# I Why is p^4 not Hermitian?

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#### vanhees71

Gold Member
We can not just use a specific eigenfunction $\psi$, for instance the radial part of ($\ell=0$) in combination with the Self_Adjoint test because we need two independent wavefunctions. See post #73

The boundery-term in your calculation in #62 for $\hat{p}^2$ does not cancel in the case of two independent wave-functions and one needs to rely on the $r^2$ factor to make it 0.
Sure, you can take any two $\ell=0$ wave functions. All belong not to the domain of $\hat{p}^2$ as an essentially self-adjoint operator. I think, Griffiths has it right here.

#### vanhees71

Gold Member
According to Arfken & Weber (10.6) the operator $p^2$ is not self-adjoint in a carthesian inner product but it is self-adjoint in a spherical radial inner product.

See the $r^2$ factor in post #69.
Sure, that's what I've shown above.

#### Dr.AbeNikIanEdL

You are misunderstanding. The problem is not the value of the integral over a function. The problem is the value of a function itself at r=0r=0r=0, which appears as a boundary term after a partial integration. In Cartesian coordinates there is simply no boundary at r=0r=0r=0, so in partial integration one does not need to worry about it.

Sure, I understand what you were saying there, but I do not think it is enough to just wave your hands and say “at infinity everything will be fine” without deriving what the terms at infinity really would be. Anyway, I was pointing the the last equation in #7. Do you agree that, if we just calculate both these integral, i.e. define the function $\Phi_n = p^4 \Psi_m$ and calculate

$\int \mathop{d^3 x} \Psi_n \Phi_m$

and

$\int \mathop{d^3 x} \Phi_n \Psi_m$

and take their difference we get a definite answer (if it is 0 $p^4$ is hermitian, otherwise not)?

Now we already know the answer if we calculate everything in spherical coordinates. So if we get a different answer in cartesian coordinates, the value of these integrals depends on what coordinate we choose to calculate them (or at least, it is not valid to go between cartesian and spherical coordinates). As these are all reasonably behaved functions, this would beg the question in what case spherical coordinates can be used anyway (though I would love to learn about subtleties).

#### Dr.AbeNikIanEdL

According to Arfken & Weber (10.6) the operator p2p2p^2 is not self-adjoint in a carthesian inner product but it is self-adjoint in a spherical radial inner product.
What is a “carthesian inner product”? The usual inner product should be

$\langle\Psi|\Phi\rangle = \int \mathop{d^3 x} \Psi^*\Phi$

and it should not matter what coordinates I choose in practice.

#### A. Neumaier

What is a “carthesian inner product”? The usual inner product should be

$\langle\Psi|\Phi\rangle = \int \mathop{d^3 x} \Psi^*\Phi$

and it should not matter what coordinates I choose in practice.
If you use a nonlinear transformation of the coordinates as new coordinates, the integral inherits an additional Jacobian determinant, and hence looks different. Thus the choice of coordinates matters;
the general form of the inner product is $\langle\Psi|\Phi\rangle = \int \mathop{d^3 x} w(x)\Psi(x)^*\Phi(x)$ with a weight $w(x)\ge0$, and this weight is different in different coordinate systems.

#### Dr.AbeNikIanEdL

Yes, but $w(x)$ is different in different coordinates systems in a way that the integral value stays the same, right? I was assuming this is implicitly contained in the notation $\mathop{d^3 x}$, i.e. in cartesian coordinates

$\mathop{d^3 x} = \mathop{dx} \mathop{dy} \mathop{dz}$

whereas in spherical coordinates

$\mathop{d^3 x} = \mathop{d\phi} \mathop{d\cos\theta} r^2 \mathop{dr}$.

In that sense I dont understand how the inner product can be “cartesian”.

#### A. Neumaier

Yes, but $w(x)$ is different in different coordinates systems in a way that the integral value stays the same, right? I was assuming this is implicitly contained in the notation $\mathop{d^3 x}$, i.e. in cartesian coordinates

$\mathop{d^3 x} = \mathop{dx} \mathop{dy} \mathop{dz}$

whereas in spherical coordinates

$\mathop{d^3 x} = \mathop{d\phi} \mathop{d\cos\theta} r^2 \mathop{dr}$.

