m cos( ml) = µ sin( ml) = tan( ml) = m µ There is a set of solutions, {α n } to this transcendental equation. The easiest way to visualize them is graphically, as shown in the figure. (34) In region, we must solve m h (x) + V h (x) = h (x) h (L + a) = h (L + a) = µh (L + a). (35) We take the ansatz h(x) = sin[ m( V )(x L a)] (36) which satisfies the D and the first boundary condition. The other boundary condition gives the transcendental equation tan[ m( V ) m( V )L] =, (37) µ which has a set of solutions {β n }, which are related to the α n via Let us, for simplicity, define β n = α n + V. (38) Φ (x) = Ψ(x) x [, L] Φ (x) = Ψ(x + L + a) in which case we can write the general solutions as x [, L], (39) Φ (x, t) = n A n sin( mα n x)e iαnt Φ (x, t) = n B n sin( mα n (x L))e iβnt (4) double-well.tex 6 January,, 7:5pm

7 3.. Properties of the igenfunctions One may worry that the peculiar boundary conditions may not be consistent with Sturm Liouville theory and therefore we may lose reality of eigenvalues and orthogonality of eigenfunctions with distinct eigenvalues. In fact one can demonstrate that one does have these properties for real µ. We consider the eigenfunctions of region, but everything works in the same way for region. Consider dx g m(x) g n (x) = m m dx gmg n L = g m(x) g n (x) dx g m(x) g n(x) = g m(l) g n (L) g m (L) g n(l) + dx g m (x) g n(x) = µ g m (L) g n (L) + µg m (L) g n (L) m n dx gmg n = (µ µ )gm(l)g n (L) m n dx gmg n. (4) In the above equation, the first term vanishes since µ = µ. If one considers the case m = n, then one finds that mm dx g m = m m dx g m = (m m ) dx g m =, (4) and therefore the eigenvalues m must be real. If one considers the master equation above for different n and m, and makes use of the reality of the eigenvalues, then ( m n ) dx g mg n =, (43) and therefore eigenfunctions of distinct eigenvalues are orthogonal. We will assume from now on that the eigenfunctions have been properly normalized. 3.3 Canonically Quantizing We canonically quantize by recognizing that H = H = { } dx Φ m Φ + m coth( ma)φ (L)Φ (L) { } dx Φ m Φ + V Φ Φ + m coth( ma)φ ()Φ () (44) double-well.tex 7 January,, 7:5pm

8 and following the usual procedure. We expand Φ and Φ in modes as (Schrodinger Picture) ˆΦ (x) = â n g n (x) ˆΦ (x) = ˆbn h n (x), (45) n n where assuming we have properly normalized g n and h n. When one plugs in, one needs to be careful of the boundary term: Ĥ = m Φ Φ (L) dx Φ m Φ + m coth( ma)φ (L)Φ (L) m = m coth( ma)φ Φ (L) + m coth( ma)φ (L)Φ (L) â m mâ n dx gm(x)g n(x) = n m,n α n â nâ n. (46) A similar relation holds for the second region, leaving one with Ĥ = n α n â nâ n Ĥ = n β nˆb nˆbn. (47) We now add the perturbation Ĥ int. = S int. = m sinh( ma) [ˆΦ (L)ˆΦ () + ˆΦ ] ()ˆΦ (L). (48) The first term is of the form Φ (L)Φ () = â mˆb n gm(l)h n () m n â L mˆb n sin( mα m L) sin( mα n L), (49) m,n and we see how the modes in one well couple to those of the other well. Note that the last line is not equality because I have not carefully normalized the modes. 3.4 The Probability to tunnel The interaction is Ĥ int = m sinh( ˆΦ (L)ˆΦ () + h.c. ma) = m sinh( â mˆb n g m (L)h n() + h.c. (5) ma) m,n double-well.tex 8 January,, 7:5pm

9 Let us consider an inital and final state of the form i = m f = n ; (5) that is, the particle tunnels from eigenstates m in the left well to eigenstate n in the right well. The matrix elements of the interaction Hamiltonian look like i Ĥint i = f Ĥint f = f Ĥint i = m sinh( ma) g m(l)h n() = γ. Let us assume that tunneling between these states dominates over tunneling between other energy levels; that is, let us assume that we can treat this as a two-state system. For this purpose, let α = α m β = β n (53) In the free energy eigenstate basis, the two-dimensional Hamiltonian becomes ( ) α γ Ĥ γ β = α + β β α σ 3 + γσ = α + β + v σ, (54) where v = ( ) γ,, β α. (55) The time-evolution operator, then, is given by Û(t) = e iĥt e i α+β t e i v σt = e i α+β t ( cos( v t) + i( σ ˆn) sin( v t) = ie i (α+β)t sin( v t) ( ) ( α β v γ = ie i (α+β)t sin( v t) γ v (5) ). (56) Thus, the amplitude to find the particle in the right well after time T, given that it starts in the left well is ( ) f Û(t) i = ie ( ) i (α+β)t sin( v t) σ ˆn ) ( ) γ = ie i (α+β)t sin( v t) γ γ + (β α) 4 α β double-well.tex 9 January,, 7:5pm

