# I Why is p^4 not Hermitian?

• Featured

#### Dr.AbeNikIanEdL

Fine, but do you agree with me that p4p4p^4 is Hermitian in Cartesian coordinates for the hydrogen atom wave functions? Technically, when you do a partial integration in Cartesian coordinates, you only need the boundary values at x,y,z→±∞x,y,z→±∞x,y,z\rightarrow\pm\infty, not at r→0r→0r\rightarrow 0 as in spherical coordinates. Since everything behaves well at infinity, all the boundary terms vanish in Cartesian coordinates so the operator is Hermitian.
I don’t think it is that simple, in cartesian coordinates, integrating by parts should give a more complicated expression. It should not really change anything in the end, the final expression e.g. in post #7 says essentially that the difference between two integrals is some number, and since a coordinate change should not change the values of the integrals, that should hold in cartesian coordinates as well.

The problem with spherical coordinates is that it puts a boundary at the place where it should not really be, that is at r→0

This might introduce additional artificial singularities, but not doing it should not change anything about the singularities actually present in the physics problem (e.g. the Coulomb potential is singular at zero, no matter what coordinates you use).

#### A. Neumaier

Fine, but do you agree with me that $p^4$ is Hermitian in Cartesian coordinates for the hydrogen atom wave functions? Technically, when you do a partial integration in Cartesian coordinates, you only need the boundary values at $x,y,z\rightarrow\pm\infty$, not at $r\rightarrow 0$ as in spherical coordinates. Since everything behaves well at infinity, all the boundary terms vanish in Cartesian coordinates so the operator is Hermitian. The problem with spherical coordinates is that it puts a boundary at the place where it should not really be, that is at $r\rightarrow 0$.
For the free particle it doesn't matter since the measure in the transformed inner product takes care of it.

But not all hydrogen atom wave functions are in the domain of $p^4$ so $p^4$ cannot be said to be self-adjoint on the space spanned by these.

#### Demystifier

2018 Award
a coordinate change should not change the values of the integrals
It should not when that change is reqular everywhere. But the change from Cartesian to spherical coordinates is not regular at $r=0$.

#### vanhees71

Gold Member
That was my first thought too, but then I realized that the problem occurs even for a free particle if you choose to write the Laplace operator in spherical coordinates.

The spherical coordinates are simply not good coordinates at $r=0$, whether you are doing quantum physics, classical physics, or pure math.

Indeed, note that the wave function itself is regular at $r=0$. So one can rewrite the wave function in Cartesian coordinates $x,y,z$ and express the momentum operator in Cartesian coordinates too. When one does that, both $p^2$ and $p^4$ become hermitian and the problem goes away. I think that's the solution of the problem.

The moral is: Using symmetry to express things in coordinates in which the problem looks "the simplest" is not always a wise thing to do.
That would mean it's the usual coordinate singularity in spherical coordinates. Of course, there many textbooks commit several sins. I've even seen an EM textbook, where they evaluated expressions with $\delta^{(3)}(\vec{x})$ using spherical coordinates and write, without any comment about the very dangerous idea $1/(r^2 \sin \vartheta) \delta(r) \delta(\vartheta) \delta(\varphi)$. Many students in the recitation I tutored at the time got very confused by getting obvious wront results using this mediocre math ;-)). Math can be abused to a certain extent but not further!

#### vanhees71

Gold Member
It should not when that change is reqular everywhere. But the change from Cartesian to spherical coordinates is not regular at $r=0$.
It's not regular along the entire polar axis. The Jacobian is $r^2 \sin \vartheta$.

#### Dr.AbeNikIanEdL

It should not when that change is reqular everywhere. But the change from Cartesian to spherical coordinates is not regular at r=0r=0r=0.
The worst bit of $p^4 \Psi_n$ at $r\to0$ behaves like $e^{-r}/r^2$, so if I write out the integrals of this part I get something like

$4 \pi \int_0^\infty \mathop{dr} e^{-r}$

or in cartesian coordinates

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathop{dx} \mathop{dy} \mathop{dz} e^{-\sqrt{x^2+y^2+z^2}}/(x^2+y^2+z^2)$

and you are saying that these are not the same? Or am I misunderstanding?

#### Hans de Vries

Gold Member
Note that $p^2$ is NOT an Hermitian MATRIX since.

