# How did the bra-ket notation become mainstream in quantum mechanics?

I noticed that Dirac bra-kets and their algebra are very much like the linear algebra.

A ket is like a vector, a bra is like the conjugate transpose of a vector, a bra-ket is like a complex inner product, a ket-bra is like an outer product, and operator is like a matrix, and operator acting on a state is like matrix product. Even we speak about eigenstates which are very much like eigenvectors.

So how is it happened, that we invented a new notation for linear algebra instead of sticking to the usual notation of vectors and matrixes as usual?

• It is not "like linear algebra" but it is exactly linear algebra. At the time of Dirac linear algebra was not taught to all undergraduates. So he invented his own notation. Physicists found them convenient for the questions they consider. – Alexandre Eremenko Apr 9 '16 at 13:30
• @AlexandreEremenko So does this mean that the usual notation of vectors and matrices we use today not existed that time? I can accept that as an answer. – Calmarius Apr 9 '16 at 16:20
• It existed but was not widely known. Bra and ket were introduced for a convenient distinction of row vectors and column vectors. In another language they are called vectors and co-vectors. It was gradually understood that this distinction is very important and physicists made great contribution to this understanding. – Alexandre Eremenko Apr 9 '16 at 17:14
• A relevant discussion is here: hsm.stackexchange.com/questions/198/… – Alexandre Eremenko Apr 9 '16 at 17:15

In modern undergraduate courses column vectors $x$ are used. They stand on the right of the matrix: $Ax$. Column vectors correspond to ket vectors in Dirac's notation: $|x>$. Then you can write $A|x>$. Row vectors will be $x^T$ or in Dirac's notation bra-vectors $<x|$. Then when we want to write the standard dot product it will be $x^Ty$ or $<x|y>$, respectively. More generally when we want to write a quadratic form, it is either $x^TAy$ or $<x|A|y>$. Here I assumed for simplicity that the vectors are real. The correct thing to use instead of $x^T$ is of course the Hermitean transpose $x^*$. Dirac's notation $<x|A|y>$ is unambiguous only if $A$ is Hermitean.