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Defining an inner product over matrices and over vectors

Physics Asked on June 25, 2021

In quantum mechanics, in Dirac notation an inner product is denoted as $langle A|Brangle$ and one fundamental postulate is given as follows:

$langle A|Brangle = langle B|Arangle ^*$

If I were to consider $|Arangle $ and $|Brangle $ as vectors represented by column matrices, then the inner product between two vectors (column matrices) is defined as $A^TB$. If the vectors are complex then, the inner product is defined to be $A^{*T}B$ where the $*$ represents complex conjugation.

I am unable to put these two definitions together, due to the following confusion.

For any vector $|Arangle$ , I take $langle A|$ to be the conjugate transpose, which would be given as $A^{*T}$.

Therefore, if I convert the postulate in dirac notation to matrix algebra, I would have the LHS to be $A^{*T}B$. Taking a transpose, I would have:

$(A^{*T}B)^T = B^TA^*$

Now taking, the complex conjugate, I would get:

$(A^{*T}B)^{*T} = B^{*T}A$

And hence, $((A^{*T}B)^{*T})^{*T} = A^{*T}B= (B^{*T}A)^{*T}$

But according to the earlier postulate, I would have:

$A^{*T}B = (B^{*T}A)^*$

It seems like I am missing a transpose. I suspect I am going wrong with my analogy to matrix representation, but I do not understand where. I’m terribly sorry about my rudimentary LaTex and formatting skills.

One Answer

The result of an inner product is a scalar--that is, a 1x1 matrix. Transposing a scalar leaves it unchanged. So, $(B^{*T}A)^{*T} = (B^{*T}A)^{*}$.

On another note, the usual symbol for the complex transpose is the dagger $^{dagger}$, dagger in LaTeX.

Answered by Mark H on June 25, 2021

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