From my notes, I have that for a unitary matrix:
$$\underline{\underline{U}}^\dagger=\underline{\underline{U}}^{-1},\qquad \underline{\underline{U}}^\dagger=\big(\underline{\underline{U}}^{T}\big)^*$$
and for unitary matrix, $\underline{\underline A}$, the scalar product is such that
$$\Big(\underline x, \,{\underline{\underline{A}}\cdot{\underline y}}\Big)=\Big({\underline{\underline{A}}^\dagger\cdot\underline x, \,{\underline y}}\Big)\tag{1}$$
then $$\Big({\underline{\underline{U}}\cdot\underline x, \,{\underline{\underline{U}}}\cdot{\underline y}}\Big)=\Big({\underline{\underline{U}}^{\dagger}\cdot{\underline{\underline{U}}}\cdot\underline x,\,{\underline y}}\Big)=(\underline x,\, \underline y)\tag{2}$$
Aside
In the notation I'm using, I denote a vector with an underline and a matrix with a double-underline, but in $(1)$ and $(2)$ I didn't want to put the dot between vectors and matrices, but without the dot the underlines merge together which looks bad, sorry about this.
My first problem is that I do not understand the logic behind equation $(1)$ (and hence $(2)$ also). As I mentioned in the first comment to my previous question – I only have access to some incomplete (written) lecture notes where $(1)$ was given without proof, I tried to find proof of this property, but am so far unsuccessful.
Instead of leaving the question at that, I'm going to put this question into context by asking 2 more related questions that utilize this property:
- Proof of linear independence of degeneracy for $C_3$ symmetry group:
Suppose we have a group $$G: C_3, C_3^2=C_3^{-1}, C_3^3=E$$
With character table
$$
\begin{array}{c|c|c|c}
C_{3} & E
& C_3 & C_3^2=C_3^{-1} \\\hline
\chi^{(1)} & 1 & 1 & 1 \\\hline
\chi^{(2)} & 1 & e^{i 2\pi/3} & e^{-i 2\pi/3}\\\hline
\chi^{(3)} & 1 & e^{-i 2\pi/3} & e^{i 2\pi/3}\\\hline
\end{array}
$$
Character table displays no degeneracy as IRREPs, $\chi$ are 1D. But,
there is actually a 2-fold degeneracy as $$\hat H \psi_n=E_n \psi_n$$
& $$\hat H \psi_n^*=E_n \psi_n^*$$ so $\psi_n$ and $\psi_n^*$ both have energy $E_n$. So if $\psi_n$, $\psi_n^*$ belong to $\chi$, $\chi^*$ then they are linearly independent. $\psi_n$ is the basis for $\chi$ and $\psi_n^*$ is the basis for $\chi^*$.
To show linear independence take inner product:
$$\bbox[5px,border:2px solid blue]{\psi_n \in \big(1,\, e^{+i\phi},\, e^{-i\phi}\big)\\
{\psi_n}^* \in \big(1,\, e^{-i\phi},\, e^{+i\phi}\big)}\tag{X}$$
where $\phi=2\pi/3$
$$\hat T_{C_3}\psi_n=e^{+i\phi}\psi_n$$
$$\hat T_{C_3}{\psi_n}^*=e^{-i\phi}{\psi_n}^*$$
Where $\hat T_{C_3}$ is a coordinate transformation operator corresponding to a $C_3$ rotation.