$p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right) ~~=~~ -\hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{\partial^2}{\partial r^2}\right)$

It is the sum of (infinite dimensional) real matrices and the righthand side tells us that one of them is anti-symmetric $(-\partial_r)$ while the other is symmetric $(-\partial^2_r)$. The diagonals of these matrices look like:

$\begin{smallmatrix} 0&1&0&0&0&0&0 \\ -1&0&1&0&0&0&0 \\ 0&-1&0&1&0&0&0 \\ 0&0&-1&0&1&0&0 \\ 0&0&0&-1&0&1&0 \\ 0&0&0&0&-1&0&1 \\ 0&0&0&0&0&-1&0 \end{smallmatrix} ~~~~~~~~~~ \begin{smallmatrix} 2&-1&0&0&0&0&0&0 \\ -1&2&-1&0&0&0&0&0 \\ 0&-1&2&-1&0&0&0&0\\ 0&0&-1&2&-1&0&0&0 \\ 0&0&0&-1&2&-1&0&0 \\ 0&0&0&0&-1&2&-1&0 \\ 0&0&0&0&0&-1&2&-1 \\ 0&0&0&0&0&0&-1&2 \end{smallmatrix}$

A real matrix must be symmetric to be Hermitian so it is the $\partial_r$ that destroys the symmetry. (The complete term $\tfrac{2}{r}\partial_r$ is not symmetric either).

Now in some cases it may be that it for fills the equation $\langle f \,|\, p^2\,g \rangle = \langle p^2\,f \,| \,g \rangle$ which is used as the definition of an Hermitian operator depending on $f$ and $g$. For instance the trivial $f=g=0$. There is no reason to assume that for less trivial examples of $f$ and $g$ this somehow should be true for $p^4$ as well.

Some good references:
Spherical Polar Laplacian: Quantum Physics, Eisberg & Resnick appendix M
Hermitian Operators: Mathematical methods for Physicists, (edit: Arfken & Weber), Chapter 10.2

Last edited:

#### vanhees71

Gold Member
Which book are you referring to? There are many with the title "Mathematical methods for Physicists". A very nice AJP article on the subject is

and another one

https://arxiv.org/abs/quant-ph/9907069

#### Hans de Vries

Gold Member
Which book are you referring to? There are many with the title "Mathematical methods for Physicists".
Did add the authors to the post: Arfken & Weber.

#### vanhees71

Gold Member
Though one should also note that "hermitian" is not enough for operators describing observables in QT. They must be self-adjoint! See the paper by Bonneau. The here discussed expample of $(\vec{p}^2)^2$ or (in position representation) $\Delta^2$ is a case, where this is obviously important!

#### Dr.AbeNikIanEdL

Note that $p^2$ is NOT an Hermitian MATRIX since.

$p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right) ~~=~~ -\hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{\partial^2}{\partial r^2}\right)$

It is the sum of (infinite dimensional) real matrices and the righthand side tells us that one of them is anti-symmetric $(\partial_r)$ while the other is symmetric $(\partial^2_r)$. The diagonals of these matrices look like:

$\begin{smallmatrix} 0&1&0&0&0&0&0 \\ -1&0&1&0&0&0&0 \\ 0&-1&0&1&0&0&0 \\ 0&0&-1&0&1&0&0 \\ 0&0&0&-1&0&1&0 \\ 0&0&0&0&-1&0&1 \\ 0&0&0&0&0&-1&0 \end{smallmatrix} ~~~~~~~~~~ \begin{smallmatrix} 2&-1&0&0&0&0&0&0 \\ -1&2&-1&0&0&0&0&0 \\ 0&-1&2&-1&0&0&0&0\\ 0&0&-1&2&-1&0&0&0 \\ 0&0&0&-1&2&-1&0&0 \\ 0&0&0&0&-1&2&-1&0 \\ 0&0&0&0&0&-1&2&-1 \\ 0&0&0&0&0&0&-1&2 \end{smallmatrix}$

A real matrix must be symmetric to be Hermitian so it is the $\partial_r$ that destroys the symmetry. (The complete term $\tfrac{2}{r}\partial_r$ is not symmetric either).

Now in some cases it may be that it for fills the equation $\langle f \,|\, p^2\,g \rangle = \langle p^2\,f \,| \,g \rangle$ which is used as the definition of an Hermitian operator depending on $f$ and $g$. For instance the trivial $f=g=0$. There is no reason to assume that for less trivial examples of $f$ and $g$ this somehow should be true for $p^4$ as well.

