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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 (3x3 matrix) will work on the vector beta (3x1 matrix) to yield a 3x1 matrix, then the adjoint of vector alpha (1x3 matrix). Thus the result is a 1x1 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 3x3 matrix) should work first on the adjoint of vector beta (a 1x3 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 3x3 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.

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2 Answers 2

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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.

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  • $\begingroup$ Thank you a lot, this has solved the problem $\endgroup$ Aug 30, 2022 at 18:16
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You made a mistake when you do the matrix representation with the adjoint operator on the left side of the braket

$\langle \hat{T} a | b \rangle = (\hat{T} a)^* b = a^* \hat{T}^* b = a^* \hat{T} b = a^* (\hat{T} b) = \langle a | \hat{T} b \rangle $

having using the assumption the operator is Hermitian, and thus the matrix self-adjoint that $\hat{T}=\hat{T}^*$.

If you're familiar to linear algebra, think at the braket as an inner product. When you evaluate the adjoint (i.e. the complex adjoint) of a matrix product (or matrix-vetcor product) remember to switch the order of the factors when transposing, i.e. $(\hat{T} a)^* = a^* \hat{T}^*$

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    $\begingroup$ Thank you for help :) $\endgroup$ Aug 30, 2022 at 18:16

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