We all know recurrence equations like e.q. Fibonacci relation

$$F_{n} = F_{n-1} + F_{n+1}$$

In order to find general expression for any $n$, we can use generating function method

$$G(x) = \sum\limits_{n=0}^{\infty}F_{n}x^{n}$$

or its variation Wikipedia. However, in quantum mechanics we try to solve the Schrodinger equation

$$i\hbar \partial_{t}\left|\psi\right\rangle = \hat{H}\left|\psi\right\rangle$$ by means of expansion of the quantum state in orthonormal basis. For definiteness, lets consider double-mode bosonic system with fixed number of particles $N$. We can expand any quantum state in a Fock state basis

$$\left|\psi\right\rangle = \sum\limits_{k=0}^{N}c_{k}(t)\left|k,N-k\right.\rangle$$

$$\left|k,N-k\right.\rangle = \frac{(\hat{a}^{\dagger})^{k}(\hat{b}^{\dagger})^{N-k}}{\sqrt{k!(N-k)!}} \left| 0,0\right\rangle$$ A double-mode bosonic system is equivalent to spin system with $s = N/2$, once we use Schwinger representation

$$\hat{S}_{x} = \frac{1}{2}(\hat{a}^{\dagger}\hat{b} + \hat{b}^{\dagger}\hat{a}), \ \ \hat{S}_{y} = \frac{1}{2i}(\hat{a}^{\dagger}\hat{b} - \hat{b}^{\dagger}\hat{a}),\ \ \hat{S}_{z} = \frac{1}{2}(\hat{a}^{\dagger}\hat{a} - \hat{b}^{\dagger}\hat{b}).$$

Consider the following hamiltonian

$$\hat{H} = \hat{S}_{x}^{2}.$$ If we use the state expansion we get the recurrence differential equation

$$\dot{c}_{k}(t) = A_{k}c_{k-2}(t) + B_{k}c_{k}(t) + A_{k+2}c_{k+2}(t),\\[3mm] A_{k} = \sqrt{k(k-1)(N-k+2)(N-k+1)},\\[3mm] B_{k} = k(N-k+1) + (k+1)(N-k).$$

Is there a general scheme for solving this kind of recurrence differential equations? Generating functions or something else? Maybe there is a whole branch of Mathematics that deals with these kind of problems?

  • $\begingroup$ The obvious solution, is to find the eigenvalues and eigenstates of the Hamiltonian $H$ first, say $|\psi_m\rangle$ with energy $E_m$. Expand your initial state $|\psi(t=0)\rangle$ as $|\psi(t=0)\rangle=\sum_m a_m |\psi_m\rangle$, then at time $t$ you just have $|\psi(t)\rangle=\sum_m a_m e^{-iE_mt}|\psi_m\rangle$. $\endgroup$ – Meng Cheng May 3 '15 at 1:11
  • $\begingroup$ How would You find eigensystem for a sparse matrix with any dimension $N$? Numerically, it is not problematic, but analytically I don't think it is possible. That's why I written this problem in terms of recurrence relation... $\endgroup$ – WoofDoggy May 3 '15 at 8:05
  • $\begingroup$ You don't, and the same to the recurrence relations. Not everything can be done analytically. $\endgroup$ – Meng Cheng May 3 '15 at 14:20
  • $\begingroup$ Some problem of this kind can be solved using generating functions. Instead of differential recurrence equation You solve partial differential equation for some function $G(x,t)$. Because this function contains information about coefficients $a_{n}$ (usually through derivatives), You have a nice "analytical" formula. $\endgroup$ – WoofDoggy May 4 '15 at 17:02

I don't think that you need some complicated apparatus. You can simply solve this problem by writing

$$ \dot{c}_k(t) = M_{kj} c_j(t). $$

Now because $M$ is symmetric and real you can find a transformation $\tilde{c}_k$ of the $c_k$ that diagonalizes $M$ and leads to trivial differential equations for the transformed functions.

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  • $\begingroup$ Of course You can do that cause $M$ is just a representation of the hamiltonian. How would You diagonalize such a matrix analytically? $\endgroup$ – WoofDoggy May 3 '15 at 8:13

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