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 Equations (IPDE)
• Historically, the first example is the so-called ODE/IM correspondence
• Many faces of the IQFT/IPDE correspondence. In particular, the appearance of the
Painlevé transcendents in the description of the monodromy of linear ODEs, say
-The correspondence between a quantum mechanical particle in a cosine potential and
Painlevé III
-The correspondence between a quasiclassical conformal block and Painlevé VI
1
3D harmonic oscillator
Any story in physics should began with the harmonic oscillator, a problem which every
physicist know. So, let me start with the three dimensional (3D) harmonic oscillator:
~2
mω 2 2
2
−
∇ + U (r) Ψ = E Ψ ,
U (r) =
r
2m
2
• U (r) =
mω 2
2
(x21 + x22 + x23 ) :
En = ~ω n1 + n2 + n3 +
• U = U (|r|) =⇒ separation of variables:
d2
`(` + 1)
2
− 2+
+z ψ = E ψ
2
dz
z
| {z }
3
2
r
2
mω
z=
|r|, E =
E
~
~ω
centrifugal potential
En = 2 (2n + `) +3
| {z }
n1 +n2 +n3
Here ` is the orbital momentum.
2
(n = 0, 1, 2, . . .)
3D anharmonic oscillator
Let us make the problem slightly less trivial and consider the “anharmonic” oscillator:
d2
`(` + 1)
− 2 + Ueff (z) ψ = E ψ ,
Ueff (z) =
+ z 2α (α > 0)
2
dz
z
Particular cases (see ref. [1])
• Harmonic oscillator: α = 1 (Hermite)
• Infinite Spherical Potential Well: α = ∞ (Bessel)
z 2α |α→+∞
• ` = 0, α =
1
2
(
0, 0≤z<1
→
∞, z>1
(Airy):
Ueff (z) = z
(constant 1D force)
Generally speaking, this is a confining potential with the discrete spectrum.
Bohr-Sommerfeld (BS) quantization condition
Ueff
E
z
p
I
z
dz
1
p(z) = n +
2π
2
Langer’s correction (1937)
Exercise: Using the BS quantization condition show that
2α
En ≈ C0 n + 41 (2` + 3) α+1 (n 1)
and find the n and `-independent constant C0 = C0 (α) explicitly. Show that for α = 1 the
BS 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 “Spectral
Determinant”.
3
Spectral determinant
• Characteristic polynomial for a linear operator (matrix):
ˆ = Det(Â)
Det(Â − λ I)
Y
(1 − λ/λn )
n
2
d
• Ĥ = − dz
2 +
`(`+1)
z2
+ z 2α :
ˆ = D(0)
D(E) ≡ Det(Ĥ − E I)
| {z }
Det(Ĥ)
∞
Y
(1 − E/En )
n=0
2α
• En ∼ n α+1 (n → ∞) =⇒ the product converges for α > 1
Calculation of the spectral determinant
Two solutions of the Shrödinger equation for the confining potential
Ueff
l(l+1)
z2
U(z)
Ueff (z) =
`(` + 1)
+ |{z}
z 2α
z2
U (z)
z
ψ(z) → z `+1
z→0
as
1
χ(z) ≈ p
exp −
U (z)
Z
z
dz
p
z 1+α U (z) = z −α/2 exp −
1+α
(z → +∞)
(WKB asymptotic)
Properties of the Wronskian W [χ, ψ] ≡ χψ 0 − χ0 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 Wronskian
W (E, `), i.e.,
ˆ
W (E, `) = const D(E) ≡ Det(Ĥ − E I)
4
Elements of Regge theory for Ueff (z|`) =
`(`+1)
z2
+ z 2α
In quantum mechanics, Regge theory is the study of the analytic properties of scattering amplitudes as functions of angular momentum, where the angular momentum is not restricted
to be an integer but is allowed to take any complex value. The nonrelativistic theory was
developed by Tullio Regge in 1959 (see § 141 in ref. [1]).
