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Matrix product states (MPS) are a way of representing a (many-body) wavefunction. The method has been described, for example, in

The density-matrix renormalization group in the age of matrix product states. U. Schollwoeck. Ann. Phys 326, 96 (2011), arXiv:1008.3477.

However, would it be possible to see a concrete example for a small system? How will the MPS look like for the following two cases?

  1. A two spin system: $\frac{1}{\sqrt{2}}(| \uparrow \downarrow \rangle + |\downarrow \uparrow \rangle)$
  2. For a three-spin system: $\frac{1}{\sqrt{2}}(| \uparrow \uparrow \uparrow \rangle + |\downarrow \downarrow \downarrow \rangle)$
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You can think of an MPS as being built up by objects with three indices. How to easily represent such an object? We can think of this as a matrix where each entry is a vector (in particular the vector will be a vector in the on-site Hilbert space, so e.g. for a spin 1/2 system it will be of the form $\alpha |\uparrow \rangle + \beta |\downarrow \rangle$).

So for a chain where the on-site Hilbert space is a spin 1/2, each site will have such an object associated to it, like this: $$ A = \left( \begin{array}{ccc} \alpha_{11} \;| \uparrow \rangle + \beta_{11} \; |\downarrow \rangle & \alpha_{12} \;| \uparrow \rangle + \beta_{12} \; |\downarrow \rangle & \cdots \\ \alpha_{21} \;| \uparrow \rangle + \beta_{21} \; |\downarrow \rangle & \alpha_{22} \;| \uparrow \rangle + \beta_{22} \; |\downarrow \rangle & \cdots \\ \vdots & \vdots & \ddots \end{array} \right)$$

If it is a translationally invariant system, we can have that every site has the same $A$ associated to it. The physical state of our system is then obtained by multiplying all these $A$ matrices (where we `multiply' the matrix elements via the tensor product structure), more precisely: $$|\psi \rangle = \textrm{Tr}(A^N)$$ where we have assumed periodic boundary conditions. Here $N$ is the number of sites. Note that this object is now purely a vector in the tensor product of the on-site Hilbert spaces. So for example in the simple case that $$ A = \left( \begin{array}{cc} |\uparrow \rangle & 0 \\ 0 & |\downarrow \rangle \end{array} \right)$$ then $$ A^N = \left( \begin{array}{cc} |\uparrow \rangle \otimes |\uparrow \rangle \otimes \cdots \otimes |\uparrow \rangle & 0 \\ 0 & |\downarrow \rangle \otimes |\downarrow \rangle \otimes \cdots \otimes |\downarrow \rangle\end{array} \right)$$ So with periodic boundary conditions we get the cat state $|\psi \rangle = |\uparrow \cdots \uparrow \rangle + |\downarrow \cdots \downarrow\rangle$.

That answers your (ii). I'll leave (i) as an exercise ;)

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  • $\begingroup$ Here is a more systematic way of doing it, but more cumbersome. This is how a computer would do it: physics.stackexchange.com/questions/251522/… $\endgroup$ – Ruben Verresen Jul 6 '16 at 15:59
  • $\begingroup$ Many thanks for the answer :) Taking cue from your comment, the solution for (i), i guess, should be $2\times 2$ matrix with elements $(0, 1/\sqrt{2}, 1/\sqrt{2}, 0)$. $\endgroup$ – Adhip A. Jul 7 '16 at 7:31
  • $\begingroup$ Based on the link you shared, the solution seems to be the one i had mentioned in my comment above. While, when i think about the same solution in the way you mentioned in the detailed answer here, there should be a $A$ matrix corresponding to each site and therefore the MPS state should be a tensor product of two matrices. Should these be, \begin{equation} |\Psi\rangle = Tr(A_1 A_2) \end{equation} where $A_1 = \begin{bmatrix} |\uparrow \rangle & 0 \\ 0 & |\downarrow \rangle \end{bmatrix}$ and $A_2 = \begin{bmatrix} |\downarrow \rangle & 0 \\ 0 & |\uparrow \rangle \end{bmatrix}$ ? Thanks. $\endgroup$ – Adhip A. Jul 7 '16 at 7:54

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