Hello there I have a problem with understanding the bracket notation of the hermitian operators.
The problem is that we know that hermitian operators must satisfies the following condition $$\langle\alpha|\hat{T}\beta\rangle=\langle\hat{T}\alpha|\beta\rangle$$
Well this seems good, but when I tried to understand the matrix representation of the previous expression I have noticed what seems to be a contradiction with matrix algebra rules. Here I will explain with the situation of three dimensions.
$$\langle\alpha|\hat{T}\beta\rangle\equiv a^\dagger Tb\\
=\begin{pmatrix}a_1^*&a_2^*&a_3^*\end{pmatrix}
\left[\begin{pmatrix}t_{11}&t_{12}&t_{13}\\t_{21}&t_{22}&t_{23}\\t_{31}&t_{32}&t_{33}\end{pmatrix}
\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}\right]$$
If we represented vector
This means that the operator (3×3 matrix) will work on the vector beta (3×1 matrix) to yield a 3×1 matrix, then the adjoint of vector alpha (1×3 matrix). Thus the result is a 1×1 matrix i.e, a scalar.
However, if we take the other side of the expression we will find:
$$\langle\hat{T}\alpha|\beta\rangle\equiv T^\dagger a^\dagger b\\
=\left[\begin{pmatrix}t_{11}&t_{21}&t_{31}\\t_{12}&t_{22}&t_{32}\\t_{13}&t_{23}&t_{33}\end{pmatrix}
\begin{pmatrix}a_1^*&a_2^*&a_3^*\end{pmatrix}\right]
\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}
$$
What does that mean? as I understand that the adjoint operator (the complex conjugate of the transformation matrix)(a 3×3 matrix) should work first on the adjoint of vector beta (a 1×3 matrix), but wait a moment! this cannot be done!! the number of columns of the first matrix is not equal to the number of the rows of the second matrix, thus matrix multiplication cannot be done.
I have thought a lot about this problem and suggested that we do the matrix multiplication form right to left, from a matrix operation view, this can be done, but the result will be a 3×3 matrix, thus even f we do that we will not get a scalar i.e, this cannot equal the left hand side of the expression $$\langle\alpha|\hat{T}\beta\rangle=\langle\hat{T}\alpha|\beta\rangle$$
please help me with this if you can.
Best Answer
Your incorrect step is the assumption that $$\langle\hat{T}\alpha|\beta\rangle\equiv T^\dagger a^\dagger b\,\,\,\,\,\,\, \text{(NOT TRUE)}$$ whereas really we have
$$\langle\hat{T}\alpha|\beta\rangle\equiv (T a)^\dagger b = a^\dagger T^\dagger b$$
because the dagger operation distributes in opposite order, just like the transpose. Then everything is consistent.