It is well known that when we consider a probe harmonic oscillators (called system) that is coupled to a reservoir of N harmonic oscillators, i.e. the Hamiltonian is written as the following, the system is actually exactlyl solvable $$H=H_S+H_R+H_{SR}= \hbar \omega_0(a^{\dagger}a+1/2)+\sum_{j=1}^{N}\hbar \omega_j(b^{\dagger}_j b_j+1/2) +\hbar(a\sum_{j=1}^{N}k^{*}_j b^{\dagger}_j+a^{\dagger}\sum_{j=1}^{N}k_j b_j)$$ Where $H_S$ describes the probe (system), and $H_R$ describes the reservoir, while $H_{SR}$ describes the coupling.

By following the procedure in this papaer, we know that this Hamiltonian is equivalent to a set of uncoupled harmonic oscillators $$H=\sum_{\mu=0}^{N}\hbar\alpha_\mu c^{\dagger}_{\mu}c_{\mu}+ C \quad C= \hbar(\frac{\omega_0}{2}+\sum_{j=1}^{N}\frac{\omega_j}{2})$$ Where $$c_{\mu}=\phi_{\mu}a+\sum_{n=1}^{N}\psi_{\mu n}b_{n} \quad (\mu=0,1,2...N)$$

I have two questions concerning this:

1. How to transform between the number states of $a/b$ and number states of $c$ ? That is, for the ladder operators $a/b/c$, we can define their number states respectively $$a|n,a>=\sqrt{n}|n-1,a> \quad b_i|n,b_i>=\sqrt{n}|n-1,b_i> \quad c_\mu|n,c_\mu>=\sqrt{n}|n-1,c_\mu>$$ How should I express $|n,a>$ in terms of $|n,c>$?

2 Could this formalism be extended to, for example, a qubit coupled to a harmonic oscillator bath? i.e.

$$H=H_S+H_R+H_{SR}= -\frac{1}{2}\hbar \omega_0\sigma_z+\sum_{j=1}^{N}\hbar \omega_j(b^{\dagger}_j b_j+1/2) +\hbar(\sigma_{-}\sum_{j=1}^{N}k^{*}_j b^{\dagger}_j+\sigma_{+}\sum_{j=1}^{N}k_j b_j)$$

Or perhaps it is impossible since qubit represents fermionic degrees of freedom, would it possible if a qubit is replaced by a spin-1 particle?


1 Answer 1

  1. You can construct them by successive application of ladder operators from the ground state. That is $$ |n, c_\mu\rangle \propto (c_\mu^\dagger)^n |0\rangle = (\phi^*_{\mu}a^\dagger+\sum_{n=1}^{N}\psi^*_{\mu n}b^\dagger_{n})^n|0\rangle \,.$$ You can then compute the desired overlap matrix elements in terms of the expansion coefficients.

  2. This Hamiltonian is generally known as the spin-boson model. Unlike the Leggett-Caldeira model for quantum brownian motion, to my knowledge there is no known exact solution for this model. However, there is a large pool of approximate and numerically exact methods for various spectral densities. Much literature on the topic is available. As a starting point, try Section III.A.4 in Breuer et al. (2016).

  • $\begingroup$ Hi, many thanks for problem 2. Let us relax the condition, and consider a spin-spin model. That is, one qubit is coupled with, e.g. 100 other qubits (while these 100 don't couple among themselves). Is this case solvable by similar methods? (i.e. transforming the Hamiltonian into 101 uncoupled qubits?) $\endgroup$
    – Tan Tixuan
    Mar 14, 2022 at 16:01
  • $\begingroup$ By the way. I would be very grateful if you can point me to more resources on this spin-boson mode. Actually I have been thinking about how to observe the decoherence in the quantum Brownian motion, this would involve trying to observe the non-diagonal elements of the density matrix of the probe harmonic oscillators. However, this seems impossible to be done in this formalism. $\endgroup$
    – Tan Tixuan
    Mar 14, 2022 at 16:17
  • $\begingroup$ @TanTixuan unfortunately the spin-spin model doesn't significantly relax the problem when one qubit is coupled with $N$ other qubits. The one-to-one problems are solvable (eg Jaynes Cummings model) and you can do $N$ qubits equally coupled to one bosonic mode (Tavis Cummings), but the general problems are all hard. Of course in finite dimensions you can diagonalize your Hamiltonian but remember that its size increases exponentially with the number of qubits $\endgroup$ Mar 16, 2022 at 17:27
  • $\begingroup$ @QuantumMechanic I have been reading Jaynes Cummings model and I think I get it. So are you aware of any formalism that could extend this Jaynes Cummings model to one qubit-two harmonic oscillator case? $\endgroup$
    – Tan Tixuan
    Mar 17, 2022 at 11:52
  • $\begingroup$ @TanTixuan if everything has the same energy spacings then it should be doable (because then conservation of energy is the same as conservation of excitations), especially for low numbers of excitations, but in general you'll have to look into numerics or approximations $\endgroup$ Mar 17, 2022 at 13:18

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