I am considering an assignment involving the Hubbard model. A state is given by $|\Phi\rangle=-|2\uparrow1\downarrow\rangle-|1\uparrow2\downarrow\rangle$ where particle 1 and 2 are electrons. The state is then written in the matrix representation $|\Phi\rangle=\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}$. I do not really know how you get to that matrix representation ? I think my problem is that I do not generally know how to go from the ket to the matrix. I know that for a spin up electron we have the matrix (1,0) and (0,1) for spin down. But for this composite systems I am not sure. \

The basis are given by the 4 states: $|1\uparrow1\downarrow\rangle$,$|2\uparrow2\downarrow\rangle$, $|1\uparrow2\downarrow\rangle$,$|1\downarrow2\uparrow\rangle$.

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    $\begingroup$ Can you provide more background on the problem? Do you know anything about the basis of your column vector? $\endgroup$ – cxx Dec 9 '18 at 20:18
  • $\begingroup$ I edited my question by adding the basis in the bottom. $\endgroup$ – Elias S. Dec 9 '18 at 20:39
  • $\begingroup$ Is your 4th basis vector correct? Seems like it should have arrows up and then down like the others. $\endgroup$ – doublefelix Dec 9 '18 at 20:44
  • $\begingroup$ It is written like that in the notes $\endgroup$ – Elias S. Dec 9 '18 at 20:49
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    $\begingroup$ I can only see this making sense if $\left|1\uparrow2\downarrow\right>=-\left|1\downarrow2\uparrow\right>$. $\endgroup$ – anonymous Dec 9 '18 at 21:03

Here is a recipe.

  1. Write down your basis kets, making sure they are orthonormal.
  2. Pick any ordering of these kets, but having picked it, stick to it.
  3. If there are $N$ basis kets you need $N$ vectors, each with $N$ components. You simply write the $N$ vectors which have just a single element 1 and the rest zero (i.e. columns of the $N \times N$ identity matrix), and assign them one by one to represent your basis kets. One can add further mathematical argument to show why this is what you have to do, but I am skipping that.

I can't understand the list of basis kets you gave, so I will have to illustrate the method by showing what happens for the standard four states for a pair of spin half particles. I have ordered them so as to try to match the vector you quoted. Thus I get $$ |\uparrow\uparrow\rangle \leftrightarrow \left[ \begin{array}{c} 1\\0\\0\\0 \end{array} \right] ,\;\; |\downarrow\downarrow\rangle \leftrightarrow \left[ \begin{array}{c} 0\\1\\0\\0 \end{array} \right] $$ $$ |\downarrow\uparrow\rangle \leftrightarrow \left[ \begin{array}{c} 0\\0\\1\\0 \end{array} \right] ,\;\; |\uparrow\downarrow\rangle \leftrightarrow \left[ \begin{array}{c} 0\\0\\0\\1 \end{array} \right] $$

One can also introduce an overall sign (or other phase factor), but that would be unusual. When working with a pair of spins I wouldn't normally pick the above order but as I say I tried to show something close to what seems to be going on in your example.

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  • $\begingroup$ I get the same for the standard fours states. But I am not sure still how to get the desired.... $\endgroup$ – Elias S. Dec 10 '18 at 18:42

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