1 What s the big deal?

Transcription

1 This note is written for a talk given at the graduate student seminar, titled how to solve quantum mechanics with x 4 potential. What s the big deal? The subject of interest is quantum mechanics in an anharmonic potential V (x) = µx + gx 4. While a complete analytic solution is not available, when µ > 0, one may try to solve it perturbatively in g. When µ < 0, one may expand around one of the two classical vacua and solve it perturbatively in g. Unfortunately, in both cases the perturbative (asymptotic) series in g has zero radius of convergence. In the former case, when there are no real instantons (more generally, when there are no instantons lying on the stationary phase path integral contour), the perturbative series can be summed up via the modified Borel transform (due to Crutchfield), and the result will reproduce the exact answer. Let us discuss this in a bit more detail. The partition function Z( ) = Dxe Sx (.) can be written in the form Z( ) = C dz πi e z/ B(z) (.) where B(z) is the modified Borel transform, given by B(z) = N i= dx i Γ( N + ) (Sx z) N + (.3) and the contour C encloses the range of Sx. Here we have used a discretized notation for B(z), though in practice it can be evaluated as ( ) B(z) = Dxe S x dt exp zt t S3 x t S 4 x + (.4) 0 where S n x stands for terms in the action Sx that are of order n in x. When we integrate x along the real contour, Sx is real and non-negative, and the contour C can be chosen appropriately. Now B(z) can be calculated perturbatively. For instance, if the S 3 x term is absent and there is only a quartic interaction term, then B(z) = ( Dx e S x S ) 4x K ( zs 4 x) (.5) z

2 Using the expansion K (u) = u + I (u) ln u u 4 k=0 ψ(k + ) + ψ(k + ) ( u k!(k + )! )k (.6) where ψ(x) = Γ (x)/γ(x) is the digamma function, one can write B(z) in the form B(z) = z + B (z) ln( z) + B (z), (.7) where B (z) and B (z) are analytic in z at z = 0. This conclusion holds more generally with nonzero S 3 x. B (z) is the standard Borel transform of the perturbative series. If B (z) is analytic on the entire z-plane, then it does not contribute to the discontinuity across the real z-axis, and therefore does not contribute to the partition function Z( ). In this case, the Borel resummation reproduces the exact path integral. In the example of µ < 0, where there are real instantons, the situation is more complicated. There are two approaches. One is to deform to a complex value with a small imaginary part, so that the path integral contour splits into a number of stationary phase contours, each of which passing through only one saddle point. In practice, this is very difficult to do in QFT of dimension greater than zero. Alternatively, one may still work with the modified Borel transform. B(z) is still analytic away from the positive real axis, but now we expect B (z) and B (z) to have singularities on the real z-axis, and branch cuts that end on them. The information of these singularities are entirely determined by the data near z = 0. In particular, B (z) is still the standard Borel transform. B (z), however, has Laurent coefficients at z = 0 of the form b n,k = Dxe Sx S 3 x n S 4 x k n+ ln S 4 x (.8) which are intrinsically nonperturbative and cannot be evaluated in terms of Feynman diagrams. It is therefore of great interest to come up with methods to compute such non-perturbative quantities in quantum mechanics and more generally QFTs. Spectral function Consider one dimensional quantum mechanics in a quartic potential V (x) = x 4. Let E k be the energy levels, k = 0,,,. The spectral function is defined as (λ) = ) ( + λek k=0 (.)

3 Its zeros are at λ = E k. Let us investigate the energy level distribution at large k, for which we can apply Born-Sommerfield quantization /4 E k E /4 k Ek x 4 dx = π Γ(5/4) E 3/4 k π(k + Γ(7/4) ) (.) and so E k k 4/3 at large k. In particular, the infinite product that defines (λ) converges absolutely. The wave functions ψ k (x) associated with the energy levels E k functions of x for even (odd) k. Let us split (λ) into ± (λ) = k even/odd Define the zeta function regularized determinants where Z ± (s) is the analytic continuation of Z ± (s) = The stationary Schrödinger equation are even (odd) ( + λ E k ). (.3) D ± (λ) = exp Z ± (0) ± (λ), (.4) k even/odd E s k. (.5) ψ (x) + (x 4 + λ)ψ(x) = 0. (.6) has normalizable solution only for λ = E k. But let us consider solutions for general λ > 0, of the WKB form { x } ψ λ (x; x 0 ) = u(x) / exp u(y)dy (.7) x 0 where u(x) is subject to the asymptotic behavior u(x) λ + x 4, x +. (.8) In the ordinary WKB approximation, one simply replaces u(x) by λ + x 4. Here u(x) is defined so as to give the exact solution with regular behavior as x +. It obeys u(x) = λ + x 4 + u (x) u(x) 3u (x) 4u(x). (.9) 3

