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I'm a bit confused with the 'bra' notation in the representation of the Schrodinger equation. For example, in the momentum representation, the state $|E_{n}\rangle$ is represented by the function $\Phi_n(p)=\langle p|E_n\rangle$, and sometimes this notation could be used to calculate expectation values and matrix elements, or appear in the projection operator. What's the meaning of bra-notation, and usually how it is used? (I know its representation in terms of linear algebra, but I'm not pretty clear how it is used in different physics scenarios.)

John Rennie
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    I think this question is overly broad and needs focus. After reading the relevant Wikipedia article, perhaps you can ask about something specific that you don’t understand. – G. Smith Sep 14 '20 at 05:18
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    Some time ago there was another person confused with bra-ket notation. Here was my answer to that person https://physics.stackexchange.com/questions/289108/bra-ket-notation/289195#289195 – physicopath Sep 14 '20 at 07:05

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A 'bra' is a linear functional. It eats a vector and spits out a number. The number it spits out is the inner product between two vectors. What you should remember is that $\langle u|v\rangle$ is just the inner product between $u$ and $v$. The difference between the bra-ket inner product and the regular dot product is that bra ket notation is more general: $u$ and $v$ can be functions and can be complex valued as well.

So why is the inner product important in QM? Let's first take a regular old vector $v$. If you want to know how much of $v$ is pointing in the x-direction you can calculate this using the dot product with $\hat x$, the normalised vector in the x-direction. $$v_x=\hat x\cdot \vec v=\hat x^T\vec v=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}v_x\\v_y\end{pmatrix}$$ Generally if you want to know how much $\vec v$ is in the $\vec n$ direction you calculate $\hat n\cdot \vec v$.

In QM the vectors represents states so the question becomes 'how much of my state $|\psi\rangle$ is in the state $|n\rangle$?'. With the answer being $\langle n|\psi\rangle$. Or, to be more precise, the chance of getting $|n\rangle$ after measuring $|\psi\rangle$ is $|\langle n|\psi\rangle|^2$.