Lecture I: Exact BS quantization condition

S. Lukyanov

Introduction: IQFT/IPDE correspondence

• Correspondence principle: when Planck constant ~→ 0

Quantum theory −→ Classical theory

• Since 1998, a new type of (mathematical) correspondence

Integrable Quantum Field Theory (IQFT) ←→ Integrable Partial Differential Equa-tions (IPDE)

• Historically, the first example is the so-called ODE/IM correspondence

• Many faces of the IQFT/IPDE correspondence. In particular, the appearance of thePainleve transcendents in the description of the monodromy of linear ODEs, say

-The correspondence between a quantum mechanical particle in a cosine potential andPainleve III

-The correspondence between a quasiclassical conformal block and Painleve VI

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3D harmonic oscillator

Any story in physics should began with the harmonic oscillator, a problem which everyphysicist know. So, let me start with the three dimensional (3D) harmonic oscillator:[

− ~2

2m∇2 + U(r)

]Ψ = E Ψ , U(r) =

mω2

2r2

• U(r) = mω2

2(x2

1 + x22 + x2

3) : En = ~ω(n1 + n2 + n3 + 3

2

)• U = U(|r|) =⇒ separation of variables:[

− d2

dz2+

`(`+ 1)

z2︸ ︷︷ ︸centrifugal potential

+z2

]ψ = E ψ

(z =

√mω

~|r|, E =

2

~ωE)

En = 2 (2n+ `)︸ ︷︷ ︸n1+n2+n3

+3 (n = 0, 1, 2, . . .)

Here ` is the orbital momentum.

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3D anharmonic oscillator

Let us make the problem slightly less trivial and consider the “anharmonic” oscillator:[− d2

dz2+ Ueff(z)

]ψ = E ψ , Ueff(z) =

`(`+ 1)

z2+ z2α (α > 0)

Particular cases (see ref. [1])

• Harmonic oscillator: α = 1 (Hermite)

• Infinite Spherical Potential Well: α =∞ (Bessel)

z2α|α→+∞ →

0 , 0 ≤ z < 1

∞ , z > 1

• ` = 0, α = 12

(Airy):Ueff(z) = z (constant 1D force)

Generally speaking, this is a confining potential with the discrete spectrum.

Bohr-Sommerfeld (BS) quantization condition

Ueff

p

E

z

z

∮dz

2πp(z) = n+

1

2

Langer’s correction (1937)

Exercise: Using the BS quantization condition show that

En ≈ C0

(n+ 1

4(2`+ 3)

) 2αα+1 (n 1)

and find the n and `-independent constant C0 = C0(α) explicitly. Show that for α = 1 theBS quantization condition turns out to be exact.

Remarkably, the problem is “integrable” in a certain sense. Namely we can find the“exact” Bohr-Sommerfeld quantization condition. It requires the notion of the “SpectralDeterminant”.

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Spectral determinant

• Characteristic polynomial for a linear operator (matrix):

Det(A− λ I) = Det(A)∏n

(1− λ/λn)

• H = − d2

dz2+ `(`+1)

z2+ z2α :

D(E) ≡ Det(H − E I) = D(0)︸ ︷︷ ︸Det(H)

∞∏n=0

(1− E/En)

• En ∼ n2αα+1 (n→∞) =⇒ the product converges for α > 1

Calculation of the spectral determinant

Two solutions of the Shrodinger equation for the confining potential

Ueff

z 2

z

U(z)l(l+1)

Ueff(z) =`(`+ 1)

z2+ z2α︸︷︷︸

U(z)

ψ(z)→ z`+1 as z → 0

χ(z) ≈ 1√U(z)

exp(−∫ z

dz√U(z)

)= z−α/2 exp

(− z1+α

1 + α

)(z → +∞)

(WKB asymptotic)

Properties of the Wronskian W [χ, ψ] ≡ χψ′ − χ′ψ• W does not depend on z, i.e., W = W (E, `)

• W (E, `) is an entire function of E (analytic in the whole complex plane)

• W (En, `) = 0 (χ ∝ ψ+ as E = En)

Exercise: Show that the spectral determinant D(E) can be identified with the WronskianW (E, `), i.e.,

W (E, `) = const D(E) ≡ Det(H − E I)

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Elements of Regge theory for Ueff(z|`) = `(`+1)z2

+ z2α

In quantum mechanics, Regge theory is the study of the analytic properties of scattering am-plitudes as functions of angular momentum, where the angular momentum is not restrictedto be an integer but is allowed to take any complex value. The nonrelativistic theory wasdeveloped by Tullio Regge in 1959 (see § 141 in ref. [1]).

