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I'm trying to write down the bath Hamiltonian for a system of dimers and trimers. Imagine each of the monomers in the excited state can interact with several phonons with given frequencies. The bath Hamiltonian has the following form $\sum\limits_{n = 1}^N {\sum\limits_{j = 1}^M {{\omega _{nj}}} } a_{nj}^\dagger {a_{nj}}$. Here N is the number of monomers and M is the number of vibrational modes. I don't understand how the matrix of vibrational modes look like for a trimer with lets say four modes. It looks like I will have a 3-by-4 matrix, where all the rows are the same since I have the same modes for all the monomers. This doesn't make sense to me, since I'm supposed to diagonalize this matrix. Another ambiguity arises when I'm trying to figure out what values the number operator takes for each element of this matrix, since it looks to me it has to be 1 for each element. Imagine, you have two or three monomers which are all coupled to the same modes.

I will appreciate your comments!

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I'm not sure where you get the 3 and the 4 from, but it looks like you're confusing the $N$-by-$M$ matrix $(\omega_{nj})_{nj}$ with the hamiltonian proper, which is an infinite-dimensional linear operator acting on $NM$ copies of the single-particle state space (i.e. the Hilbert space $\mathcal H=(L_2(\mathbb R))^{\otimes NM}$).

The hamiltonian itself is already in a 'diagonal' form, and you can construct its eigenstates from the number basis of each mode of each monomer. Thus, if you denote by $|m_{nj}\rangle_{nj}$ the $m$th eigenstate of the number operator of the $j$th mode of the $n$th monomer, then the tensor product state $$ |m_{11}\rangle_{11}\cdots|m_{1M}\rangle_{1M}|m_{21}\rangle_{21}\cdots|m_{2M}\rangle_{2M}\cdots|m_{N1}\rangle_{N1}\cdots|m_{NM}\rangle_{NM} $$ is an eigenstate of your bath hamiltonian, with eigenvalue $$ E_{m_{11}\cdots m_{NM}}=\sum_{n=1}^N\sum_{j=1}^M\omega_{nj}m_{nj}. $$

If you had linear couplings between the different modes, then you would have to perform a small-matrix diagonalization procedure, on the matrix $H_{nj,n'j'}$ whose diagonal elements are the frequencies $H_{nj,nj}=\omega_{nj}$. However, this is not the case for your situation.

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  • $\begingroup$ What if I have a dimer or trimer then? Do I have to write Hamiltonian for any of them separately? What values the number operator gets for each element of diagonal matrix? I'm not quite sure that I understand the notation you used like when you talked about the N by M matrix what do you exactly mean by ${({\omega _{nj}})_{nj}}$? I'm sorry for my naive questions. I have never taken a course in second quantization and I'm learning everything on my own. Can you kindly give me some references with numerical examples on the format of spin-boson systems? How does the Hamiltonian matrix look like? $\endgroup$
    – Miranda
    Jan 24, 2014 at 21:53
  • $\begingroup$ I also think that I'm confused how $({\omega _{nj}})$ is addressed. I understand that each $\omega$ is equivalent to the energy state it belongs to. But, don't you think refering to $({\omega _{nj}})$ as the jth mode of monomer n can be confusing as one tries to write down the Hamiltonian operator? On the other hand doesn't n refers to the number of row and j to the number of column in the Hamiltonian matrix? $\endgroup$
    – Miranda
    Jan 24, 2014 at 22:03

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