Change of basis and the transformation matrix
Contents:
# Change of basis
Take two incompatible operators $\hat A, \hat B$ $\implies [\hat A, \hat B] \neq 0$.
The entire ket space, as we know, can be spanned by eigenkets and their linear combinations. In this case, it can be $\lbrace|a'\rangle\rbrace$ or $\lbrace|b'\rangle\rbrace$.
What transformation takes us from $\lbrace|a'\rangle\rbrace$ to $\lbrace|b'\rangle\rbrace$?
Changing the set of base kets ($\lbrace|a'\rangle\rbrace$ to $\lbrace|b'\rangle\rbrace$ for eg.) is a change of basis. We want to figure out some transformation operator to connect these two.
We define a “unitary” operator $U$ as:
$$ U = \sum_k |b^{(k)}\rangle\langle a^{(k)}| $$
Then applying $U$ to $|a^{(l)}\rangle$ gives:
$$ U|a^{(l)}\rangle = \sum_k |b^{(k)}\rangle\langle a^{(k)}|a^{(l)}\rangle $$
Now since the $\lbrace|a'\rangle\rbrace$ are orthonormal, only $k = l$ survives:
$$ U|a^{(l)}\rangle = |b^{(l)}\rangle\underbrace{\langle a^{(l)}|a^{(l)}\rangle}_{1} $$
$$ \boxed{U|a^{(l)}\rangle = |b^{(l)}\rangle} $$
$\to$ that’s the transform.
Further, using the adjoint of an outer product, $U^\dagger = \sum_l |a^{(l)}\rangle\langle b^{(l)}|$:
$$ U^\dagger U = \sum_l\sum_k |a^{(l)}\rangle\langle b^{(l)}|b^{(k)}\rangle\langle a^{(k)}| $$
and since $\lbrace|b'\rangle\rbrace$ is orthonormal,
$$ \boxed{U^\dagger U = \sum_k |a^{(k)}\rangle\langle a^{(k)}|} $$
and we know from before that this is the identity
$$ \implies \boxed{U^\dagger U = 1} $$
# Transformation matrix
Let’s look at the matrix representation of $U$ in the $\lbrace|a'\rangle\rbrace$ basis. This basis is now the “old” basis, and $\lbrace|b'\rangle\rbrace$ is now the “new” basis.
$$ \langle a^{(k)}|U|a^{(l)}\rangle = \langle a^{(k)}|b^{(l)}\rangle $$
$\therefore$ the matrix elements of $U$ are made of the inner products of old basis bras and new basis kets.
Now consider rotating the axes, $(\hat x, \hat y, \hat z)$ to $(\hat x', \hat y', \hat z')$:
$$ R^{(3)} = \begin{pmatrix} \hat x\cdot\hat x' & \hat x\cdot\hat y' & \hat x\cdot\hat z' \\ \hat y\cdot\hat x' & \hat y\cdot\hat y' & \hat y\cdot\hat z' \\ \hat z\cdot\hat x' & \hat z\cdot\hat y' & \hat z\cdot\hat z' \end{pmatrix} $$
Same idea: old basis vectors dotted with the new ones.
So, the square matrix made of $\langle a^{(k)}|U|a^{(l)}\rangle$ is the transformation matrix from $\lbrace|a'\rangle\rbrace$ to $\lbrace|b'\rangle\rbrace$.
# Coefficients in the new basis
Consider
$$ |\alpha\rangle = \sum_{a'}|a'\rangle\langle a'|\alpha\rangle $$
(in the old basis). To get the expansion coefficients in the new basis, hit it with $\langle b^{(k)}|$:
$$ \begin{aligned} \langle b^{(k)}|\alpha\rangle &= \sum_l \langle b^{(k)}|a^{(l)}\rangle\langle a^{(l)}|\alpha\rangle \\ &= \sum_l \langle a^{(k)}|U^\dagger|a^{(l)}\rangle\langle a^{(l)}|\alpha\rangle \end{aligned} $$
since $\langle b^{(k)}| = \langle a^{(k)}|U^\dagger$ (the dual of $U|a^{(k)}\rangle = |b^{(k)}\rangle$).
That’s applying the transformation matrix to the coefficients in the old basis!
$$ \boxed{(\text{new}) = U^\dagger(\text{old})} $$
# Trace
$$ \text{tr}(X) = \sum_{a'}\langle a'|X|a'\rangle $$
$\to$ the sum of its diagonal elements.