Figure 1: TullioRegge 1931-2014

Here are some facts from Regge theory for Ueff(z|`) = `(`+1)z2

+ z2α

• χ(z|`)→ 0 (z →∞) : χ(z|`) = χ(z| − 1− `)is an entire function of the complex variable `

• ψ(z|`)→ z`+1 (z → 0) is a meromorphic function of `.

Only simple (Regge) poles are allowed for <e(`) < −12

• ψ+(z) ≡ ψ(z|`) , ψ−(z) = ψ(z| − ` − 1) – two linear independent solutions:(ψ−ψ

′+ − ψ+ψ

′−)|z→0 = 2`+ 1

• χ(z) = 12`+1

(W+ ψ−(z)−W− ψ+(z)

), W± = W [χ, ψ±]

• W−(E, `) = W+(E,−`− 1)

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Remarkable symmetry of the anharmonic oscillator

Let z 7→ q z, then

− d2

dz2+`(`+ 1)

z2+ z2α − E 7→ 1

q2

[− d2

dz2+`(`+ 1)

z2+ q2α+2 z2α − q2E

]

α+1

πq = 12α+2

α+1

πi

eq =

z qz

z

If i.e. :

Re(z)

Im(z)

Ω : z 7→ q z , E 7→ q−2E(q = e

iπα+1)

is a symmetry

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Derivation of the “Quantum Wronskian” relation [2]

• ψ+(z)→ z`+1: Ωψ+(z) = q`+1 ψ+(z) , Ωψ−(z) = q−` ψ−(z)

• χ(z) = 12`+1

[W+(E) ψ−(z)−W−(E) ψ+(z)

]:

Ωχ(z) = 12`+1

[W+(q−2E) q−`ψ−(z)−W−(q−2E) q`+1ψ+(z)

]

• W [χ, Ωχ] = 12`+1

[q−`W−(E)W+(q−2E)− q`+1W+(E)W−(q−2E)

]• χ(z)→ z−α/2 exp

(− z1+α

1+α

): W [χ, Ωχ]|z→+∞ = 2

q−`W−(E)W+(q−2E)− q`+1W+(E)W−(q−2E) = 2(2`+ 1)

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Derivation of the exact BS quantization condition

• q−`W−(E)W+(q−2E)− q`+1W+(E)W−(q−2E) = 2(2`+ 1)

• q−`W−(q2E)W+(E)− q`+1W+(q2E)W−(E) = 2(2`+ 1) (E 7→ q2E)

•

q−`W−(En)W+(q−2En) = 2(2`+ 1)

W+(En) = 0 :

−q`+1W+(q2En)W−(En) = 2(2`+ 1)

•W+(q−2En)

W+(q2En)= −q2`+1 (q = e

iπα+1 )

• W+(E) = D(E) ≡ D(0)∏∞

n=0

(1− E/En

)D(q−2En)

D(q2En)= −q2`+1

(does not depend on D(0)!

)• Q(E) = E

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(2`+1) D(E):

Q(q−2En)

Q(q2En)= −1 i.e.

1

2πilog

(Q(q−2En)

Q(q2En)

)= Nn +

1

2

where Nn are some integers.

• For α = 1: Nn = n

1

2πilog

(Q(q−2En)

Q(q2En)

)= n+

1

2

(q = e

iπα+1 , n = 0, 1, 2 . . .

)• In the WKB approximation

n+1

2=

1

2πilog

(Q(q−2En)

Q(q2En)

)≈∮

dz

2πp(z)

• It allows one to develop a systematic large-n expansion:

En (4n+ 2`+ 3)2αα+1

(C0(α) + C1(α)

12`2 + 12`− 2α + 1

(4n+ 2`+ 3)2+O(1/n4)

)• Numerical procedure: M equations for E0,. . .EM−1,

Q(E) ≈ const E`+ 12

M−1∏n=0

(1− E/En

) ∞∏n=M

(1− E/E(WKB)

n

)

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“Monster” potentials [3]

z 7→ q z, q = eπiα+1 : − d2

dz2+ Ueff(z)− E 7→ 1

q2

[− d2

dz2+ q2Ueff(qz)− q2E

]• Symmetry: Ueff(qz) = q−2 Ueff(z)

• Asymptotic behavior:

Ueff(z)→

`(`+1)z2

+ o(1) as z → 0

z2α + o(1) as z →∞

• For any E all solutions of(− d2

dz2+Ueff(z)−E

)ψ = 0 are monodromy free everywhere

(so that only poles are allowed, no branch points) except for z = 0 and z =∞.

