Proof that parity operator is Hermitian in 3-D

[dyn]

Hi.
I have been looking at the proof that the parity operator is hermitian in 3-D in the QM book by Zettili and I am confused by the following step
∫ d³r φ*(r) ψ(-r) = ∫ d³r φ*(-r) ψ(r)
I realize that the variable has been changed from r to -r. In 3-D x,y,z this is achieved by taking the modulus of the Jacobian which obviously gives a positive value as opposed to working in I-D when a minus sign arises. But on the LHS of the equation the triple integrals all had +∞ at the top and -∞ at the bottom but on the RHS changing the variable results in -∞ at the top of the integrals and +∞ at the bottom. Is this correct ?

 

[blue_leaf77]

obviously gives a positive value

I don't think so. The reflection with respect to the origin in spherical coordinate amounts to the transformation rule
$$
r' = r \\
\theta' = \pi - \theta \\
\phi' = \pi + \phi
$$
The Jacobian matrix of this transformation is a diagonal matrix with its two diagonal elements equal to 1 and the last one equal to -1. So the determinant is -1.

But on the LHS of the equation the triple integrals all had +∞ at the top and -∞ at the bottom but on the RHS changing the variable results in -∞ at the top of the integrals and +∞ at the bottom. Is this correct ?

Correct, but there is also a -1 from the Jacobian which allows you to revert the integral limits hence removing the minus sign.
EDIT: Thanks to vanhees, I forgot that the Jacobian used in the integral transformation is indeed the modulus of it.

 

[vanhees71]

The transformation rule for the volume elements uses the modulus of the Jacobian determinant of the transformation. That's intuitive since volumes are positive.

 

[dyn]

So I still end up with the all the integral limits reversed ?

 

[dyn]

In the 1-D case if you just make a straightforward substitution you end up with a negative sign and the limits of integration reversed so they just cancel each other out but this doesn't work in the 3-D case ( or a 2-D case ) as the modulus is taken.
Which raises another question - the formula for change of variables where the modulus is taken doesn't state that it doesn't work in 1-D but in the 1-D case you can end up with a minus sign but a modulus is always positive !

 

[vanhees71]

Ok, let's do this carefully. In the 1D case you have
$$\langle \phi| \hat{P} \psi \rangle=\int_{-\infty}^{\infty} \mathrm{d} x \phi^*(x) \psi(-x).$$
Now substitute $y=-x$. Of course you have $\mathrm{d} x=-\mathrm{d} y$. This gives
$$\langle{\phi}|\hat{P} \psi \rangle=-\int_{+\infty}^{-\infty} \mathrm{d} y \phi^*(-y) \psi(y) = + \int_{-\infty}^{\infty} \mathrm{d} y \phi^*(-y) \psi(y) = \langle \hat{P} \phi|\psi \rangle,$$
and thus $\hat{P}$ is self-adjoint.

The very same calculation goes through for $\mathbb{R}^3$. There you have, with $\vec{y}=-\vec{x}$,
$$\mathrm{d}^3 \vec{x} = \left | \det \frac{\partial (x_1,x_2,x_3)}{\partial (y_1,y_2,y_3)} \right|=|-1|=1,$$
and thus again immediately the self-adjointness of $\hat{P}$.

 

[dyn]

Ok, let's do this carefully. In the 1D case you have
$$\langle \phi| \hat{P} \psi \rangle=\int_{-\infty}^{\infty} \mathrm{d} x \phi^*(x) \psi(-x).$$
Now substitute $y=-x$. Of course you have $\mathrm{d} x=-\mathrm{d} y$. This gives
$$\langle{\phi}|\hat{P} \psi \rangle=-\int_{+\infty}^{-\infty} \mathrm{d} y \phi^*(-y) \psi(y) = + \int_{-\infty}^{\infty} \mathrm{d} y \phi^*(-y) \psi(y) = \langle \hat{P} \phi|\psi \rangle,$$
and thus $\hat{P}$ is self-adjoint.