Figure 1: Tullio
Regge 1931-2014
Here are some facts from Regge theory for Ueff (z|`) =
`(`+1)
z2
+ z 2α
• 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(`) < − 21
• ψ+ (z) ≡ ψ(z|`) ,
ψ− (z) = ψ(z| − ` − 1) – two linear independent solutions:
0
0
(ψ− ψ+ − ψ+ ψ− )|z→0 = 2` + 1
•
χ(z) =
1
2`+1
W+ ψ− (z) − W− ψ+ (z) , W± = W [χ, ψ± ]
• W− (E, `) = W+ (E, −` − 1)
5
Remarkable symmetry of the anharmonic oscillator
Let z 7→ q z, then
d2
`(` + 1)
1
− 2+
+ z 2α − E 7→ 2
2
dz
z
q
d2
`(` + 1)
− 2+
+ q 2α+2 z 2α − q 2 E
dz
z2
z
Im(z)
z
iπ
If
q
2α+2
=1
i.e.
q= e
α+1
π
α+1
:
Re(z)
Ω̂ :
iπ
z 7→ q z , E 7→ q −2 E
q = e α+1
6
is a symmetry
qz
Derivation of the “Quantum Wronskian” relation [2]
• ψ+ (z) → z `+1 :
• χ(z) =
1
2`+1
Ω̂χ(z) =
Ω̂ψ− (z) = q −` ψ− (z)
h
i
W+ (E) ψ− (z) − W− (E) ψ+ (z) :
1
2`+1
• W [χ, Ω̂χ] =
Ω̂ψ+ (z) = q `+1 ψ+ (z) ,
h
i
W+ (q −2 E) q −` ψ− (z) − W− (q −2 E) q `+1 ψ+ (z)
1
2`+1
h
i
−`
−2
`+1
−2
q W− (E)W+ (q E) − q W+ (E)W− (q E)
1+α • χ(z) → z −α/2 exp − z1+α :
W [χ, Ω̂χ]|z→+∞ = 2
q −` W− (E)W+ (q −2 E) − q `+1 W+ (E)W− (q −2 E) = 2(2` + 1)
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Derivation of the exact BS quantization condition
• q −` W− (E)W+ (q −2 E) − q `+1 W+ (E)W− (q −2 E) = 2(2` + 1)
• q −` W− (q 2 E)W+ (E) − q `+1 W+ (q 2 E)W− (E) = 2(2` + 1) (E 7→ q 2 E)
•
q −` W− (En )W+ (q −2 En ) = 2(2` + 1)
W+ (En ) = 0 :
−q `+1 W+ (q 2 En )W− (En ) = 2(2` + 1)
•
W+ (q −2 En )
= −q 2`+1
W+ (q 2 En )
Q
• W+ (E) = D(E) ≡ D(0) ∞
n=0 1 − E/En
D(q −2 En )
= −q 2`+1
D(q 2 En )
iπ
(q = e α+1 )
does not depend on D(0)!
1
• Q(E) = E 4 (2`+1) D(E):
Q(q −2 En )
= −1
Q(q 2 En )
1
log
2πi
i.e.
Q(q −2 En )
Q(q 2 En )
= Nn +
1
2
where Nn are some integers.
• For α = 1:
1
2πi
Nn = n
log
Q(q −2 En )
Q(q 2 En )
=n+
1
iπ
q = e α+1 , n = 0, 1, 2 . . .
2
• In the WKB approximation
1
1
n+ =
log
2
2πi
Q(q −2 En )
Q(q 2 En )
I
≈
dz
p(z)
2π
• It allows one to develop a systematic large-n expansion:
2
2α
12`
+
12`
−
2α
+
1
4
En (4n + 2` + 3) α+1 C0 (α) + C1 (α)
+ O(1/n )
(4n + 2` + 3)2
• Numerical procedure: M equations for E0 ,. . . EM −1 ,
Q(E) ≈ const E
`+ 12
M
−1
Y
1 − E/En
n=0
∞
Y
n=M
8
1 − E/En(WKB)
“Monster” potentials [3]
z 7→ q z,
q=e
πi
α+1
d2
1
: − 2 + Ueff (z) − E 7→ 2
dz
q
d2
− 2 + q 2 Ueff (qz) − q 2 E
dz
• Symmetry: Ueff (qz) = q −2 Ueff (z)
• Asymptotic behavior:
(
Ueff (z) →
`(`+1)
z2
2α
z
+ o(1)
+ o(1)
as z → 0
as z → ∞
d2
• For any E all solutions of − dz
2 + Ueff (z) − E ψ = 0 are monodromy free everywhere
(so that only poles are allowed, no branch points) except for z = 0 and z = ∞.