Ueff(z) =`(`+ 1)

z2+ z2α − 2

d2

dz2

L∑k=1

log(z2α+2 − zk

),

where zk, k = 1, 2 . . . L satisfies a certain system of L algebraic equations.

1

2πilog

(Q(q−2En)

Q(q2En)

)= Nn +

1

2(Nn ∈ Z)

Monster potential ↔ set of integers Nn.

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ExercisesExercise I.1. Show that the change of variables z = ey, ψ = e

y2 ψ brings the[

− d2

dz2+`(`+ 1)

z2+ z2α

]ψ = E ψ

to the form[− d2

dy2+ e2(1+α)y − E e2y

]ψ = −4k2 ψ , where k = 1

4(2`+ 1) .

Exercise I.2. Let us define the functions

A±(E) ≡ W±(E, `)

W±(0, `)

andf(z, E) =

√−4k2 + Ez − z1+α

(k = 1

4(2`+ 1)

).

In order to make f(z, E) a single-valued function of the variable z, we introduce a branchcut along the segment z ∈ [0, z∗], f(z∗) = 0 and set f(z + i0) > 0, z ∈ [0, z∗].1

Show that in the WKB approximation

A±(E) ≈ exp

(i

∫C±

dz

2z

(f(z, E)− f(z, 0)

)),

where the contour C+(C−) starts at the point z = 0, goes below (above) the cut and thento z → +∞. In this case

q2`+1 A+(Eq2)

A+(Eq−2)≈ exp

(− i

∫C

dz

2zf(z, E)

),

here the contour C starts at z = z∗ above the cut, goes around the segment [0, z∗] andreturns to z = z∗. At the same time,

A+(E)

A−(E)≈ exp

(− i

∫C

dz

2z

(f(z, E)− f(z, 0)

)),

here the contour C starts at z = 0 below the cut, goes around the segment [0, z∗] and returnsto the z = 0.

Exercise I.3. (a) Using the result of the previous exercise show that for α > 1

A+(E) = (−E)−k D−10 exp

(π

cos( π2α

)(−E/C0)

α+12α +o(1)

)as |=m(−E)| < π , E →∞ ,

where

C0 =

[2√πΓ(3

2+ 1

2α)

Γ(1 + 12α

)

] 2αα+1

.

1Here, for simplicity, we assume that <e(`+ 1

2

)= 0 and =m(E) = 0.

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The E-independent constant D0 remains undetermined within the WKB approximation.Notice that it can be naturally identified with the (regularized) functional determinant

D0 = Det(reg)(H) , H = − d2

dz2+`(`+ 1)

z2+ z2α .

(b) Derive the BS quantization condition:

En ≈ C0

(n+ 1

4(2`+ 3)

) 2αα+1 (n 1) .

Exercise I.4. (a) Show that, in the case α = 1,

A±(E) ≡ W±(E, `)

W±(0, `)=

Γ(12± k) e

E4γE

Γ(12± k − E

4)

where γE = 0.5772 . . . stands for the Euler constant and k = 14

(2`+ 1).

(b) Show that for α→∞

limα→∞

A±(E) = Γ(1± 2k) (√E/2)∓2k J±2k(

√E) ,

where Jν(z) is the conventional Bessel function.

Exercise I.5. Given the spectral set En∞n=0 it is useful to introduce the followingfunction

Θk(ω) =

√π 21+iω Γ( i(1+α)ω

2α)

Γ( iω2α

)Γ(−12

+ iω2

)

∞∑n=0

(En)−iω(1+α)

2α ,

where

2k = `+ 12.

(a) Show that Θk(ω) is an analytic function in the half-plane =m(ω) < −1 and

logA+(E) = − i

4π32

∫Cω

dω

ωΓ(1− iω(1+α)

2α

)Γ( iω

2α

)Γ( iω−1

2

)2−iω Θk(ω) .

Here the integration contour goes along the line =m(ω) = −1− ε with arbitrary small ε > 0.(b) Using the result of Exercise I.2, show that for any =m(ω) < −1

Θk(ω) = (1 + α)−1 k1−iω(1 + o(1)

)as k → +∞ .

