Quantum Damped Harmonic Oscillator

In this chapter we treat the quantum damped harmonic oscillator, and study mathematical structure of the model, and construct general solution with any initial condition, and give a quantum counterpart in the case of taking coherent state as an initial condition. This is a simple and good model of Quantum Mechanics with dissipation which is important to understand real world, and readers will get a powerful weapon for Quantum Physics.


Introduction
In this chapter we introduce a toy model of Quantum Mechanics with Dissipation. Quantum Mechanics with Dissipation plays a crucial role to understand real world. However, it is not easy to master the theory for undergraduates. The target of this chapter is eager undergraduates in the world. Therefore, a good toy model to understand it deeply is required.
The quantum damped harmonic oscillator is just such a one because undergraduates must use (master) many fundamental techniques in Quantum Mechanics and Mathematics.
They are "jewels" in Quantum Mechanics and Mathematics. If undergraduates master this model, they will get a powerful weapon for Quantum Physics. I expect some of them will attack many hard problems of Quantum Mechanics with Dissipation.
The contents of this chapter are based on our two papers [3] and [6]. I will give a clear and fruitful explanation to them as much as I can.
· · · From Yokohama with Love · · · 2 Some Preliminaries In this section let us make some reviews from Physics and Mathematics within our necessity.

From Physics
First we review the solution of classical damped harmonic oscillator, which is important to understand the text. For this topic see any textbook of Mathematical Physics.
Solutions (with complex form) are well-known to be so the general solution is given by where α is a complex number. If γ/2ω is small enough we have an approximate solution Next, we consider the quantum harmonic oscillator. This is well-known in textbooks of Quantum Mechanics. As standard textbooks of Quantum Mechanics see [2] and [11] ( [2] is particularly interesting).

For the Hamiltonian
where q = q(t), p = p(t), the canonical equation of motion readṡ From these we recover the equation See (2.1) with q = x and λ = 0.
Next, we introduce the Poisson bracket. For A = A(q, p), B = B(q, p) it is defined as where {, } c means classical. Then it is easy to see Now, we are in a position to give a quantization condition due to Dirac. Before that we prepare some notation from algebra. and by use of (2.7), and if we set by use of (2.7). As a result we obtain a well-known form Here we used an abbreviation 1/2 in place of (1/2)1 for simplicity.
If we define an operator N = a † a (which is called the number operator) then it is easy to see the relations Then we can easily prove and moreover can prove both the orthogonality condition and the resolution of unity For the proof one can use for example a 2 (a † ) 2 = a(aa † )a † = a(N + 1)a † = (N + 2)aa † = (N + 2)(N + 1) by (2.11), therefore we have The proof of the resolution of unity may be not easy for readers (we omit it here).
As a result we can define a Fock space generated by the generator {a † , a, N} This is just a kind of Hilbert space. On this space the operators (= infinite dimensional matrices) a † , a and N are represented as by use of (2.12).
Note We can add a phase to {a, where θ is constant. Then we have another Heisenberg algebra Next, we introduce a coherent state which plays a central role in Quantum Optics or Quantum Computation. For z ∈ C the coherent state |z ∈ F is defined by the equation a|z = z|z and z|z = 1.
The annihilation operator a is not hermitian, so this equation is never trivial. For this state the following three equations are equivalent : (1) a|z = z|z and z|z = 1, The remaining part of the proof is left to readers.
From the equation (3) in (2.16) we obtain the resolution of unity for coherent states The proof is reduced to the following formula dxdy π e −|z| 2z m z n = n! δ mn (z = x + iy).
The proof is left to readers. See [14] for more general knowledge of coherent states.

From Mathematics
We consider a simple matrix equation A standard form of linear differential equation which we usually treat is is a vector and C is a matrix associated to the vector. Therefore, we want to rewrite (2.20) into a standard form.
For the purpose we introduce the Kronecker product of matrices. For example, it is defined as for A and B above. Note that recently we use the tensor product instead of the Kronecker product, so we use it in the following. Here, let us list some useful properties of the tensor where E is the unit matrix. The proof is left to readers. [9] is recommended as a general introduction.

Then the equation (2.20) can be written in terms of components as
x 11 x 12 x 21 x 11 x 12 x 21 where T is the transpose, then we obtain a standard form Similarly we have a standard form From these lessons there is no problem to generalize (2.23) and (2.24) based on 2 × 2 matrices to ones based on any (square) matrices or operators on F . Namely, we have where I is the identity E (matrices) or 1 (operators).

Quantum Damped Harmonic Oscillator
In this section we treat the quantum damped harmonic oscillator. As a general introduction to this topic see [1] or [16].

Model
Before that we introduce the quantum harmonic oscillator. The Schrödinger equation is given by In the following we use ∂ ∂t instead of d dt . Now we change from a wave-function to a density operator because we want to treat a mixed state, which is a well-known technique in Quantum Mechanics or Quantum Optics.
If we set ρ(t) = |Ψ(t) Ψ(t)|, then a little algebra gives This is called the quantum Liouville equation. With this form we can treat a mixed state where u j ≥ 0 and N j=1 u j = 1. Note that the general solution of (3.1) is given by We are in a position to state the equation of quantum damped harmonic oscillator by use of (3.1).
Definition The equation is given by where µ, ν (µ > ν ≥ 0) are some real constants depending on the system (for example, a damping rate of the cavity mode) 1 .
Note that the extra term is called the Lindblad form (term). Such a term plays an essential role in Decoherence.

