Unavoided crossing of energy levels in PT -symmetric Natanzon-class - PowerPoint PPT Presentation

. Unavoided crossing of energy levels in PT -symmetric Natanzon-class potentials G´ eza L´ evai Institute for Nuclear Research, Hungarian Academy of Sciences (MTA Atomki), Debrecen, Hungary AAMP19, Prague, 6-9 June 2016

Important novelties in PT QM There can be two sets of normalizable wave functions Discriminated by the q quasi-parity quantum number These states merge when the spectrum is complexified Then they are related by PT ψ ( q ) n ( E ) = ψ ( − q ) ( E ∗ ) n They can undergo unavoided crossing Then the two wave functions become dependent This is very different from the Hermitian setting

The first examples The PT harmonic oscillator MZ PLA 1999 G V ( x ) = x 2 − 2i cx + x ∈ ( −∞ , ∞ ) ( x − i c ) 2 Unavoided crossing of energy levels E ( q ) n n = 4 n + 2 − 2 qα ψ ( q ) L ( qα ) (( x − ic ) 2 ) n ( x ) expressed via n The PT Coulomb potential MZ, GL PLA 2000 Similar results The background: ( z ) ∼ z j L ( j ) L ( − j ) n − j ( z ) for integer j n But is α = ± j legal?

A further, more recent example The PT Scarf II potential Ahmed et al. PLA 2015 Unavoided crossing of energy levels The background: ( z ) ∼ (1 − z ) j P ( j,β ) P ( − j,β ) n − j ( z ) for integer j n Very similar results But is α = ± j legal? Are there further exactly solvable examples?

The Book ∗ said... * M. Abramowitz and I. A. Stegun, Handbook of Mathematical functions (New York, Dover, 1970), Eq. 22.3.1

...but temptation was too strong...

...which led to expulsion from Paradise

A reminder on the exact solutions of the Schr¨ odinger equation An old variable transformation method Bhattacharjie and Sudarshan 1962 Schr¨ odinger eq. = ⇒ differential equation of special function F d 2 ψ d x 2 + ( E − V ( x )) ψ ( x ) = 0 insert ψ ( x ) = f ( x ) F ( z ( x )) and compare with d 2 F d z 2 + Q ( z )d F d z + R ( z ) F ( z ) = 0 to get � 2 � z ′′ ( x ) � � E − V ( x ) = z ′′′ ( x ) 2 z ′ ( x ) − 3 R ( z ( x )) − 1 d Q ( z ) − 1 + ( z ′ ( x )) 2 4 Q 2 ( z ( x )) . 4 z ′ ( x ) 2 d z Schwartzian derivatve terms E and the main potential terms

New features of PT -symmetric potentials Unusual trajectories off the real x axis Reconsidered boundary conditions A relatively simple case: Imaginary coordinate shift (i.e integration constant in x ( z ) ) δ = − i c Avoid singularities Note: Introducing i c does not change the spectrum ...so it cannot introduce spontaneous PT breaking In some cases the problem has to be defined on a more general trajectory The boundary conditions cannot be satisfied on y = x − i c This is the case for the Coulomb and Morse potentials Duplication of bound states: quasi-parity q = ± 1 Complex energy eigenvalues: spontaneous breakdown of PT symmetry PT ψ ( q ) n ( x ) = ψ ( − q ) ( x ) n

Apply the method to the Jacobi polynomials: F ( z ) = P ( α,β ) ( z ) n � 2 � z ′′ ( x ) + ( z ′ ( x )) 2 � � � � 2 z ′ ( x ) − 3 z ′′′ ( x ) n + α + β n + α + β E − V ( x ) = + 1 1 − z 2 ( x ) 4 z ′ ( x ) 2 2  � 2  � 2 ( z ′ ( x )) 2 � α + β � α − β +  1 − −  (1 − z 2 ( x )) 2 2 2 − 2 z ( x )( z ′ ( x )) 2 � α + β � � α − β � . (1 − z 2 ( x )) 2 2 2 The solutions are β +1 α +1 ψ ( x ) ∼ ( z ′ ( x )) − 1 2 (1 − z ( x )) 2 P ( α,β ) 2 (1 + z ( x )) ( z ( x )) . n The yet unknown z ( x ) can be obtained from a differential equation � 2 φ ( z ) ≡ � 2 p I (1 − z 2 )+ p II + p III z � � d z d z = C . (1 − z 2 ) 2 d x d x by direct integration � φ 1 / 2 ( z )d z = C 1 / 2 x + ǫ . ǫ : integration constant, coordinate shift