Some good references:
Spherical Polar Laplacian: Quantum Physics, Eisberg & Resnick appendix M
Hermitian Operators: Mathematical methods for Physicists, (edit: Arfken & Weber), Chapter 10.2

How do you obtain these matrices?

#### vanhees71

Gold Member
Note that $p^2$ is NOT an Hermitian MATRIX since.

$p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right) ~~=~~ -\hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{\partial^2}{\partial r^2}\right)$

It is the sum of (infinite dimensional) real matrices and the righthand side tells us that one of them is anti-symmetric $(\partial_r)$ while the other is symmetric $(\partial^2_r)$. The diagonals of these matrices look like:

$\begin{smallmatrix} 0&1&0&0&0&0&0 \\ -1&0&1&0&0&0&0 \\ 0&-1&0&1&0&0&0 \\ 0&0&-1&0&1&0&0 \\ 0&0&0&-1&0&1&0 \\ 0&0&0&0&-1&0&1 \\ 0&0&0&0&0&-1&0 \end{smallmatrix} ~~~~~~~~~~ \begin{smallmatrix} 2&-1&0&0&0&0&0&0 \\ -1&2&-1&0&0&0&0&0 \\ 0&-1&2&-1&0&0&0&0\\ 0&0&-1&2&-1&0&0&0 \\ 0&0&0&-1&2&-1&0&0 \\ 0&0&0&0&-1&2&-1&0 \\ 0&0&0&0&0&-1&2&-1 \\ 0&0&0&0&0&0&-1&2 \end{smallmatrix}$

A real matrix must be symmetric to be Hermitian so it is the $\partial_r$ that destroys the symmetry. (The complete term $\tfrac{2}{r}\partial_r$ is not symmetric either).

Now in some cases it may be that it for fills the equation $\langle f \,|\, p^2\,g \rangle = \langle p^2\,f \,| \,g \rangle$ which is used as the definition of an Hermitian operator depending on $f$ and $g$. For instance the trivial $f=g=0$. There is no reason to assume that for less trivial examples of $f$ and $g$ this somehow should be true for $p^4$ as well.

Some good references:
Spherical Polar Laplacian: Quantum Physics, Eisberg & Resnick appendix M
Hermitian Operators: Mathematical methods for Physicists, (edit: Arfken & Weber), Chapter 10.2
That's not a valid argument since you forgot that the scalar product in this case is (omitting the angular piece, i.e., concentrating on wave functions that are only $r$ dependent (particularly of course the $\ell=0$ energy eigenstates of spherical systems fulfill this constraint)
$$\langle \psi |\phi \rangle=\int_{0}^{\infty} \mathrm{d} r r^2 \psi^*(r) \phi(r).$$
Now consider
$$p^2=-\frac{1}{r^2} \partial_r (r^2 \partial_r)$$
Let's check for hermiticity (as a necessary constraint for self-adjointness) [edit: corrected in view of #63]
$$\langle \psi |p^2 \phi \rangle=-\int_0^{\infty} \mathrm{d} r \psi^*(r) \partial_r (r^2 \partial_r \phi(r)) \\ =-r^2 \psi^*(r) \partial_r \phi(r))|_{r=0}^{\infty} + \int_0^{\infty} \mathrm{d} r r^2 \partial_r \psi^*(r) \partial_r \phi(r) \\ =r^2 [\partial_r \psi^*(r) \phi(r)-\psi^*(r) \partial_r \phi(r))]|_{r=0}^{\infty} - \int_0^{\infty} \mathrm{d} r \partial_r[r^2 \partial_r \psi^*(r)] \phi(r) \\ =r^2[\psi^*(r) \partial_r \phi(r)-\psi^*(r) \partial_r \phi(r)]|_{r=0}^{\infty} + \langle p^2 \psi|\phi \rangle.$$
So $p^2$ is Hermitean indeed if the boundary terms vanish.

This holds for the $\ell=0$ wave functions (see also the quotes from Griffiths's book in the first few postings of the thread, though he forgot to take the conjugate complex of the left wave function, which is unimportant for his case though since there you can choose the eigenstates as real functions).

Last edited:

#### Dr.AbeNikIanEdL

Isn’t there a $\partial_r$ too many in your first boundary term (although not changing the conclusion)?

#### vanhees71

Gold Member
Of course you are right. I edited the posting accordingly.

