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Change of basis

  • Thread starter Qubix
  • Start date
  • #1
82
1
I have two possible bases (a,b) and (a',b'). If I also have the transformation matrix P, such that
P(a,b)=(a',b'), am I correct in assuming that I can change an operator A, from the (a,b) basis to the (a',b') basis by applying

A' = P_transposed * A * P ?
 

Answers and Replies

  • #2
52
7
Since ##(a',b')## is a basis we can write ##a = c_1a' + c_2b'## for some scalars ##c_1, c_2##.
Similarly, ##b = d_1a' + d_2b'##. The the change of basis matrix, ##P##, from ##(a,b)## to ##(a', b')## is given by:

##P = \left( \begin{array}{cc}
c_1 & d_1 \\
c_2 & d_2 \\ \end{array} \right) ##.

Then ## A' = PAP^{-1} ##.

There are times when you can get away with using the transpose instead of the inverse, but that is only when the transpose is actually equal to the inverse. This is a very special case.
 
  • #3
82
1
So considering I have two bases (a,b) and (a', b'), with

a' = 1/sqr(2) ( a + b)
b' = 1/sqr(2) (a - b)

am I correct in saying that the unitary transformation between them is

U = 1/sqr(2) ## \left( \begin{array}{cc}
1 & 1 \\
1 & -1 \\ \end{array} \right) ##. ?

and then

Then ## A' = UAU^{-1} ##.
 
  • #4
52
7
So considering I have two bases (a,b) and (a', b'), with

a' = 1/sqr(2) ( a + b)
b' = 1/sqr(2) (a - b)

am I correct in saying that the unitary transformation between them is

U = 1/sqr(2) ## \left( \begin{array}{cc}
1 & 1 \\
1 & -1 \\ \end{array} \right) ##. ?

and then

Then ## A' = UAU^{-1} ##.
That looks great. In this case, not only is ##U^T = U^{-1}##. But you have that both of those are ##U## itself.
 

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