The very same calculation goes through for $\mathbb{R}^3$. There you have, with $\vec{y}=-\vec{x}$,
$$\mathrm{d}^3 \vec{x} = \left | \det \frac{\partial (x_1,x_2,x_3)}{\partial (y_1,y_2,y_3)} \right|=|-1|=1,$$
and thus again immediately the self-adjointness of $\hat{P}$.

Taking the modulus in the 3-D case means we have d³x = d³y meaning we are missing the minus sign that appears in the 1-D case but the limits are still reversed so that doesn't appear to prove P is Hermitain

 

[vanhees71]

No, in 3D the signs of the boundaries are by definition not reversed. The measure in the integral is always positive.

In this simple case of the transformation of integration variables you can, by the way, get to the right result using the theorem by Beppo-Levi in Cartesian coordinates, using the calculation as in the 1D, because
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} = \int_{-\infty}^{\infty} \mathrm{d} x_1 \int_{-\infty}^{\infty} \mathrm{d} x_2 \int_{-\infty}^{\infty} \mathrm{d} x_3.$$

 

[blue_leaf77]

After reviewing my old notes and some online articles, it seems that, as vanhees suggested, the Jacobian is taken to be its modulus in the transformed integral so that the limits are chosen such that the more negative values serves as the lower limit. This way the integral will always be positive.

in 3D the signs of the boundaries are by definition not reversed.

I think the definition of modulus Jacobian as utilized in variable transformation in an integral holds not only for 3D but for any dimension.

 

[vanhees71]

Sure, all this is valid in any (finite) dimension.

 

[DrClaude]

For the benefit of everyone, I would like to point out this other thread by the OP touching on the same subject:
https://www.physicsforums.com/threads/change-of-variable-and-jacobians.905474/

 

[dyn]

Let me check that I understand this. If I just make the substitution in 1-D I get a minus sign and the integral limits reverse. But if I use the modulus of the Jacobian in any dimension including 1-D I don't reverse the order of the integral limits ?
This seems strange ! But as long as it works !

 

[vanhees71]

Have you carefully read my posting #6? There it's explained in detail. In 1D you get the same result as with the general formula with the modulus of the Jacobian: Substituting $y=-x$ leads to $\mathrm{d} x=-\mathrm{d}y$, and the integral runs from $+\infty$ to $-\infty$. Interchanging the boundaries back to the usual order means that you have to take another factor $(-1)$. The general formula for arbitrary dimensions tells you to take the modulus of the Jacobian and take the volume element always positive (including the boundaries when calculating the integral by a concrete parametrization of the volume).

 

[dyn]

Yes I have read your post #6. I totally understand the 1-D situation. I understand taking the modulus of the Jacobian for arbitrary dimensions. The only bit I don't understand and have not encountered before is that the limits don't reverse. If I have a triple integral over x.y.z with the upper limit +∞ and I have made the substitution x→ -x , y→ -y , z→ -z then I don't understand why the upper limits are not -∞ .

 

[Jilang]

Well they are, but -1 to the power of 3 is -1.

 

[vanhees71]

Well, in this case you can even do the calculation without using the Jacobian, using the Beppo-Levi theorem. We want to calculate the integral
$$I=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \phi^*(\vec{x}) \psi(\vec{x}).$$
Now according to the theorem by Beppo and Levi you can write this in Cartesian coordinates as
$$I=\int_{-\infty}^{\infty} \mathrm{d} x \int_{-\infty}^{\infty} \mathrm{d} y \int_{-\infty}^{\infty} \mathrm{d} z \phi^*(x,y,z) \psi(-x,-y,-z).$$
Now you substitute successively $x=-x'$, then $y=-y'$, then $z=-z'$. You repeat three times the calculation we did for the 1D case, and you get
$$I=\int_{-\infty}^{\infty} \mathrm{d} x' \int_{-\infty}^{\infty} \mathrm{d} y' \int_{-\infty}^{\infty} \mathrm{d} z' \phi^*(-x',-y',-z') \psi(x',y',z').$$

 

[dyn]

Thanks to everyone who replied in this thread