# College Math Teaching

## August 17, 2011

### Quantum Mechanics and Undergraduate Mathematics XIV: bras, kets and all that (Dirac notation)

Filed under: advanced mathematics, applied mathematics, linear albegra, physics, quantum mechanics, science — collegemathteaching @ 11:29 pm

Up to now, I’ve used mathematical notation for state vectors, inner products and operators. However, physicists use something called “Dirac” notation (“bras” and “kets”) which we will now discuss.

Recall: our vectors are integrable functions $\psi: R^1 \rightarrow C^1$ where $\int^{-\infty}_{\infty} \overline{\psi} \psi dx$ converges.

Our inner product is: $\langle \phi, \psi \rangle = \int^{-\infty}_{\infty} \overline{\phi} \psi dx$

Here is the Dirac notation version of this:
A “ket” can be thought of as the vector $\langle , \psi \rangle$. Of course, there is an easy vector space isomorphism (Hilbert space isomorphism really) between the vector space of state vectors and kets given by $\Theta_k \psi = \langle,\psi \rangle$. The kets are denoted by $|\psi \rangle$.
Similarly there are the “bra” vectors which are “dual” to the “kets”; these are denoted by $\langle \phi |$ and the vector space isomorphism is given by $\Theta_b \psi = \langle,\overline{\psi} |$. I chose this isomorphism because in the bra vector space, $a \langle\alpha,| = \langle \overline{a} \alpha,|$. Then there is a vector space isomorphism between the bras and the kets given by $\langle \psi | \rightarrow |\overline{\psi} \rangle$.

Now $\langle \psi | \phi \rangle$ is the inner product; that is $\langle \psi | \phi \rangle = \int^{\infty}_{-\infty} \overline{\psi}\phi dx$

By convention: if $A$ is a linear operator, $\langle \psi,|A = \langle A(\psi)|$ and $A |\psi \rangle = |A(\psi) \rangle$ Now if $A$ is a Hermitian operator (the ones that correspond to observables are), then there is no ambiguity in writing $\langle \psi | A | \phi \rangle$.

This leads to the following: let $A$ be an operator corresponding to an observable with eigenvectors $\alpha_i$ and eigenvalues $a_i$. Let $\psi$ be a state vector.
Then $\psi = \sum_i \langle \alpha_i|\psi \rangle \alpha_i$ and if $Y$ is a random variable corresponding to the observed value of $A$, then $P(Y = a_k) = |\langle \alpha_k | \psi \rangle |^2$ and the expectation $E(A) = \langle \psi | A | \psi \rangle$.