How to get the operator in Dirac bra-ket notation?

Question

Suppose I have an operator given as $$T_a \psi(x) = \psi(x+a)$$ Is there a way I can get the operator in Dirac bra-ket notation. I am a newbie to QM. Please do give me hints.

Answer 1

This is using physicist math rather than rigorous math. The reason it is not rigorous is because when we work with "position space" we are typically working with a continuous rather than discrete Hilbert space (because the possible positions form a continuum). We'll just treat the math as if the Hilbert space was discrete, this is the physicist hack.

For a discrete Hilbert space we have the resolution of the identity for a complete set of states:

$$ \boldsymbol{I} = \sum_i |i\rangle \langle i| $$

For position space the sum turns into an integral and we write

$$ \boldsymbol{I} = \int |x\rangle\langle x|dx $$

Consider an arbitrary state $|\psi\rangle$:

$$ |\psi\rangle = \boldsymbol{I}|\psi\rangle = \int |x\rangle\langle x|\psi\rangle dx = \int \psi(x)|x\rangle dx $$

Where I've defined the position space wavefunction

$$ \psi(x) = \langle x|\psi\rangle $$

We have found the position space representation of $|\psi\rangle$ by acting $\boldsymbol{I}$ on the left and identifying $\psi(x)$ as the amplitude of the $|x\rangle$ basis vector. We play a similar trick to find the position space representation of arbitrary operator $T$.

$$ |\phi\rangle = T|\psi\rangle = \boldsymbol{I} T \boldsymbol{I} |\psi\rangle = \int |x\rangle \langle x|T|y\rangle\langle y |\psi\rangle dx dy $$

We see that we can identify the weight next to $|x\rangle$ in the integral as the position space representation (the wavefunction) of $|\phi\rangle$:

$$ \phi(x) = \int \langle x |T |y \rangle \psi(y) dy $$

So we see that the position space representation of $T$ is related to

$$ \langle x|T|y\rangle $$

We can express $T$ in the position basis playing some similar tricks with the resolution of the identity.

\begin{align} T =& \boldsymbol{I}T\boldsymbol{I} = \int |x\rangle\langle x|T |y\rangle\langle y|dxdy\\ =& \int \langle x|T|y\rangle |x\rangle\langle y| dx dy \end{align}

This technique allows us to express any operator in terms of a certain basis. We can think of objects like $|x\rangle \langle y|$ as "basis vectors" for operators and the "matrix elements" $\langle x|T|y\rangle$ as the coefficients of those basis vectors

For the translation operator in question we have

\begin{align} T_a =& \int \langle x|T_a|y\rangle |x\rangle\langle y|dydx\\ =& \int \langle x|y+a\rangle |x\rangle \langle y|dydx\\ =& \int |y+a\rangle \langle y| dy \end{align}

This expresses $T_a$ in bra-ket notation.