(c) Show that for α = 1, Θk(ω) is an entire function of ω.Hint: Check that in this case Θk(ω) can be expressed in terms of the Hurwitz ζ-function

ζ(s, q) =∑∞

n=01

(q+n)s:

Θk(ω) = 22iω−1 (iω − 1) ζ(iω, k + 1

2

).

Excercise∗ I.6. Show that Θk(ω) is an entire function of ω for any α > 0.

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Hint: The proof is based on the so called DDV equation (see Appendix A in ref. [4]).

Exercise I.7. Using the result of Exercise I.7, show that for |=m(−E)| < π, E →∞,

A+(E) = D−10 (−E)−k exp

(π

cos( π2α

)(−E/C0)

α+12α + C−1 (−E/C0)−

α+12α + o

(E−

α+12α

)),

where the constant C0 is the same as in Exercise I.3, whereas

logD0 = − 2αk

1 + α

[log(4/e) + i ∂ω log Θk(0)

]C−1 = − Θk(i)

sin( π2α

).

Notice that the constant is the so-called zeta-regularized functional determinant

D0 = Det(reg)(H) , H = − d2

dz2+`(`+ 1)

z2+ z2α .

Excercise I.8. Show that2

(a) Θk(0) = 11+α

k

(b) i ∂ω log Θk(0) = 1+α2αk

log(

1√1+α

Γ(1+2k)

Γ(1+ 2k1+α

)

)− 1

αlog((1 + α)(2/e)α

)(c) Θk(i) = − 1

24+ k2

1+α.

(d) Θk

(i(2m− 1)

)= 1

1+αPm(k2) (m = 1, 2, 3 . . .), where Pm is a polynomial of degree

m in k2 such that Pm(k2) = k2m − m(2m−1)24

(1 + α) k2m−2 + . . . .

Excercise I.9. The substitution u = 2(1 + α)y − 2 log 2(1 + α) brings[− d2

dy2+ e2(1+α)y − E e2y

]ψ = −4k2 ψ

to the form[− d2

du2+eu+δU(u)

]ψ = −ρ2 ψ , where δU(u) = −E

(2+2α

)− 2α1+α e

u1+α , ρ =

k

1 + α.

Consider the Lipman-Schwinger equation

χ(u) = K2ρ(eu)−

∫ ∞−∞

du′ G(u, u′)χ(u′) ,

whereG(u, u′) is the Green’s function for the last ODE with δU = 0 subject to the asymptoticcondition limu→∞G(u, u′) = 0.

(a) Show that1

A+(E)= 1−

∫ ∞−∞

du I2ρ

(eu)δU(u) χ(u) .

2If you cannot prove these equations, please check them for the case α = 1.

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(b) Calculate the value Θ(− 2iα1+α

).

Here Kρ(z) and Iρ(z) denote the conventional modified Bessel functions.

Excercise I.10. Let zjLj=1 be a set of complex numbers satisfying the system of Lalgebraic equations

L∑m=1m 6=j

zj(z2j + (3 + α)(1 + 2α)zjzm + α(1 + 2α)z2

m

)(zj − zm)3

− α zj4(1 + α)

+ ∆ = 0 ,

where ∆ = (2`+1)2−4α2

16(α+1). Show that all solutions of the ODE(

− d2

dz2+ Ueff(z)− E

)ψ = 0

with

Ueff(z) =`(`+ 1)

z2+ z2α − 2

d2

dz2

L∑k=1

log(z2α+2 − zk

),

are monodromy free everywhere except for z = 0 and z =∞.Hint: See Appendix B in ref. [4].

References

[1] L. D. Landau and E. M. Lifshitz, “Quantum Mechanics” ( Volume 3 of A Course ofTheoretical Physics )

[2] V. V. Bazhanov, S. L. Lukyanov and A. B. Zamolodchikov, J. Statist. Phys. 102, 567(2001) [arXiv:hep-th/9812247].

[3] V. V. Bazhanov, S. L. Lukyanov and A. B. Zamolodchikov, Adv. Theor. Math. Phys. 7,711 (2003) [arXiv:hep-th/0307108].

[4] V. V. Bazhanov, S. L. Lukyanov and A. B. Zamolodchikov, Commun. Math. Phys. 190,247 (1997) [arXiv:hep-th/9604044].

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