The trace of an operator is independent of its base kets.
$$ \begin{aligned} \sum_{a'}\langle a'|X|a'\rangle &= \sum_{a'}\sum_{b'}\sum_{b''}\langle a'|b'\rangle\langle b'|X|b''\rangle\langle b''|a'\rangle \\ &= \sum_{b'}\sum_{b''}\langle b''|b'\rangle\langle b'|X|b''\rangle \\ &= \sum_{b'}\langle b'|X|b'\rangle \end{aligned} $$
(closure over $a'$ for the second line, orthonormality of the $|b'\rangle$ for the third).
We can also prove the following:
$$ \text{tr}(XY) = \text{tr}(YX), \qquad \text{tr}(U^\dagger XU) = \text{tr}(X) $$
$$ \text{tr}(|a'\rangle\langle a''|) = \delta_{a'a''}, \qquad \text{tr}(|b'\rangle\langle a'|) = \langle a'|b'\rangle $$
The first one is just inserting closure in the middle,
$$ \text{tr}(XY) = \sum_{a'}\sum_{a''}\langle a'|X|a''\rangle\langle a''|Y|a'\rangle = \sum_{a''}\langle a''|YX|a''\rangle = \text{tr}(YX) $$
and the second one follows from it: $\text{tr}(U^\dagger XU) = \text{tr}(XUU^\dagger) = \text{tr}(X)$.
# Diagonalization
Assume we know $\langle a''|B|a'\rangle$ $\to$ the matrix elements of $\hat B$ in the old basis. How do we find its eigenvalues and eigenkets now? Find a unitary matrix ($U^\dagger U = 1$) that diagonalizes $\hat B$.
We want $b'$ and $|b'\rangle$ from
$$ \hat B|b'\rangle = b'|b'\rangle $$
First we rewrite this for matrices (insert closure and hit with $\langle a''|$):
$$ \sum_{a'}\langle a''|B|a'\rangle\langle a'|b'\rangle = b'\langle a''|b'\rangle $$
$$ \begin{pmatrix} B_{11} & B_{12} & B_{13} & \cdots \\ B_{21} & B_{22} & B_{23} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix} \begin{pmatrix} C_1^{(l)} \\ C_2^{(l)} \\ \vdots \end{pmatrix} = b^{(l)}\begin{pmatrix} C_1^{(l)} \\ C_2^{(l)} \\ \vdots \end{pmatrix} $$
with
$$ B_{ij} = \langle a^{(i)}|B|a^{(j)}\rangle, \qquad C_k^{(l)} = \langle a^{(k)}|b^{(l)}\rangle $$
This is the matrix rep. Non-trivial solutions for $C_k^{(l)}$ are possible iff
$$ \boxed{\det(B - \lambda I) = 0} $$
$\to$ solving this gives the eigenvalues $b^{(l)}$. Knowing $b^{(l)}$, we can find $C_k^{(l)}$.
For this to work, $B$ must be Hermitian ($B = B^\dagger$).
Notice that $C_k^{(l)} = \langle a^{(k)}|b^{(l)}\rangle$ are exactly the transformation matrix elements $\langle a^{(k)}|U|a^{(l)}\rangle$. So finding the eigenvectors is finding $U$.
But what does diagonalizing a matrix mean?
$$ A \to PDP^{-1} $$
$\to$ this is diag., where $P \to$ eigenvectors (as columns) and $D \to$ eigenvalues (on the diagonal).
# Unitary equivalent observables
Theorem: Consider two sets of orthonormal bases $\lbrace|a'\rangle\rbrace$ and $\lbrace|b'\rangle\rbrace$, with $\hat U$ telling us the transform. With this we may construct a “unitary transform” of $\hat A$ as $UAU^{-1}$. Then $\hat A$ and $UAU^{-1}$ are said to be unitary equivalent observables.
The eigenvalue equation for $\hat A$:
$$ \hat A|a^{(l)}\rangle = a^{(l)}|a^{(l)}\rangle $$
$$ \implies UAU^{-1}U|a^{(l)}\rangle = a^{(l)}U|a^{(l)}\rangle $$
But wtf is $U^{-1}$? $\to$ Turns out that for unitary operators, $U^\dagger = U^{-1}$.
$$ \implies (UAU^\dagger)|b^{(l)}\rangle = a^{(l)}|b^{(l)}\rangle $$
$\implies |b^{(l)}\rangle$ is the eigenket of $UAU^\dagger$, and the $a^{(l)}$ are the eigenvalues of $UAU^\dagger$ (the same as $\hat A$’s).
And we know by definition
$$ \hat B|b^{(l)}\rangle = b^{(l)}|b^{(l)}\rangle $$
$\implies \hat B$ and $UAU^\dagger$ are simultaneously diagonalizable.
But is $UAU^{-1}$ (or $UAU^\dagger$) $= \hat B$? For physical cases, yes. Always? No.
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