Calculating Matrix Representation of 11><11 for 2 Qubits

[Getterdog]

Forming the matrix representation of say 1><1 is no problem but how does one calculate the matrix representation of 11><11 ? Is it
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
Any help? thanks jack

 

[jk22]

i suppose 1> is (1 0)t then 11 is (1 0 0 0)t and the matrix is 1 in the upper left corner and all the rest 0.

 

[Getterdog]

Clarification

I take that 11>. Is (0 1 0 1 ) and < 10. Refers to ( 0 1 1 0 ) . The outer product as a matrix has more than 1 non zero entry.,so I'm still stuck. Any clarification on the correct way to do this.? Thanks

 

[Getterdog]

Unless 11< refers to ( 0 0 0 1) and 10< refers to (0 1 0 0 ) ,01< to (0 0 1 0) Is this it??

 

[Fredrik]

The matrix elements depend on the order of the basis vectors, so you need to choose a way to order them, e.g. (|00>,|01>,|10>,|11>). The matrix of |11><11| with respect to this ordered basis is
\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} This could be anticipated from the fact that |11><11| is a projection operator for a 1-dimensional subspace of the vector space, which is 4-dimensional.

The matrix of any operator T with respect to this ordered basis is
\begin{pmatrix}\langle 00|T|00\rangle & \langle 00|T|01\rangle &\dots & &\\ \langle 01|T|00\rangle & \langle 01|T|01\rangle & \dots \\ \vdots & \vdots & \ddots \\ \end{pmatrix}
For more information, see the https://www.physicsforums.com/showthread.php?t=694922 about the relationship between linear operators and matrices.

 

[Getterdog]

If my previous reply didn't register, thanks,my problem was that I didn't order the basis vectors correctly. Now I can proceed , although something bugs me about ordering of the basis, nature doesn't seem to care about humans need to order anything. Thanks