In that sense I dont understand how the inner product can be “cartesian”.
To a Cartesian coordinate system corresponds the weight $w(x)=1$. Note that $x$ is just a dummy variables and can as well be the Cartesian $(x_1,x_2,x_3)$ as the spherical $(r,\phi,\theta)$; for the latter, $w(x)=r^2\sin\theta$.

#### Dr.AbeNikIanEdL

Note that xxx is just a dummy variables and can as well be the Cartesian (x1,x2,x3)(x1,x2,x3)(x_1,x_2,x_3) as the spherical (r,ϕ,θ)(r,ϕ,θ)(r,\phi,\theta); for the latter, w(x)=r2sinθw(x)=r2sin⁡θw(x)=r^2\sin\theta.
Which are exactly the expressions I wrote above. My point is that

$\langle\Psi|\Phi\rangle = \int \mathop{dx}\mathop{dy}\mathop{dz} \Psi^*(x,y,z)\Phi(x,y,z) = \int \mathop{d\phi}\mathop{d\theta}\mathop{dr} r^2 \sin\theta \Psi^*(r,\theta,\phi)\Phi(r,\theta,\phi)$

so what sense does it make to call the inner product “cartesian” or “spherical”? It is just “integral over $R^3$”, no matter what coordinates I choose to perform that integral.

#### A. Neumaier

My point is that

$\langle\Psi|\Phi\rangle = \int \mathop{dx}\mathop{dy}\mathop{dz} \Psi^*(x,y,z)\Phi(x,y,z) = \int \mathop{d\phi}\mathop{d\theta}\mathop{dr} r^2 \sin\theta \Psi^*(r,\theta,\phi)\Phi(r,\theta,\phi)$

so what sense does it make to call the inner product “cartesian” or “spherical”? It is just “integral over $R^3$”, no matter what coordinates I choose to perform that integral.
It is the Lebesgue integral over $R^3$, but only if you say that $x$ denote Cartesian coordinates.

#### Dr.AbeNikIanEdL

Ok, so this is supposed to be an cartesian inner product over spherical coordinates, i.e.

$\langle\Psi|\Phi\rangle = \int \mathop{d\phi}\mathop{d\theta}\mathop{dr} \Psi^*(r,\theta,\phi)\Phi(r,\theta,\phi)$

that then is somehow only defined on the subspace where the integral converges? What am I learning from this other than that it is a strange definition of the inner product?

#### A. Neumaier

Ok, so this is supposed to be an cartesian inner product over spherical coordinates, i.e.

$\langle\Psi|\Phi\rangle = \int \mathop{d\phi}\mathop{d\theta}\mathop{dr} \Psi^*(r,\theta,\phi)\Phi(r,\theta,\phi)$

that then is somehow only defined on the subspace where the integral converges? What am I learning from this other than that it is a strange definition of the inner product?
Rhi is a valid inner product defining a Hilbert space in which id/dr is self adjoint. It is equivalent to the physical inner product when one rescales the wave function by the square root of the weight obtained by the substitution rule.

#### Demystifier

2018 Award
Anyway, I was pointing the the last equation in #7. Do you agree that, if we just calculate both these integral, i.e. define the function $\Phi_n = p^4 \Psi_m$ and calculate

$\int \mathop{d^3 x} \Psi_n \Phi_m$

and

$\int \mathop{d^3 x} \Phi_n \Psi_m$

and take their difference we get a definite answer (if it is 0 $p^4$ is hermitian, otherwise not)?
I agree.

Now we already know the answer if we calculate everything in spherical coordinates. So if we get a different answer in cartesian coordinates, the value of these integrals depends on what coordinate we choose to calculate them (or at least, it is not valid to go between cartesian and spherical coordinates). As these are all reasonably behaved functions, this would beg the question in what case spherical coordinates can be used anyway (though I would love to learn about subtleties).
I think I should do a careful calculation by myself, before that I cannot tell anything definite.