4 Shifting x 0, of course, only affects the solution by an overall constant factor. We would like to take the limit x 0 + in order to get a canonical wave function that is independent of the choice of a base point. The exponent would be ill defined as the integration of u(y) diverges at y. We can however define a regularized expression { ψ λ (x) = u(x) / exp u(y) } λ + y 4 dy + U(x, /) (.0) where U(x, s) is defined as x U(x, s) = x (λ + y 4 ) s dy (.) for Re(s) > /, and U(x, /) in particular is defined by the analytic continuation of U(x, s) to s = /. In the x limit, ψ λ (x) has the leading asymptotic behavior ψ λ (x) x exp x3 3 The following important identities were proven in Voros 983: (.) D (λ) = ψ λ (0), D + (λ) = ψ λ(0). (.3) We will not reproduce the proof here, which utilizes the representation of Green s function kernel in terms of a pair of recessive solutions. It suffices to note here that the LHS and RHS have the same set of zeros, namely at λ = E k, k even for D + and odd for D, and are entire analytic functions of λ. 3 Wronskian identity Given ψ λ (x), we can generate another complex solution ψ λ (x) = ψ α λ(α / x). (3.) where α e πi/3. The Wronskian of the two solutions ψ and ψ W (x) = ψ λ (x) ψ λ(x) ψ λ (x)ψ λ(x) (3.) is a constant. By matching it at x = 0, W (0) = α / D (λ)d + (α λ) + D (α λ)d + (λ) (3.3) 4

5 and at x = +, W (+ ) = lim x + = α /. we arrive at the identity x e x3 /3 d ( ) α / x e x3 /3 α / x e x3 /3 d ( ) x e x3 /3 dx dx (3.4) α /4 D + (α λ)d (λ) α /4 D + (λ)d (α λ) = i. (3.5) There are two equations obtained by replacing λ αλ and λ α λ. Define D = D + D, and D P = D + /D. We have then α /4 D P (α λ) D P (λ) α /4 D P (λ) D P (α λ) = i D(α λ)d(λ) (3.6) Or equivalently, α /4 D P (α λ) D P (λ) = exp i arcsin D(λ)D(α λ) (3.7) It follows that i = α /4 = exp and so sin D P (α λ) D P (λ) i arcsin arcsin For x, y, z obeying we have α /4 D P (α λ) D P (α λ) α/4 D P (λ) D P (α λ) + i arcsin D(λ)D(αλ) D(αλ)D(α λ) + i arcsin D(α λ)d(λ) (3.8) + arcsin D(λ)D(αλ) D(αλ)D(α λ) + arcsin =, D(α λ)d(λ) (3.9) It then follows that D(λ) obeys the identity sin(arcsin x + arcsin y + arcsin z) =, (3.0) x + y + z + xyz =. (3.) D(λ)D(αλ)D(α λ) = D(λ) + D(αλ) + D(α λ) +. (3.) 5

6 4 TBA equation and Y-system Define S ab (θ) = a+b p= a b +, step {p}, a, b =,, 3, {p} = sinh( θ + i π(p ) ) sinh( θ + i π(p+) ) 8 8 sinh( θ i π(p ) ) sinh( θ i π(p+) ), 8 8 φ ab (θ) = i θ ln S ab (θ), (4.) S ab (θ) are factorized scattering phases of 3 species of particles. Consider a TBA-like equation of the form ε a (θ) = m a e θ π where stands for the convolution f(θ) g(θ) = 3 φ ab (θ) ln( + e εb(θ) ) (4.) b= In the driving term, (m, m, m 3 ) = (,, )m, with dθ f(θ θ )g(θ ). (4.3) m = πγ( 5 4 ) Γ( 7 4 ). (4.4) The φ ab s obey the following remarkable relation. First, consider their Fourier transform Then the matrix has the inverse where the matrix l ab is φ ab (k) = dθ e ikθ φ ab (θ). (4.5) M ab = δ ab π φ ab (k) (4.6) ( M ) ab = δ ab ( M ) ab has Fourier transform back to θ, l ab cosh(kπ/4) (4.7) 0 0 l ab = 0. (4.8) 0 0 N ab (θ) = δ ab δ(θ) 6 l ab π cosh(θ). (4.9)

