Can a 2N by 2N matrix written in terms of N by N matrices?

  • Thread starter sokrates
  • Start date
  • #1
sokrates
483
2
I posted this question over at the QM page,

https://www.physicsforums.com/showthread.php?t=714076

but I realized I am really looking for a

hard Mathematical proof ...

A description of a numerical way of proving this would also be very helpful for me.

or a reference covering the subject.

Many thanks in advance,
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
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As a warmup, would you know how to prove that ##\{ I_{2}, \sigma_i\}## is a basis for Hermitian ##2\times 2## matrices? The result for ##2N\times 2N## will follow by writing the matrix in block form and using the basis as explained by wle in that thread.
 
  • #3
sokrates
483
2
Yes - I can do the 2x2 proof I guess.

Because any 2x2 Hermitian matrix can be written as:

[tex]
H=\begin{bmatrix}
a & c -i \ d \\
c + i \ d & b
\end{bmatrix}
[/tex]

where a,b,c,d are all real numbers.

Then H can be uniquely defined in terms of Pauli matrices:
[tex]
\frac{1}{2}\left[ (a+b) \ I_{2\times 2} + (a-b) \ \sigma_z + 2 \ c \ \sigma_x + 2 \ d \ \sigma_y\right]
[/tex]

But how to extend this to 2N by 2N ?
 
  • #4
sokrates
483
2
Yes, I got it ...

Just write it out explicitly and choose A,B,C,D accordingly to get the random 2N by 2N matrix.

Many thanks for directing me to that.
 

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