#### Hans de Vries

Gold Member
What would be $P$ to get $P^2= -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right)$?
Actually $P$ (or better $\hat{p}_r$) here is given by the radial part of the Spherical Polar form of the Dirac equation. The following is from Paul Strange's book

Look at (8.9) and concentrate on the essential radial part of $\hat{p}$:

$$\hat{p}_r ~=~i\tilde{\gamma}_5\tilde{\sigma}_r\left(\hbar\dfrac{\partial}{\partial r}+\dfrac{\hbar}{r}\right)$$

If we square this radial part then we get $p_r^2$ because the $\tilde{\sigma}_r$ anti-commute but the two terms commute.

$$\hat{p}_r^2 ~=~\left[~i\tilde{\gamma}_5\tilde{\sigma}_r\left(\hbar\dfrac{\partial}{\partial r}+\dfrac{\hbar}{r} \right)~\right]^2 ~=~ -\hbar^2\left(\dfrac{\partial^2}{\partial r^2}+\dfrac{2}{r}\dfrac{\partial}{\partial r} \right)~=~ -\dfrac{\hbar^2}{r^2}\dfrac{\partial}{\partial r}\left( r^2\dfrac{\partial}{\partial r} \right)$$

The definition of $\tilde{K}$ which contains the angular parts of $\hat{p}$ using the angular momentum operators is:

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#### Hans de Vries

Gold Member
What would be $P$ to get $P^2= -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right)$?
So, with the (correct) version of the post above we may write more general for $P=\hat{p}_r$:

$\hat{p}_r~~=~~ i\hbar\left(\dfrac{\partial}{\partial r} +\dfrac{1}{r}\right)$

For an arbitrary power $\hat{p}_r^n$ we can write

$\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{\partial}{\partial r} +\dfrac{n}{r}\right)\left(\dfrac{\partial}{\partial r}\right)^{n-1}$

or alternatively:

$\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{1}{r^n}\dfrac{\partial}{\partial r}r^n\right)\left(\dfrac{\partial}{\partial r}\right)^{n-1}$

#### Demystifier

2018 Award
Now we already know the answer if we calculate everything in spherical coordinates. So if we get a different answer in cartesian coordinates, the value of these integrals depends on what coordinate we choose to calculate them (or at least, it is not valid to go between cartesian and spherical coordinates). As these are all reasonably behaved functions, this would beg the question in what case spherical coordinates can be used anyway (though I would love to learn about subtleties).
Ah, now I found the error in my argument. When one does partial integration in Cartesian coordinates, one encounters sub-integrals of the form
$$\int_{-\infty}^{\infty}dx \, \partial_x F(x,y,z)=F(\infty,y,z)-F(-\infty,y,z)$$
Naively I thought that such terms vanish because $F$ exponentially vanishes for $x\rightarrow\pm\infty$. But that's not necessarily true for $y,z\rightarrow 0$, because $F(x,y,z)$ diverges for $y,z\rightarrow 0$. So now I agree with older statements by @vanhees71 and @A. Neumaier that the source of the problem is divergence of the potential at $r=0$. If the potential has been regularized for small $r$, then the Hamiltonian eigenfunctions would have well defined derivatives at $r=0$ and the problem would disappear.

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#### A. Neumaier

So, with the (correct) version of the post above we may write more general for $P=\hat{p}_r$:

$\hat{p}_r~~=~~ i\hbar\left(\dfrac{\partial}{\partial r} +\dfrac{1}{r}\right)$

For an arbitrary power $\hat{p}_r^n$ we can write

$\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{\partial}{\partial r} +\dfrac{n}{r}\right)\left(\dfrac{\partial}{\partial r}\right)^{n-1}$

or alternatively:

$\hat{p}^n_r~~=~~ (i\hbar)^n\left(\dfrac{1}{r^n}\dfrac{\partial}{\partial r}r^n\right)\left(\dfrac{\partial}{\partial r}\right)^{n-1}$
But these powers have different domains, which causes the problems discussed in the present thread.

#### A. Neumaier

This shows explicitly that the domain changed; it is no longer an operator on the physical Hilbert space.

#### vanhees71

Gold Member
It is the Lebesgue integral over $R^3$, but only if you say that $x$ denote Cartesian coordinates.
The integral is over Euclidean $\mathbb{R}^3$ and as such the volume element is independent of the choice of coordinates,
$$\mathrm{d}^3 x = \epsilon_{ijk} \partial_i \vec{x} \partial_j \vec{x} \partial_k \vec{x} \mathrm{d}^3 q.$$
Of course, it's this specific integral measure to be used in the Hilbert space, because we are dealing with a representation/realization of the Galilei group, where space is Euclidean.

"Why is p^4 not Hermitian?"

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