So inner product is $\langle \psi_n |{\psi_n}^*\rangle=\langle \psi_n |\hat T_{C_3}^{-1}\hat T_{C_3}|{\psi_n}^*\rangle$, where $\bbox[yellow,5px]{\hat T_{C_3} \hat T_{C_3}^{-1}=\hat 1}$, where $\hat 1$ is the unit operator, since $\bbox[5px,border:2px solid green]{\hat T_g^{-1}=\hat T_g^\dagger}$ for group element $g$. So
$$\langle \psi_n |\hat T_{C_3}^{-1}\hat T_{C_3}|{\psi_n}^*\rangle\stackrel{\color{red}{?}}{=}\langle\hat T_{C_3}\psi_n | \hat T_{C_3}{\psi_n}^*\rangle =\langle e^{+i\phi}\psi_n |e^{-i\phi}{\psi_n}^*\rangle\stackrel{\color{blue}{?}}{=}{\big(e^{+i\phi}\big)}^*e^{-i\phi}\langle\psi_n|\psi_n^*\rangle\tag{3}$$$$\implies\langle\psi_n|\psi_n^*\rangle=e^{-2i\phi}\langle\psi_n|\psi_n^*\rangle\implies \langle\psi_n|\psi_n^*\rangle=0 \quad\blacksquare$$
For the red question mark, I have no idea how that equality follows. With regards to the blue question mark, I'm not sure, but, I think this may be a property of the Dirac notation; whereby the Bras are complex row vectors such that, $\langle A |=\big(a_1^*,a_2^*,a_3^*,\cdots\big)$, so to take the exponential outside requires taking the complex conjugate of the exponential. Can someone please confirm whether this logic is correct and explain to me how the red equality follows?
- Part of the derivation of the matrix element in selection rules.
For matrix element $$O_{\alpha\, i\, j}=\langle \psi_i |\hat O_\alpha|\phi_j\rangle=\langle\psi_i|\color{red}{\hat T_g^{-1}}\hat T_g \hat O_\alpha \hat T_g^{-1}\hat T_g | \phi_j\rangle\stackrel{?}{=}\langle\hat T_g \psi_i | \hat T_g \hat O_\alpha \hat T_g^{-1}|\hat T_g \phi_j\rangle\tag{4}$$
In the above, I think the author is inserting the unit operator twice, used earlier in the green box, (where it was used for a particular group element, $C_3$ – highlighted yellow). In the last equality with a question mark above it, I can see that the author has distributed the $\hat T_g$ into the Ket (which by my understanding are just column vectors with no complex numbers). But how did the $\hat T_g^{-1}$ (marked red) end up operating on the far left of the Bra?
Update:
I have been given a nice swift proof of equation $(1)$; $(x,Ay) = x^\dagger Ay = (A^\dagger x)^\dagger y = (A^\dagger x, y)$ with thanks to @BenGrossmann. I understand this simple proof and I had hoped it would allow me to understand instances 1. and 2. But there is a problem.
Am I to interpret inner product, $(x,Ay)$ as $\langle x|A y\rangle$ in the Dirac notation?
The problem is that in $(3)$ and $(4)$ we are not computing an inner product, but instead a matrix element, which by my understanding is written generally as $$\langle \phi |\hat Q|\psi \rangle\equiv \int\phi^*(x)\hat Q\psi(x)dx$$ and this can be seen on page 1 of this
In the case of $(3)$ the unit operator is being used, $$\langle \psi_n |\hat 1|{\psi_n}^*\rangle\langle \psi_n |\hat T_{C_3}^{-1}\hat T_{C_3}|{\psi_n}^*\rangle\stackrel{\color{red}{?}}{=}\langle\hat T_{C_3}\psi_n | \hat T_{C_3}{\psi_n}^*\rangle$$
Whereas in the case of $(4)$
$$O_{\alpha\, i\, j}=\langle \psi_i |\hat O_\alpha|\phi_j\rangle=\langle\psi_i|\hat 1 \hat O_\alpha \hat 1| \phi_j\rangle=\langle\psi_i|\color{red}{\hat T_g^{-1}}\hat T_g \hat O_\alpha \hat T_g^{-1}\hat T_g | \phi_j\rangle\stackrel{?}{=}\langle\hat T_g \psi_i | \hat T_g \hat O_\alpha \hat T_g^{-1}|\hat T_g \phi_j\rangle
$$
Since neither $(3)$ or $(4)$ are inner products I cannot utilize the now proven $(1)$. So could someone please explain how the equalities (with question marks above them) follow?
Best Answer
The bra-ket notation "in the abstract" represents a Hermitian inner product on the Hilbert space of states, where by definition linearity holds in second argument and swapping the arguments conjugates the result: $\langle \phi | (z+w) \psi \rangle=z\langle \phi | \psi \rangle+w\langle \phi | \psi \rangle$ and $\langle \phi | \psi \rangle=\overline{\langle \psi | \phi \rangle}$. This implies $\langle (z+w) \phi | \psi \rangle=\overline{z+w}\langle \phi | \psi \rangle$, so the product is conjugate-linear in the first argument. (Note that this is a matter of convention; mathematicians often choose the Hermitian inner product to be linear in the first entry and conjugate-linear in the second one. This does not affect things in any substantial way, but I think both your sources use the "linear in second entry" convention, so we'll stick to that).