7 Define ϕ(θ) = cosh(θ). (4.0) So we have N ab (θ) δ bc δ(θ) π φ bc(θ) = δ ac δ(θ), (4.) or π N ab(θ) φ bc (θ) = δ ac δ(θ) N ac (θ) = l ab π ϕ(θ), (4.) Applying this to the TBA equation, we obtain or N ab (θ) ε b (θ) = m b N ab (θ) e θ + π Define Y a (θ) = e εa(θ). We have 3 l ab ϕ(θ) ln( + e εb(θ) ) (4.3) b= ε a m a e θ = l ab ϕ ε b m b e θ + ln( + e ε b ) π (4.4) ln Y a (θ) m a e θ = π ϕ(θ) b = π ϕ(θ) b l ab m b e θ + ln b m a e θ + ln b ( + Y b (θ)) l ab ( + Y b (θ)) l ab. (4.5) Shifting the variable θ, we have ln Y a (θ + πi 4 ) m ae πi/4 e θ = πi ϕ(θ + π 4 ) ln Y a (θ πi 4 ) m ae πi/4 e θ = πi ϕ(θ π 4 ) and adding them up gives ln Y a (θ + πi 4 ) + ln Y a(θ πi 4 ) m a e θ = m a e θ + ln b m a e θ + ln b ( + Y b (θ)) l ab ( + Y b (θ)) l ab,, (4.6) πi πi ϕ(θ + ) + ϕ(θ ) 4 4 m a e θ + ln π b (4.7) Naively, ϕ(θ + πi πi ) + ϕ(θ ) is zero. But we have cheated a bit. What we are really 4 4 doing is to shift θ after taking the product, and hence we should replace ϕ(θ + πi πi ) + ϕ(θ ) 4 4 π π dke ikθ πik (e 4 + e πik 4 ) = δ(θ). (4.8) cosh(kπ/4) 7 ( + Y b (θ)) l ab

8 This results in the Y -system equations Y a (θ + iπ 4 )Y a(θ iπ 4 ) = b ( + Y b (θ)) l ab. (4.9) More explicitly, we have Y (θ + iπ 4 )Y (θ iπ 4 ) = + Y (θ), Y (θ + iπ 4 )Y (θ iπ 4 ) = ( + Y (θ)). (4.0) where we used Y = Y 3. The Y a (θ) s also obey periodicity condition Now if we make the identification Y a (θ + 3πi ) = Y a(θ). (4.) Y (θ) = Y 3 (θ) = D(e 4θ/3 ). (4.) The Wronskian identity of D(λ) would then follow from the Y -system equations. To show that D(λ) obtained this way is indeed the spectral function, we need to match the behavior with Y a (θ) at λ = 0, or θ. This is why the driving term is proportional to e θ rather than cosh θ as in standard TBA equations. To fix the relative normalization between Y (θ) and D(λ = e 4θ/3 ), we need to examine the zeroes of Y (θ) at large θ. By the first equation of (4.0), the zeroes of Y are related to the locus of Y (θ iπ/4) =. At large θ, we have and so the solutions to Y (θ iπ/4) = are given by With the identification λ = e 4θ/3, this is equivalent to ln Y (θ) = ε (θ) m e θ, (4.3) m e θ iπ/4 πi(k + ). (4.4) m ( λ 3/4 ) π(k + ), (4.5) This is precisely the asymptotic location of energy levels at large k as given by Born- Somerfield quantization, which determines m, as claimed. 8

9 5 Solving the TBA equation The TBA equation can be solved recursively efficiently with numerics when θ is real and positive, or more generally, when e θ has sufficiently large positive real part. This cannot be used to find the zeroes of D(λ) directly, as the zeroes correspond to θ with Imθ = 3π/4. But we can use the Wronskian identity (equivalently, Y -system equations) to rewrite D(λ) = 0 as D(αλ) + D(α λ) + = 0, (5.) or Y (θ + iπ ) + Y (θ iπ ) + = 0. (5.) Writing θ = ρ + 3πi/4, and using the periodicity of Y (θ), we can write the above equation as Y (ρ πi 4 ) + Y (ρ + πi ) + = 0. (5.3) 4 This equation can now be solved by solving Y a (θ) = exp m a e θ π 3 b= ( φ ab (θ) ln + ) (5.4) Y b (θ) numerically. For the ground state energy, the solution is E 0 = Accuracy within 0.% is obtained in the first 4 recursive steps. 9