Method of Solution
First we solve the Lindblad equation : Interesting enough, we can solve this equation completely.
Let us rewrite (3.3) more conveniently using the number operator N ≡ a † a where we have used aa † = N + 1.

Then (3.4) becomes
where we have used a T = a † from the form (2.15), so that the solution is formally given by In order to use some techniques from Lie algebra we set then we can show the relations This is just the su(1, 1) algebra. The proof is very easy and is left to readers.
The equation (3.8) can be written simply as so we have only to calculate the term which is of course not simple. Now the disentangling formula in [4] is helpful in calculating (3.11).
If we set {k + , k − , k 3 } as then it is very easy to check the relations That is, {k + , k − , k 3 } are generators of the Lie algebra su(1, 1). Let us show by SU(1, 1) the corresponding Lie group, which is a typical noncompact group.
Since SU(1, 1) is contained in the special linear group SL(2; C), we assume that there exists an infinite dimensional unitary representation ρ : From (3.11) some algebra gives ≡ ρ e tA (3.13) and we have The proof is based on the following two facts.
Note that The remainder is left to readers.
The Gauss decomposition formula (in SL(2; C))  gives the decomposition and moreover we have Since ρ is a group homomorphism (ρ(XY Z) = ρ(X)ρ(Y )ρ(Z)) and the formula ρ e Lk = e Ldρ(k) (k = k + , k 3 , k − ) holds we obtain As a result we have the disentangling formula by (3.13).
In the following we set for simplicity Readers should be careful of this "proof", which is a heuristic method. In fact, it is incomplete because we have assumed a group homomorphism. In order to complete it we want to show a disentangling formula like For the purpose we set For t = 0 we have A(0) = B(0) = identity anḋ Next, let us calculateḂ(t). By use of the Leibniz rulė where we have used relations The proof is easy. By comparing coefficients ofȦ(t) andḂ(t) we have Note that the equationḟ is a (famous) Riccati equation. If we can solve the equation then we obtain solutions like Unfortunately, it is not easy. However there is an ansatz for the solution, G, F and E. That is, in (3.15). To check these equations is left to readers (as a good exercise). From this for all t and we finally obtain the disentangling formula e t{νK + +µK − −(µ+ν)K 3 } = e G(t)K + e −2 log(F (t))K 3 e E(t)K − (3. 16) with (3.15).
Therefore (3.8) becomes with (3.15). Some calculation by use of (2.22) gives where we have used N T = N and a † = a T . By coming back to matrix form by use of (3.6) we finally obtain

(3.18)
This form is very beautiful but complicated !

General Solution
Last, we treat the full equation (3.2) From the lesson in the preceding subsection it is easy to rewrite this as Then it is easy to see from (3.9), which is left to readers. That is, K 0 commutes with all {K + , K 3 , K − }. Therefore so that the general solution that we are looking for is just given by by use of (3.17) and (3.18).

Quantum Counterpart
In this section we explicitly calculate ρ(t) for the initial value ρ(0) given in the following.

Case of ρ(0) = |0 0|
Noting a|0 = 0 (⇔ 0 = 0|a † ), this case is very easy and we have because the number operator N (= a † a) is written as n|n n| =⇒ N|n = n|n , see for example (2.15). To check the last equality of (4.1) is left to readers. Moreover, ρ(t) can be written as see the next subsection.
The proof is divided into four parts. For simplicity we set z = αe γ and calculate the term Since |z = e −|z| 2 /2 e za † |0 we have G(t) n |n n| ez a = e −|z| 2 e za † e log G(t)N ez a by (4.1). Namely, this form is a kind of disentangling formula, so we want to restore an entangling formula.
For the purpose we use the disentangling formula Therefore (u → z, v → log G(t), w →z) so by noting we have By the way, from (3.15) we finally obtain [Fourth Step] In last, let us give the proof to the disentangling formula (4.4) because it is not so popular as far as we know. From (4.4) we have Then, careful calculation gives the disentangling formula The proof is simple. For example, e sa e ta † = e sa e ta † e −sa e sa = e te sa a † e −sa e sa = e t(a † +s) e sa = e st e ta † e sa .
The remainder is left to readers.
We finished the proof. The formula (4.3) is both compact and clear-cut and has not been known as far as we know. See [1] and [16] for some applications.
In last, let us present a challenging problem. A squeezed state |β (β ∈ C) is defined as |β = e 1 2 (β(a † ) 2 −βa 2 ) |0 . (4.6) See for example [4]. For the initial value ρ(0) = |β β| we want to calculate ρ(t) in (3.21) like in the text. However, we cannot sum up it in a compact form like (4.3) at the present time, so we propose the problem, Problem sum up ρ(t) in a compact form.

Concluding Remarks
In this chapter we treated the quantum damped harmonic oscillator, and studied mathematical structure of the model, and constructed general solution with any initial condition, and gave a quantum counterpart in the case of taking coherent state as an initial condition.
It is in my opinion perfect.
However, readers should pay attention to the fact that this is not a goal but a starting point. Our real target is to construct general theory of Quantum Mechanics with Dissipation.
In the papers [7] and [8] (see also [13]) we studied a more realistic model entitled "Jaynes-Cummings Model with Dissipation" and constructed some approximate solutions under any initial condition. In the paper [5] we studied "Superluminal Group Velocity of Neutrinos" from the point of view of Quantum Mechanics with Dissipation.
Unfortunately, there is no space to introduce them. It is a good challenge for readers to read them carefully and attack the problems.