Generator for parity?

Question

The unitary translation operator, $\hat{T}(\lambda) = e^{i\hat{p}\lambda/\hbar}$, is generated from the Hermitian operator $\hat{p}$.

The unitary rotation operator, $\hat{R}_z(\alpha)=e^{-i\hat{L_z}\alpha/\hbar}$, is generated from the Hermitian operator $\hat{L}_z$.

The unitary parity operator $\hat{P}: \Psi(x) \rightarrow \hat{P}\Psi(x)=\Psi(-x)$, is generated from which Hermitian operator?

Answer 1

This is a slightly counterintuitive approach, but very easy: take the harmonic oscillator eigenfunctions $\psi_n(x)$. It is known that the functions $\psi_{2k}(x)$ are even under parity and the functions $\psi_{2k+1}(x)$ are odd. Now the time evolution operator with the harmonic oscillator Hamiltonian is

$$ U(t) = \exp\left(\frac{-it}{2\hbar}\,\left(-\partial_x^2 + x^2\right) \right)\,, $$ where I set to $1$ both the mass and the frequency. Acting on such eigenfunctions it yields obviously

$$ U(t) \,\psi_n(x) = e^{-it\left(n+\frac12\right)}\psi_n(x)\,, $$ because the energy of $\psi_n$ is $\hbar\left(n+\tfrac12\right)$. Now if we evolve for a time of $\pi$ and define $$ \tilde{U}(\pi) \equiv e^{i\frac{\pi}{2}}U(\pi)\,, $$ it is immediate to see $$ \tilde{U}(\pi)\,\psi_n(x) = e^{-i\pi n} \psi_n(x) = \psi_n(-x)\,. $$ In the last step I used the parity of $\psi_n(x)$ according to the parity of $n$. Namely if $n$ is odd the phase factor is $-1$ and $\psi_n(-x) = - \psi_n(x)$, while if $n$ it's even the phase is $1$ and $\psi_n(-x)=\psi_n(x)$.

Now it doesn't matter that the theory you are studying is not the H.O., $\tilde{U}(\pi)$ is a linear operator and the $\psi_n(x)$ form a basis of $L^2(\mathbb{R})$, so on any function in $L^2(\mathbb{R})$ it will be true that $$ \tilde{U}(\pi) \,\psi(x) = \psi(-x)\,, $$ because in principle you could expand $\psi(x)$ in the H.O. eigenfunctions (even if you don't actually need to do it).

Answer 2

The parity operator does not have a generator in the way that the translation or rotation operators do.

Notice how you gave the translation operators and the rotation operators a parameter, like $\hat{T}(\lambda)$. There isn't just one translation operator, but a whole family of translation operators that form a group. What's more, the operator family is continuous in the sense that $\hat{T}(\lambda)$ and $\hat{T}(\lambda')$ are close if $\lambda$ and $\lambda'$ are close. In this case, Stone's theorem proves that we can take the derivative of this family: $$ \hat{t} \equiv i\times \lim_{h\rightarrow 0} \frac{\hat{T}(h)-I}{\lambda}$$ and that this "derivative operator" $\hat{t}$ is a hermitian operator, and that $\hat{T}(\lambda) = \exp(-i\lambda \hat{t})$. We call $\hat{t}$ the generator of the group $\hat{T}(\lambda)$

On the other hand, the parity operator is just a single operator. There is no continuous family of parity operators, and hence no way to take a derivative or define a generator.

In the language of group theory, a group of transformations that is continuous like translations or rotations forms a "Lie group." Lie groups have all sorts of special structure because of their continuity, including Lie group generators, where we can say things like $g(\lambda) = \exp(\lambda X)$ where $g$ is an element of a group and $X$ is an element of a different mathematical object called a Lie algebra.

Discrete groups, like the combined group of CPT, or the discrete symmetries of a crystal, do not have Lie group generators, or the concept of the exponential, or Lie algebras. Group theory does talk about "generators" for these groups. In this case the generators are elements of the group that can be multiplied to make any other element of the group. For example, 90 degree clockwise rotations can be chained together to make a 180 degree rotation and a 90 degree counterclockwise rotation.

Answer 3

I gather you want the "seat of the pants" beastie: $$\bbox[yellow]{\hat P= \exp \left ( \frac{-\pi}{2\hbar}(\hat {x}\hat p+\hat {p} \hat {x}) \right )}.$$

This is clearly hermitean, $\hat P ^\dagger = \hat P$, but also unitary, $\hat P ^{-1}=\hat P ^\dagger =\hat P$ : compose the exponentials in $\hat P ^2= \exp \left ( \frac{-\pi}{\hbar}(\hat {x}\hat p + \hat p \hat {x} ) \right )=1\!\! 1 $.

Given $[\hat x \hat p, \hat x]=-i\hbar \hat x $, it is evident that $$ e^{-\pi \hat x \hat p /\hbar} f(\hat x) ~e^{\pi \hat x \hat p /\hbar} = f(e^{-\pi \hat x \hat p /\hbar} ~\hat x ~e^{\pi \hat x \hat p /\hbar} )= f(e^{-[\pi \hat x \hat p /\hbar ,~\bullet}~~\hat x)=f(e^{i\pi} \hat x) =f(-\hat x).$$ $[A,\bullet \equiv \operatorname{ad}_A$ so that $e^A B e^{-A}= e^{[A,~\bullet} ~ B\equiv B+[A,B]+[A,[A,B]]/2!+...$, the Hadamard identiy. It should be apparent that the same works with the full hermitean exponent, and for arbitrary functions of $\hat p$ as well. The $\hat P ^2$ expression plugged in preserves all operators.

Recall that $\hat {p}|z\rangle= i\hbar \partial_z |z\rangle$, so the resulting pseudo-dilatation operator merely flips the sign of the space argument of the ket, $$\hat P |z\rangle=\exp (-i\pi z\partial_z +i\pi /2)|z\rangle=i|-z\rangle,$$ the i phase being an immaterial consequence of the conventions adopted here.

In operator language, in the x-representation, the pseudodilatation presents as a trivial change of variable $x=\log z$ application of Lagrange's shift operator, $\exp (a\partial_x) f(x)=f(a+x)$, namely $$ \hat P f(z) \hat P^\dagger = e^{i\pi ~ z\partial_z } f(z) = e^{i\pi \frac{\partial}{\partial \log z}} f(e^{\log z })=f(e^{i\pi +\log z })= f(-z). $$

• PS: There is an alternative expression, the Weyl-ordering of the above in terms of a double parametric integral, but presumably you'd have no use for it here. An alternative, circular, formal representation of it is also $\hat P= \int dx ~|-x\rangle \langle x|$ .