L
d2 X
`(` + 1)
2α
+z −2 2
log z 2α+2 − zk ,
Ueff (z) =
2
z
dz k=1
where zk , k = 1, 2 . . . L satisfies a certain system of L algebraic equations.
1
2πi
log
Q(q −2 En )
Q(q 2 En )
= Nn +
Monster potential ↔ set of integers {Nn }.
9
1
2
(Nn ∈ Z)
Exercises
y
Exercise I.1. Show that the change of variables z = ey , ψ = e 2 ψ̃ brings the
d2
`(` + 1)
2α
− 2+
ψ =Eψ
+z
dz
z2
to the form
d2
2(1+α)y
2y
− 2 +e
−Ee
ψ̃ = −4k 2 ψ̃ ,
dy
where k =
1
4
(2` + 1) .
Exercise I.2. Let us define the functions
A± (E) ≡
and
f (z, E) =
W± (E, `)
W± (0, `)
√
−4k 2 + Ez − z 1+α
k=
1
4
(2` + 1) .
In order to make f (z, E) a single-valued function of the variable z, we introduce a branch
cut 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
Z
dz
f (z, E) − f (z, 0)
A± (E) ≈ exp i
,
C± 2z
where the contour C+ (C− ) starts at the point z = 0, goes below (above) the cut and then
to z → +∞. In this case
Z
2
dz
2`+1 A+ (Eq )
≈ exp − i
f (z, E) ,
q
A+ (Eq −2 )
C 2z
here the contour C starts at z = z ∗ above the cut, goes around the segment [0, z ∗ ] and
returns to z = z ∗ . At the same time,
Z
dz
A+ (E)
≈ exp − i
f (z, E) − f (z, 0)
,
A− (E)
C̃ 2z
here the contour C̃ starts at z = 0 below the cut, goes around the segment [0, z ∗ ] and returns
to the z = 0.
Exercise I.3. (a) Using the result of the previous exercise show that for α > 1
α+1
π
−k
−1
2α
A+ (E) = (−E) D0 exp
(−E/C0 ) +o(1)
as |=m(−E)| < π , E → ∞ ,
π
cos( 2α
)
where
1
2α
√
1 α+1
)
2 πΓ( 23 + 2α
C0 =
.
1
Γ(1 + 2α
)
Here, for simplicity, we assume that <e ` +
1
2
= 0 and =m(E) = 0.
10
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) (Ĥ) ,
Ĥ = −
d2
`(` + 1)
+
+ z 2α .
2
dz
z2
(b) Derive the BS quantization condition:
En ≈ C0 n + 41 (2` + 3)
2α
α+1
(n 1) .
Exercise I.4. (a) Show that, in the case α = 1,
E
Γ( 12 ± k) e 4 γE
W± (E, `)
A± (E) ≡
=
W± (0, `)
Γ( 12 ± k − E4 )
where γE = 0.5772 . . . stands for the Euler constant and k =
1
4
(2` + 1).
(b) Show that for α → ∞
√
√
lim A± (E) = Γ(1 ± 2k) ( E/2)∓2k J±2k ( E) ,
α→∞
where Jν (z) is the conventional Bessel function.
Exercise I.5. Given the spectral set {En }∞
n=0 it is useful to introduce the following
function
√ 1+iω i(1+α)ω ∞
iω(1+α)
π2
Γ( 2α ) X
Θk (ω) =
(En )− 2α ,
iω
1
iω
Γ( 2α )Γ(− 2 + 2 ) n=0
where
2k = ` +
1
2
.
