24.3. Asymmetric well potential It is interesting to see what happens when the bosonic potential has two asymmetric wells

| a and | b: in this case the asymptotic behaviours of the kink at x±∞are different and therefore we should expect to find two different spectra piling up on the two vacua. The purely bosonic model was discussed in Section 23.6, which showed that there could exist a range of the coupling constant where one vacuum (say | a ) has no bound states on top of it, while the other vacuum | b can have instead one or more. Let us now consider how the presence of the fermion may change this picture. In opportune units, as bosonic potential U(ϕ) of this model we choose U(ϕ) = 1 2

(ϕ +a)2(ϕ −b)2(ϕ2 +c), while as potential term V(ϕ) we take V(ϕ) = gϕ(x).

a. Compute the curvature at ϕ(0)

0

=−a and ϕ(0)

0

= b and show that a =

b, the two wells are asymmetric. From now on choose a > b.

b. Show that in this model there is a zero-mode ψ(0)

ab corresponding to the kink configuration ϕab(x) and another one ψ(0)

ba relative to the anti-kink ϕba(x), with the behaviour at x→∓∞of one equal to that one of the other at x→±∞, so that

ψ(0)

ab (x) = A ˆψ exp−  x x0 Vab(t) dt → A ˆψ evax 1+

n=1 tn enωax, x →−∞

ψ(0)

ba (x) = A ˆψ exp−  x x0 Vba(t) dt → A ˆψ evbx 1+

n=1 tn enωbx, x→−∞

where va ≡ −V(−a)=ga m λ , vb ≡ V

(b) =gbm λ .

c. Posing ηk = vk/(πM) and ξk = ωk/(πM) (k = a,b), where M is the mass of the kink, show that there are two different fermion spectra on the two vacua mFk (nk) = 2Msinπ
2 (ηk +(nk −1)ξk), nk = 1, 2, . . .1−ηk ξk .

d. Show that varying the coupling constants λ and g results in many curious situations at the two vacua as, for instance, no bosonic bound state on the vacuum | a and only one on the vacuum | b but a complete reverse situation for what concerns the fermionic spectrum!

1. Semi-classical energy spectrum Consider the semi-classical quantization condition  p(x)dx = n+ 1 2  h, where p(x) = 2m(E −V(x)), is the momentum of the classical particle with energy E equal to the energetic level En.
The integral is along the periodic orbit of the particle.
Compute the semi-classical energy levels for a potential V(x) = λ|x|a.
2. Variational principle Using the completeness relation ∞

n=0 |nn| = 1, show that for any quantum Hamiltonian the energy of the ground state satisfies the inequality E0 ≤
ψ | H | ψ
ψ | ψ .

3. Expectation values in the harmonic oscillator Consider the wave function ψω(x) of the ground state of an harmonic oscillator with frequency ω, here used as variational parameter ψω(x) = mω
¯hπ

1/4 e−mω
2¯h x2 .
Show that it holdsWe need to compute the matrix element of H, on this state as function of ω
ψω | p2 2m | ψω = ¯hω
4 , ψω | |x|a | ψω = √λ
π
a+1 2  ¯h mω

a/2 .