Operator Transformation under Change of Basis

[truth is life]

Homework Statement

Consider the three operators defined by $$\left(S_i\right)_{jk} = -i\epsilon_{ijk}$$ in the x-y-z space and the basis vectors given in x-y-z space as $$e^{\left(1\right)} = -\frac{1}{\sqrt{2}}\left(e_x + ie_y\right), e^{\left(0\right)} = e_z, e^{\left(-1\right)} = \frac{1}{\sqrt{2}}\left(e_x - ie_y\right)$$ Show that in the $e^{\left(1\right)}, e^{\left(0\right)}, e^{\left(-1\right)}$ basis the operators $S$ take on a particular given form (which happens to be the form of the spin-1 operators in quantum mechanics, and which I am therefore not writing out as it would be rather tedious)

Homework Equations

This is a change of basis problem, so the relevant equation is that for the change of basis for operators, $$B = Q^{-1}AQ$$ (where B is the transformed matrix)

The Attempt at a Solution

I found a Q which transformed the basis vectors from their $e^{\left(1\right)}$-type forms to (1,0,0), etc. correctly and which was also invertible (in fact Hermitian), $$ Q = \begin{bmatrix} -\frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} & 0 \end{bmatrix}$$

However, when I attempted to apply it to the S-matrices, I got something significantly different from the (provided) correct answer, to wit when I tried it with the $S_3$ matrix $$S_3 = -i\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ instead of the correct answer $$T_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$ I found $$T_3 = \begin{bmatrix} 0 & 0 & \frac{i}{\sqrt{2}} \\ 0 & 0 & \frac{-1}{\sqrt{2}} \\ \frac{-i}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{bmatrix}$$

 

[voko]

How did you find Q?

 

[truth is life]

How did you find Q?

It was essentially a guess-and-check-type method. I knew that in the old basis the new basis vectors had the particular form in the problem; I also knew that in the new basis they should take on a known different form ($e^{\left(1\right)} = \left(1, 0, 0\right)$, for example). I then constructed matrices for each basis vector which would take them to the correct representation in the new basis, then added those matrices together. Each matrix corresponds to a different row in the Q I wrote down.

I also (re)checked my linear algebra book, which stated that the columns of Q should be equivalent to the column vectors of the new basis in the old basis; this happens to be equivalent to the Hermitian conjugate of the Q I wrote down. Checking against how the book defines Q (taking new basis vectors to the old basis), it certainly does what it is supposed to. In the end, all this amounts to is swapping $Q^{-1}$ and $Q$ in the change of basis equation, which I just tried and which failed completely to give the correct answer, in fact giving something similar in form to $T_3$. This is exceptionally bewildering because it seems to indicate that I have the correct Q (or at least its Hermitian conjugate), and I've been testing with the known form of $S_3$, yet I am getting a known wrong answer. Obviously this suggests that I'm doing the matrix multiplication wrong, but I've tried this several times and I'm usually pretty good at matrix math.

 

[voko]

You have lost me here. What basis is "old" and what basis is "new"?

 

[truth is life]

You have lost me here. What basis is "old" and what basis is "new"?

The old basis is the standard x-y-z Cartesian basis. The new basis uses the $e^{\left(x\right)}$ vectors as the basis. So, in the old basis $$e^{\left(0\right)} = \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}$$ whereas in the new basis it would be $$e^{\left(0\right)} = \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$$ Similarly, in the old basis $$e^{\left(1\right)} = \begin{bmatrix} \frac{-1}{\sqrt{2}} \\ \frac{-i}{\sqrt{2}} \\ 0 \end{bmatrix}$$ while in the new basis $$e^{\left(1\right)} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ and $$e^{\left(-1\right)} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{-i}{\sqrt{2}} \\ 0 \end{bmatrix}$$ in the old basis and $$e^{\left(-1\right)} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ in the new.

 

[voko]

This I agree with it. But then Q's columns are definitely not the new basis in the old basis.

 

[truth is life]

This I agree with it. But then Q's columns are definitely not the new basis in the old basis.

Well, the book (which is not the book the problem comes from, to be clear) defines $Q$ slightly differently; if $e$ is the vector in the old basis and $E$ the vector in the new basis, then $$e = QE$$ whereas I was calculating $Q$ in the straightforward way as $$E = Qe$$ Both are valid (bases are bases, after all), but (calling the $Q$ defined by the book $R$) $Q = R^{-1}$ and the formula is predicated on using $R$ instead of $Q$, so you have to reverse the order of operations to $B = QAQ^{-1}$

This is all a little bit beside the point as either way the multiplication doesn't work out, though.

 

[vela]

With your Q, I found QS₃Q^-1 gives the correct result.

 

[voko]

If there is some "old" basis $u_i$ and there is some "new" basis $v_j$, and $v_j = \sum a_{ij} u_i$, then matrix $A = {a_{ij}}$ transforms any vector x from the v-basis (new) representation $x_v$ into the u-basis (old) $x_u = A x_v$. The columns of A are the components of the v-basis in the u-basis.

Any operator, whose matrix is O in the u-basis (old), becomes $A^{-1} O A$ in the new basis.

So it seems to me, you are doing this backwards.

 

[voko]

vela beat me to it!

 

[truth is life]

If there is some "old" basis $u_i$ and there is some "new" basis $v_j$, and $v_j = \sum a_{ij} u_i$, then matrix $A = {a_{ij}}$ transforms any vector x from the v-basis (new) representation $x_v$ into the u-basis (old) $x_u = A x_v$. The columns of A are the components of the v-basis in the u-basis.

Any operator, whose matrix is O in the u-basis (old), becomes $A^{-1} O A$ in the new basis.

So it seems to me, you are doing this backwards.

Which I noted...twice...in my own replies.

And it turns out the whole problem was...I was doing the multiplications wrong :facepalm: Everything else was right(-ish), I just messed up the basic arithmetic. Gets you every time.