#### Hans de Vries

Gold Member
How do you obtain these matrices?
They are just the difference (discrete) versions of the first and second order differentiation. When you consider a function $f(x)$ then its discrete version becomes $f(n)$ where $n$ is an integer. The discrete version of the first and second order derivatives may be given by:

$\Delta(n) = f(n+1)-f(n)~~~~~~\mbox{and}~~~~~~\Delta^2(n) = f(n+1)-2f(n)+f(n-1)$

These two sequences "slide" along the diagonal of the matrices because a differentiation is in principle a convolution.

Thus $\partial_x$ is a convolution with $\tfrac{\partial \delta(x)}{\partial x}$ and $\partial^2_x$ is a convolution with $\tfrac{\partial^2 \delta(x)}{\partial x^2}$ where $\delta(x)$ is the Dirac function.

This is where we get the two sequences from:

$\begin{array}{lrrrrrrrrr} \mbox{The discrete delta function:} & ~~0&~~0&~~0&~~0&~~1&~~0&~~0&~~0&~~0 \\ 1^{st}~\mbox{difference of delta function:} & 0&0&0&1&-1&0&0&0&0 \\ 2^{nd}~\mbox{difference of delta function:} & 0&0&0&1&-2&1&0&0&0 \\ \end{array}$

Note we used f(n+1)-f(n-1) for the first order difference in the matrices. Technically this avoids the extra unwanted shift of an 1/2 lattice step

Last edited:

#### Dr.AbeNikIanEdL

ok, I see what you did but not really how it applies here. I would have assumed that, to get a matrix representation, I would have to calculate something like

$(p^2)_{mn} = \langle\Psi_m|p^2|\Psi_n\rangle$

and this would be symmetric $(p^2)_{mn} = (p^2)_{nm}$?

#### vanhees71

Gold Member
$$(p^2)_{mn}=(p^2)_{nm}^*.$$
What's to prove is that this is not the case for the $\ell=0$ bound states of the hydrogen atom for the operator $p^4$, which is shown in #62.

The conclusion is that the $\ell=0$ bound states of the hydrogen atom are not in the domain of the operator $p^4$.

#### Dr.AbeNikIanEdL

(p2)mn=(p2)∗nm.(p2)mn=(p2)nm∗.​

(p^2)_{mn}=(p^2)_{nm}^*.
Sure, but in this specific case everything is real anyway.

What's to prove is that this is not the case for the ℓ=0ℓ=0\ell=0 bound states of the hydrogen atom for the operator p4p4p^4, which is shown in #62.

The conclusion is that the ℓ=0ℓ=0\ell=0 bound states of the hydrogen atom are not in the domain of the operator p4p4p^4.
I know, and I think we have arrived at these conclusions already multiple times during this thread. With my latest posts I was trying to understand @Hans de Vries comments about $p^2$, since I am not familiar with this “getting matrices from discretised versions of functions and operators”.

#### Hans de Vries

Gold Member
... the scalar product in this case is (omitting the angular piece, i.e., concentrating on wave functions that are only $r$ dependent ...
1) The trivial Self-Adjoint, Hermitian Operators.

The Laplacian $\nabla^2$ is symmetric while $p^2$ is the part that acts on the radial part of the eigenstates. We have in a trivial way the following (integral) equations:

$\langle\,\bar{g}\,|\,\nabla^2 f\,\rangle ~~=~~ \langle\,\overline{\nabla^2 g}\,|\,f\,\rangle~~=~~ \langle\,\nabla^2 \bar{g}\,|\,f\,\rangle$

because for each individual point of the fields $f$ and $g$ the following holds.

$g(\nabla^2 f)~~=~~ ((\nabla^2)^\intercal g) f ~~=~~ (\nabla^2 g) f$

The integral expressions are trivial in this case because the expressions above hold for every point so their integral over any bounded area holds as well. An operator here is classified as Self-Adjoined / Hermitian based on the integral equations. This is one step beyond qualifying them based on their (infinite-dimensional) matrix properties. This is a key difference.

2) The original question in the OP

Now if $p^2$ was real-symmetric then $p^4$ would be as well. They both would classify as (infinite dimensional) Self-Adjoined Hermitian operators in a trivial way just like $\nabla^2$. This seems to be the expectation in the OP. He expects $p^4$ to be Hermitian operator but based on the wrong assumption that $p^2$ is an (infinite dimensional) Self-Adjoint, Hermitian matrix, but this is not the case.