The Scarf II potential p II = p III = 0  � 2 � 2  � α + β � α − β � β + α � � β − α � 1 − 1  + 2i sinh x V ( x ) = − + cosh 2 x cosh 2 x  2 2 4 2 2 Relations for the parameters: PT symmetry: = ⇒ α , β are real or imaginary α ↔ β : = ⇒ V ( x ) ↔ V ( − x ) V ( x ) invariant under α ↔ − α = ⇒ qα ≡ ± α quasi-parity qα 2 + 1 β 2 + 1 ψ ( q ) n ( x ) = C ( q ) 4 P ( qα,β ) n (1 − i sinh( x + i ǫ )) 4 (1 + i sinh( x + i ǫ )) (i sinh( x + i ǫ )) n n ( q ) < − [Re( qα + β ) + 1] / 2 Normalizable if The second set corresponds to resonances in the Hermitian setting ( α ∗ = β ) � 2 � n + qα + β + 1 E ( q ) = − n 2 Complex conjugate pairs if α is imaginary Spontaneous breakdown of PT symmetry “Sudden” mechanism: all the E ( q ) turn complex at the same time n

The transition to complex energy eigenvalues Reparametrize to V ( x ) = − v r V R ( x ) + i v i V I ( x ), v r fixed, v i varied v r = 15 . 1, v i = 12 . 7 v i = 0 → 20 Unavoided crossings! Some are at the same v i . Complexification occurs for | v i | = v r + 1 / 4 = 15 . 35 for all n Sudden mechanism of PT symmetry breaking Note: ψ ( − ) ( x ) normalizable for v i ≥ 3 . 886, ψ ( − ) ( x ) for v i ≥ 10 . 858, ψ ( − ) ( x ) for v i ≥ 15 . 803. 0 1 2 ψ (+) ( x ) exists for v i ≤ 12 . 325. 3

α = 1 . 9 Re ψ Im ψ ψ ( − α,β ) ψ ( α,β ) ( z ) ( z ) 0 1

α = 1 . 5 Re ψ Im ψ ψ ( − α,β ) ψ ( α,β ) ( z ) ( z ) 0 1

α = 1 . 2 Re ψ Im ψ ψ ( − α,β ) ψ ( α,β ) ( z ) ( z ) 0 1

α = 1 . 0 Re ψ Im ψ ψ ( − α,β ) ψ ( α,β ) ( z ) ( z ) 0 1

A Natanzon-class example: the generalized Ginocchio potential with 2+2 parameters G. L´ evai et al. J. Phys. A 36 (2003) 7611 p I = γ 2 + 1, p II = 0, p III = 4 Take γ = 1: generalized P¨ oschl–Teller limit Implicit z ( x ) function: r ≡ x − i ε = 1 ( γ 2 + sinh 2 u ) − 1 � tanh − 1 � � 2 sinh u γ 2 2 ( γ 2 + sinh 2 u ) − 1 + ( γ 2 − 1) ( γ 2 − 1) 1 1 2 tan − 1 � �� 2 sinh u . ε � = 0 to avoid singularity at x = 0 coth 2 u − γ 4 ( s ( s + 1) + 1 − γ 2 ) + γ 4 λ ( λ − 1) V ( r ) = γ 2 + sinh 2 u γ 2 + sinh 2 u − 3 γ 4 ( γ 2 − 1)(3 γ 2 − 1) 5 γ 6 ( γ 2 − 1) 2 + 4( γ 2 + sinh 2 u ) 3 , 4( γ 2 + sinh 2 u ) 2 The solutions and eigenvalues energy E nq = − γ 4 µ 2 nq depend on the q = ± 1 quasi-parity ( γ 2 + sinh 2 u ) 1 / 4 (cosh u ) − 2 n − 1 − µ nq − q ( λ − 1 1 2 + q ( λ − 1 2 ) (sinh u ) 2 ) ψ nq ( x ) ∼ ( q ( λ − 1 2 ) , − 2 n − 1 − µ nq − q ( λ − 1 2 )) × P (cosh(2 u )) . n  � 2 � 1 / 2  µ nq = 1 2 n + 1 + q ( λ − 1 � γ 2 ( s + 1 2 n + 1 + q ( λ − 1 � � � 2) 2 + (1 − γ 2 )  − 2) + 2)  γ 2