This explains the blue question mark in (3).
In the case when $\psi$ and $\phi$ are column vectors in some $\mathbb{C}^n$, the "usual" or "standard" inner product given by $\langle \phi | \psi \rangle= \sum_{j=1}^n \overline{\phi_j}\psi_j$ (where each of the components $\phi_j$ and $\psi_j$ is a complex number). This has the properties above, that is, it's an example of a Hermitian inner product (making $\mathbb{C}^n$ into a Hilbert space). It can also be written in matrix notation as $\langle \phi | \psi \rangle=\bar{\phi}^T\psi=\psi^T\overline{\phi}$.
Suppose now that $A$ is a (bounded, linear) operator on our Hilbert space then its Hermitian conjugate $A^\dagger$ is DEFINED by $\langle \phi, A \psi \rangle=\langle A^\dagger \phi, \psi \rangle$ (this is uniquely defined since knowing product of $A^\dagger \phi$ with all vectors specifies $A^\dagger \phi$ uniquely, and one can check that the resulting map $A^\dagger$ is linear and bounded). Thus your equation (1) is a definition of $A^\dagger$.
We have $(XY)^\dagger=Y^\dagger X^\dagger$. Indeed, from definitions $\langle (XY)^\dagger u, v\rangle=\langle u, XYv \rangle=\langle X^\dagger u, Yv \rangle=\langle Y^\dagger X^\dagger u, Yv \rangle$ for all $u,v$, so $(XY)^\dagger=Y^\dagger X^\dagger$. This extends to products of more than one operators, so that, for example, $(ABCD)^\dagger=D^\dagger C^\dagger B^\dagger A^\dagger$.
If the Hilbert space is $\mathbb{C}^n$, then dagger becomes "conjugate transpose" operation, since it is easy to check that in that case $\langle \phi, A \psi \rangle=\overline{\phi}^T A\psi=(A^T\overline{\phi})^T\psi=\langle \overline{A}^T \phi, \psi\rangle$.
The bra-ket notation is then extended to be $\langle \phi | A|\psi\rangle=\langle \phi | A\psi\rangle$ by definition. We have from above that this equals $\langle A^\dagger \phi | \psi\rangle$.
Unitary matrices can be defined EITHER as ones with $\langle U\phi| U\psi\rangle=\langle \phi| \psi \rangle$ OR as ones for which $U^\dagger=U^{-1}$. It is immediate from definition of $\dagger$ that these are equivalent.
Now all the operations in your question marks are explained: Suppose we have an expression like $\langle\phi |ABCD|\psi\rangle$. By definition this is $\langle\phi |ABCD\psi\rangle$. This is also equal to $\langle\phi|ABC |D\psi\rangle$ or $\langle\phi|AB |CD\psi\rangle$ etc. This means that we can move the righttmost operator inbetween-bars to the right. It is also, as we noted above, by the definition adjoint, equal to $\langle (ABCD)^\dagger \phi |\psi\rangle$. As established before, this is the same as $\langle D^\dagger C^\dagger B^\dagger A^\dagger \phi |\psi\rangle$. and the same as $\langle (BCD)^\dagger A^\dagger \phi |\psi\rangle=\langle A^\dagger \phi |(BCD)\psi\rangle=\langle A^\dagger \phi|BCD |\psi\rangle$. This means that we can move the lefttmost operator inbetween-bars to the left with a dagger.
If the operator is unitary, then the dagger becomes inverse.
As a final note, when $\psi$ and $\phi$ are not finite-dimensional, but, instead lie in the space of complex-valued $L^2$ functions (on some space $X$) then the Hermitian product is the L^2 inner product $\langle \phi| \psi \rangle=\int_X \overline{\phi(x)}\psi(x) dx$ and our extended definition becomes $\langle \phi|A|\psi\rangle=\int_X \overline{\phi(x)}A(\psi(x)) dx$ agreeing with the formula in your post.