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Is there a way to describe interactions between systems with particles of different species, that is to say with different statistics?

For example: I am placing a boson next to a free fermion gas. How can I write down interaction Hamiltonians that make sense? By interaction Hamiltonians I mean Hamiltonians that contain products of fermion and boson operators?

Somewhat I am puzzled with this question, since both the boson and fermion operators act on different Hilbert spaces.

Might bosonization or fermionization a solution (at least for 1D systems)?

Best.

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There is nothing inherently wrong with having products of operators "acting on different Hilbert spaces". That's really an expression for saying the total Hilbert space is a direct product space and the operators act on different factors. Think of a much simpler system: two quantum spins $\vec{S}_1,\vec{S}_2$. The full Hilbert space has dimension $(2s_1+1)(2s_2+1)$. Now does a Hamiltonian like $H= -\mu \vec{S}_1\cdot\vec{S}_2$ make sense? ($\mu$ is just some constant with appropriate units) Yes it does, despite $\vec{S}_1$ and $\vec{S}_2$ acting on different factors of the Hilbert space. –  Michael Brown Oct 11 '13 at 14:05
    
Related to @MichaelBrown 's comments physics.stackexchange.com/questions/60409/…. –  joshphysics Oct 11 '13 at 16:51
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This is really just an expansion of Michael Brown's comment.

Let's say someone hands us any two quantum systems described by Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$. Then we might be curious to know how we can write down an interaction Hamiltonian for these systems. This includes, as a subcase, systems consisting of particles with different statistics.

To write down a Hamiltonian for the composite system that, in particular, can incorporate interactions between the two systems, we first need to decide what the Hilbert space of the composite system will be. The natural choice is called the tensor product of $\mathcal H_1$ and $\mathcal H_2$ written as $\mathcal H_1\otimes \mathcal H_2$. See the following physics.SE post for more information on why this choice is natural:

Should it be obvious that independent quantum states are composed by taking the tensor product?

Once we have chosen this as our composite Hilbert space, then there is no problem including interactions between the systems we started with. Let's examine Michael Brown's example in a bit more notational detail. Suppose the two systems consisted of one a particle of spin $s_1$ and a particle of spin $s_2$, and suppose that we want to write a spin-spin interaction term in the Hamiltonian. To do this, we appeal to a certain kind of product of two operators called the tensor product (not to be confused with the tensor product of the two Hilbert spaces). In particular, if $S_1^i$ are the operators representing the components of the first spin (which are operators on $\mathcal H_1$), and if $S_2^i$ are the operators representing the components of the second spin (which are operators on $\mathcal H_2$), then we first form operators $S_1^i\otimes I_2$ and $I_1\otimes S_2^i$ which act on the total Hilbert space $\mathcal H_1\otimes\mathcal H_2$ and represent "copies" of the original operators acting on the total Hilbert space. The notation here is that $I_1$ and $I_2$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$ respectively. Once we have done this, we can now write down the spin-spin interaction Hamiltonian as \begin{align} H_\mathrm{int} = \sum_i (S_1^i\otimes I_2)(I_1\otimes S_2^i) \end{align} As I wrote in the comments above, the following physics.SE post has more detail on what this tensor product notation means:

How to tackle 'dot' product for spin matrices

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