(a) Show that Θk (ω) is an analytic function in the half-plane =m(ω) < −1 and
Z
iω iω−1 −iω
dω
i
Γ(1 − iω(1+α)
Γ( 2α Γ( 2 2
Θk (ω) .
log A+ (E) = − 3
2α
4π 2 Cω ω
Here the integration contour goes along the line =m(ω) = −1 − with arbitrary small > 0.
(b) Using the result of Exercise I.2, show that for any =m(ω) < −1
Θk (ω) = (1 + α)−1 k 1−iω 1 + o(1)
as k → +∞ .
(c) Show that for α = 1, Θk (ω) is an entire function of ω.
Hint:P
Check that in this case Θk (ω) can be expressed in terms of the Hurwitz ζ-function
1
ζ(s, q) = ∞
n=0 (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 → ∞,
α+1
α+1
α+1 π
−1
−k
−
−
A+ (E) = D0 (−E) exp
(−E/C0 ) 2α + C−1 (−E/C0 ) 2α + o E 2α
,
π
)
cos( 2α
where the constant C0 is the same as in Exercise I.3, whereas
i
2αk h
log(4/e) + i ∂ω log Θk (0)
log D0 = −
1+α
Θk (i)
C−1 = −
.
π
sin( 2α
)
Notice that the constant is the so-called zeta-regularized functional determinant
D0 = Det(reg) (Ĥ) ,
Ĥ = −
d2
`(` + 1)
+
+ z 2α .
2
2
dz
z
Excercise I.8. Show that2
(a) Θk (0) =
1
1+α
k
(b) i ∂ω log Θk (0) =
1+α
2αk
log
Γ(1+2k)
√1
2k
1+α Γ(1+ 1+α
)
−
1
α
log (1 + α)(2/e)α
2
k
1
+ 1+α
.
(c) Θk (i) = − 24
1
(d) Θk i(2m − 1) = 1+α
Pm (k 2 ) (m = 1, 2, 3 . . .), where Pm is a polynomial of degree
m in k 2 such that Pm (k 2 ) = k 2m − m(2m−1)
(1 + α) k 2m−2 + . . . .
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Excercise I.9. The substitution u = 2(1 + α)y − 2 log 2(1 + α) brings
d2
2(1+α)y
2y
− 2 +e
−Ee
ψ̃ = −4k 2 ψ̃
dy
to the form
d2
u
− 2 + e + δU (u) ψ̃ = −ρ2 ψ̃ ,
du
where δU (u) = −E 2 + 2α
Consider the Lipman-Schwinger equation
Z
u
χ(u) = K2ρ (e ) −
2α
− 1+α
u
e 1+α ,
ρ=
k
.
1+α
∞
du0 G(u, u0 ) χ(u0 ) ,
−∞
where G(u, u0 ) is the Green’s function for the last ODE with δU = 0 subject to the asymptotic
condition limu→∞ G(u, u0 ) = 0.
(a) Show that
Z ∞
1
=1−
du I2ρ eu δU (u) χ(u) .
A+ (E)
−∞
2
If you cannot prove these equations, please check them for the case α = 1.
12
2iα
(b) Calculate the value Θ(− 1+α
.
Here Kρ (z) and Iρ (z) denote the conventional modified Bessel functions.
Excercise I.10. Let {zj }Lj=1 be a set of complex numbers satisfying the system of L
algebraic equations
L
2
X
zj zj2 + (3 + α)(1 + 2α)zj zm + α(1 + 2α)zm
α zj
+∆=0 ,
−
3
(zj − zm )
4(1 + α)
m=1
m6=j
where ∆ =
(2`+1)2 −4α2
.
16(α+1)
Show that all solutions of the ODE
−
with
d2
dz 2
+ Ueff (z) − E ψ = 0
L
d2 X
`(` + 1)
2α+2
2α
log
z
−
z
+
z
−
2
,
Ueff (z) =
k
z2
dz 2 k=1
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 of
Theoretical 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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