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.

  • $\begingroup$ What exactly do you mean by "get"? Do you want to express the same expression that you have expressed using Dirac notation? Or you want to get the explicit form of the operator in Dirac notation? $\endgroup$
    – Dvij D.C.
    May 9 '20 at 8:11
  • $\begingroup$ @DvijD.C. wanted to get the explicit form of the operator in Dirac notation. Sorry, if that was not clear. $\endgroup$
    – Suraj S
    May 9 '20 at 8:50
  • 1
    $\begingroup$ Thanks for the clarification, no maybe it was clear and I just got confused somehow. 😅 $\endgroup$
    – Dvij D.C.
    May 9 '20 at 8:51
  • 1
    $\begingroup$ See this one. $\endgroup$ May 9 '20 at 12:49

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.


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