Unbroken PT symmetry: real µ qn , s , λ The real (left panel) and imaginary (right panel) component of the PT -symmetric generalized Ginocchio potential for ε = 0 . 3, γ = 1 . 75, s = 8 . 1 and λ = 1 . 25 (solid line) and its supersymmetric partners V (+1) ( x ) + (dashed line) and V ( − 1) ( x ) (dotted line). + Normalisable states are found at E 0 +1 = − 171 . 313, E 1 +1 = − 106 . 160, E 2 +1 = − 46 . 679, E 3 +1 = − 5 . 666; E 0 − 1 = − 218 . 913, E 1 − 1 = − 154 . 978, E 2 − 1 = − 90 . 379, E 3 − 1 = − 33 . 993 and E 4 − 1 = − 1 . 061. The spectrum of V ( q ) + ( x ) is the same, with the exception of the E 0 q level, which is missing from its spectrum. The equivalence of ψ ( − q ) ( x ) and ψ ( q ) n − j ( x ) can be proven for α = j integer n

Another 2+2-parameter subset of the Natanzon class G. L´ evai, J. Phys. A 45 (2012) 444020 p I = 1, p II = δ , p III = 0 It contains all SI potentials as special case : δ = 0 Scarf II δ → ∞ and C/δ = const. Rosen–Morse I V R ( x ) and V I ( x ) for C = − 1, δ = 0 . 3, Σ = 11 . 0624 and Λ = 1 . 26.

Implicit potential: only x ( z ) is known in closed form. . . . . . nevertheless, everything can be evaluated exactly � 2 ( δ + 1 − z 2 ( x )) 2 + 5 Cδ 2 � n + α + β + 1 − 3 Cδ (3 δ + 2) ( δ + 1) E − V ( x ) = C ( δ + 1 − z 2 ( x )) 3 2 4 4 C Σ 2 C Λ z ( x ) − δ + 1 − z 2 ( x ) − δ + 1 − z 2 ( x ) � 2 � 2 � 2 � � α + β � α − β n + α + β + 1 − 1 Σ = δ − δ + + 2 2 2 4 Λ = α + β α − β α = α n , β = β n 2 2 Ψ n ( x ) = N n ( δ + 1 − z 2 ( x )) 1 / 4 (1 − z ( x )) α/ 2 (1 + z ( x )) β/ 2 P ( α,β ) ( z ( x )) n A four-parameter (2+2) potential: C and δ control the variable transformation Σ and Λ set the coupling coefficients Λ = 0: symmetric Ginocchio case Energy eigenvalues determined from the roots of the spectral equation for ω = ( α + β ) / 2: � δ 4(2 n + 1) 2 − δ − Σ − 1 � ( δ + 1) ω 4 + δ (2 n + 1) ω 3 + ω 2 + Λ 2 = 0 4

How does the spectrum evolve if Σ is fixed and Λ is varied? Complexification occurs at Λ crit0 = 6 . 298, Λ crit1 = 7 . 542, Λ crit2 = 8 . 234 Now at different locations when α = ± M integer. Unavoided crossings! The q quasi-parity does not appear explicitly. ( x ) and ψ ( q ) The equivalence of ψ ( − q ) n − j ( x ) can be proven for α = j integer n

The effect of ( n, α = − j ) ⇔ ( n − j, α = j ) Equations determining E n for Natanzon-class potentials � 2 � n + α 2 + β + 1 − 1 E 4 + s I − p I C = 0 2 1 − α 2 2 − β 2 E 2 + s II − p II C = 0 − α 2 2 + β 2 E 2 + s III − p III C = 0 Now take α = j The equations are invariant with respect to ( n, α = − j ) ⇔ ( n − j, α = j )