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?

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    $\begingroup$ Not an expert, but intuitively translations and rotations are continuous symmetries, they form Lie groups which are smooth manifolds hence they have a Lie algebra which contains generators and an exponential map. Parity on the other hand is a discontinuous symmetry, so I wouldn't expect something similar to generators to exist. But do wait for an input from somebody who knows more group theory. $\endgroup$ Commented Aug 9, 2019 at 12:37
  • $\begingroup$ That makes sense! $\endgroup$
    – Travis Lee
    Commented Aug 9, 2019 at 12:40

3 Answers 3


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).

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    $\begingroup$ Nice. Up to redefinitions, and phases, the algebra in optical phase-space ($a^\dagger, a$) dances in lockstep with that of my answer. $\endgroup$ Commented Aug 9, 2019 at 16:20
  • $\begingroup$ This is a beautiful example of how a question with a very simple and straightforward answer can also have a very interesting and subtle answer. $\endgroup$
    – tparker
    Commented Aug 10, 2019 at 19:02
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    $\begingroup$ If you leave the mass and frequency explicit, then you'll find that the final operator $\tilde{U}(\pi)$ only depends on them (and on $\hbar$) through the combined quantity $\sqrt{\hbar/m\omega}$ which sets a length scale $l$ for the problem: $$\tilde{U}(\pi/\omega) = \exp\left[ -i \frac{\pi}{2} \left( -(l \partial_x)^2 + \left( \frac{x}{l} \right)^2 - 1\right) \right].$$ How can I see that this operator does not actually depend on $l$? It's not obvious. $\endgroup$
    – tparker
    Commented Aug 10, 2019 at 19:11
  • $\begingroup$ I think I may have figured it out. Conjugating any logarithm of a matrix $M$ by a matrix that commutes with $M$ always yields another logarithm of $M$. So logarithms of degenerate operators like the parity operator are highly non-unique: conjugating one log by any parity-conserving operator gives another log. In your case, rescaling the value of $l$ corresponds to a change a basis that conserves parity, so it yields another log. I guess Cosmas Zachos's very different-looking logarithm is also similar to yours via a parity-preserving conjugation. Would you agree with this reasoning? $\endgroup$
    – tparker
    Commented Aug 11, 2019 at 4:29
  • 2
    $\begingroup$ Yes. In this case, I guess, you'd be conjugating by the operator that rescales $x \to x/l$, since if you do this change of variables your modified $\tilde{U}$ reduces to mine. $\endgroup$
    – MannyC
    Commented Aug 11, 2019 at 4:39

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.

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    $\begingroup$ There is however a way out. You did not prove that there is not a selfadjoint operator $K$ such that $(e^{iK}\psi)(x) = \psi(-x)$. The existence of $K$ is independent from the existence of a continuous one-parameter subgroup of $O(3)$ which includes the parity $P$ as an elements. (This cannot exist since $\det(P) =-1$ whereas $\det(I)=1$ and the subgroup must preserve the sign of the determinant it being continuous.) Even if the subgroup does not exist it does not automatically mean that $K$ does not exist as well. $\endgroup$ Commented Aug 9, 2019 at 13:36
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    $\begingroup$ There's a classical analogue to this as well. Remember that Noether's theorem only guarantees the existence of conserved quantities for continuous (rather than discrete) symmetries. These conserved quantities are the same symmetry generators as in the quantum case. $\endgroup$
    – tparker
    Commented Aug 9, 2019 at 13:54
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    $\begingroup$ @ValterMoretti Good point. I suppose there are two different questions, "what is the generator of parity" and "what is the logarithm of parity." My answer to the first is "no such thing." The other answers answer the second question. $\endgroup$ Commented Aug 9, 2019 at 16:42
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    $\begingroup$ To clarify a potential confusion from @ValterMoretti 's comment: you can't have a continuous one-parameter subgroup of $\mathrm{O}(3)$ that continuously rotates space to reach the 3D parity operator, for the reason he discusses. But you can have a continuous one-parameter group of unitary operators on the Hilbert space, that connects the identity operator to the parity operator that spatially flips wavefunctions. Manny C.'s answer gives an explicit example of such a group. Since the Hilbert space is complex, the eigenvalues can continuously go from 1 to -1 through the complex plane. $\endgroup$
    – tparker
    Commented Aug 10, 2019 at 23:35

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 $\hat P= \int dx ~|-x\rangle \langle x|$ .
  • $\begingroup$ Why is exp(−𝜋ℏ(𝑥̂ 𝑝̂ +𝑝̂ 𝑥̂ )) = 1? Also, what is the dot in the second offset line, and is there a missing close bracket? $\endgroup$
    – tparker
    Commented Aug 10, 2019 at 19:37
  • $\begingroup$ Also, as the exponential of a Hermitian operator, shouldn't this expression be positive-definite? $\endgroup$
    – tparker
    Commented Aug 10, 2019 at 20:22
  • $\begingroup$ @tparker I used the physicists' convention, $[A,\bullet = \operatorname {ad}_A$, as many physicists are not aware of the standard latter one. So the exponential of $\operatorname {ad}_A$ means power-series-wise operation of commutation with A on the left. Using it with 2π as in the "given..." argument yields the identity on each quantum variable, and, intercalation wise, on all operators. $\endgroup$ Commented Aug 10, 2019 at 20:37
  • $\begingroup$ @tparker You mean for P? there are a few i s and branches of the complex logarithm flying loose (as this was transcribed from phase-space QM), and I am shaky on infinite spaces like Hilbert / Fock space and the arsenal of kludges involved. Unlike the phase-space rotator answer, this is "seat of the pants", as advertised. $\endgroup$ Commented Aug 10, 2019 at 21:26

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