10 The probability, then, is simply The tunneling time is given by and the probability to tunnel is = ie i (α+β)t sin( v t). (57) + (β α) 4γ P (t) = T tunnel = = P tunnel = sin v t + (β α) 4γ. (58) π v π 4γ + (β α), (59) + (β α) 4γ. (6) When the energy levels are equal there is a resonance and the probability converges to unity. Of course, this is only an approximate result for our system, since we have neglected tunneling into other energy levels. Let us consider the case where α = β. Then the rate (one over the tunneling time) is given by Γ tunnel = = T tunnel π γ = π m sinh( ma) g m(l) h n(). (6) For the above problem with V =, this becomes Γ tunnel = g n (L) π m sinh( ma), (6) where The integral, we can do: g n (L) = sin ( mα n L) sin ( mα n x)dx. (63) sin ( mα n x)dx = L sin( mα n L) cos( mα n L) mα n = L tan( mαn L) mα n tan ( mα n L) + = L + µ mαn L mα n mα n L + µ = L ( ) µ + mα n L + µ double-well.tex January,, 7:5pm

11 = L L ( ) m coth( ma) + mα n L + m coth ( ma) ( ) + m L (64) Furthermore, we can approximate which gives Finally, to leading order, we have where ħ has been reintroduced. sin ( mα n L) = tan ( ) tan ( ) + Γ tunnel n = mα n mα n + µ α n α n + α n n π ml, (65) g m (L) n π ml 3. (66) 4n π e ma ħ ħ (67) ml 3 m 4 Comparison with Other Methods Now consider solving the quantum mechanics problem in the double well potential. Reflecting one well into the other commutes with the Hamiltonian, and so the energy eigenstates can have definite parity. Consider the symmetric and antisymmetric states which in each well look like the nth energy level. We call these states S n A n. (68) We can define L n = ( S n + A n ) R n = ( S n A n ). (69) The S states will have slightly different energy levels from the A state, and thus they evolve at slightly different rates: Û(t) L = ( e i S t S + e i At A ) double-well.tex January,, 7:5pm

12 = e ist ( S + e i( S A )t A ), (7) Thus, we see that T tunneling = π. (7) We need some way of estimating the difference in energy between the symmetric and the anti-symmetric state. 4. stimating the nergy Levels To begin consider the position-space energy eigenvalue equation: m ψ (x) + V (x)ψ(x) = Ψ(x). (7) We write the solution in the form A sin( [ mx) x (, L) m ( [ Ψ(x) = B cosh x (L + a m ))] ( + C sinh x (L + a ))] x (L, L + a) [ m ( ) ] D sin x (L + a) x (L + a, L + a). (73) Demanding that Ψ C (, L + a) imposes the following conditions (after assuming ) A sin( ml) = B cosh( m a) C sinh( m a) A cos( ml) = B sinh( m a) + C cosh( m a) D sin( ml) = B cosh( m a) + C sinh( m a) (74) D cos( ml) = B sinh( m a) + C cosh( m a). The symmetric solution must have C =, which leads to the energy eigenvalue equation S tan( m S L) = coth( m a ). (75) Similarly, the antisymmetric solution must have B =, which leads to the energy eigenvalue equation We can use the approximations tan( m A L) = A tanh( m a ). (76) tanh x e x coth x + e x x. (77) Let ε = e ma, (78) double-well.tex January,, 7:5pm

13 and then the two eigenvalue equations are tan( ml) = ( ± ε). (79) We solve the problem, perturbatively, to first order in ε. The zeroeth order result is degenerate. We write n = n + ε n. (8) Suppose tan( mnl) = n. (8) xpanding the left-hand side to first order in ε, one finds tan( m n L) = tan( m(n + εn)l) ( ) = tan m n( + ε n )L n ( m = tan m nl + εn = xpanding the right-hand side: n ) L tan( m nl) + tan(ε n m n L) tan( m nl) tan( m n ε nl) tan( mnl) + ε m n L n = tan( mnl) m ε nl n [ = tan( ] [ m mnl) + εn L + tan( m m nl) n = tan( m mnl) + εn L n n ] εnl ( + tan ( ) mnl). (8) ( ± ε) = n + εn ( ± ε) ( ) = n + ε n ( ± ε) n ( ) = n + ε n ± ε. (83) n Setting the two sides equal to each other one finds tan( m ( mnl) + εn L + tan ( ) ( ) m nl) = n + ε n ± ε n n double-well.tex 3 January,, 7:5pm

14 n m n ( ) ( ) L + n = n n ±. (84) n sing the approximation that n, one gets n m n L = n, (85) and therefore the first-order shifts in the energy-levels are given by ε n = n L = n L n π ml 3 m ε ma m e m e ma (86) This suggests that the energy difference between the symmetric and anti-symmetric states is given by = n π ma ml 3 m e, (87) and Γ = n π ma ml 3 m e. (88) This exactly matches the estimate from the previous method. double-well.tex 4 January,, 7:5pm