$p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right) ~~=~~ -\hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{\partial^2}{\partial r^2}\right)$

The first order derivative term is real anti-symmetric. Therefore the transpose of $p^2$ is:

$\Big(p^2\Big)^\intercal~~=~~ \hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}-\dfrac{\partial^2}{\partial r^2}\right)$

Nevertheless $p^2$ becomes a Self-Adjoint, Hermitian operator according to the integral definition, not in the trivial way as $\nabla^2$ does.

3) Non trivial Self-Adjoint, Hermitian Operators.

We consult the excelent Arfken & Weber to see under which conditions a second order differential operator $\mathcal{L}$ becomes a Self-Adjoined operator. In other words when $\langle\,v\,|\,\mathcal{L}\,u\,\rangle = \langle\,\mathcal{L}\,v\,|\,u\,\rangle$, starting with the simpler case $\langle\,u\,|\,\mathcal{L}\,u\,\rangle = \langle\,\mathcal{L}\,u\,|\,u\,\rangle$

It is Arfken & Weber (10.6) which tells us if $\mathcal{L}$ is self-adjoint.

4) Operator $p^2$ as a Self-Adjoint Operator.

At first it seems that acording to (10.6) that $p^2$ is not a Self Adjoined operator.

$p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right)$

but if we multiply it by a factor $r^2$ then $r^2 p^2$ becomes a Self Adjoined operator. This factor $r^2$ is readily explained when we look at the calculations of Hendrik van Hees a few post back. What he does is identical to the steps in Arfken & Weber but now for the specific case of $p^2$.

Now consider
$$p^2=-\frac{1}{r^2} \partial_r (r^2 \partial_r)$$
Let's check for hermiticity (as a necessary constraint for self-adjointness) [edit: corrected in view of #63]
$$\langle \psi |p^2 \phi \rangle=-\int_0^{\infty} \mathrm{d} r \psi^*(r) \partial_r (r^2 \partial_r \phi(r)) \\ =-r^2 \psi^*(r) \partial_r \phi(r))|_{r=0}^{\infty} + \int_0^{\infty} \mathrm{d} r r^2 \partial_r \psi^*(r) \partial_r \phi(r) \\ =r^2 [\partial_r \psi^*(r) \phi(r)-\psi^*(r) \partial_r \phi(r))]|_{r=0}^{\infty} - \int_0^{\infty} \mathrm{d} r \partial_r[r^2 \partial_r \psi^*(r)] \phi(r) \\ =r^2[\partial_r \psi^*(r) \phi(r)-\psi^*(r) \partial_r \phi(r)]|_{r=0}^{\infty} + \langle p^2 \psi|\phi \rangle.$$
So $p^2$ is Hermitean indeed if the boundary terms vanish.
The factor $r^2$ is explained by the integration. We integrate over spheres so we need a factor of $r^2$ to account for the surface of the spheres depending on $r$. The multiplication of $p^2$ by $r^2$ thus makes it into a self-adjoint operator according to (10.6)

There seems to be no reason that a combination of $p^4$ and $r^2$ would also make a valid self-adjoint operator as in the trivial cases.

Last edited:

#### Demystifier

2018 Award
The worst bit of $p^4 \Psi_n$ at $r\to0$ behaves like $e^{-r}/r^2$, so if I write out the integrals of this part I get something like

$4 \pi \int_0^\infty \mathop{dr} e^{-r}$

or in cartesian coordinates

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathop{dx} \mathop{dy} \mathop{dz} e^{-\sqrt{x^2+y^2+z^2}}/(x^2+y^2+z^2)$

and you are saying that these are not the same? Or am I misunderstanding?
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 $r=0$, which appears as a boundary term after a partial integration. In Cartesian coordinates there is simply no boundary at $r=0$, so in partial integration one does not need to worry about it.

#### A. Neumaier

The first order derivative term is real anti-symmetric.
Only in the Lebesgue inner product. But the transformation to spherical coordinates changes the inner product.

Thus you are working in the wrong inner product!

#### vanhees71

Gold Member
There seems to be no reason that a combination of $p^4$ and $r^2$ would also make a valid self-adjoint operator as in the trivial cases.
All these operators are self-adjoint or more precisely essentially self-adjoint. The important lesson to be learnt is

(a) Hermiticity is not sufficient for an operator to represent an observable; it must be an essentially self-adjoint operator
(b) In QT the operators describing observables with a continuous or partially continuous spectrum have a domain and co-domain which is smaller than the entire Hilbert space. For position and momentum you can use some Schwartz space of quickly falling functions, which is dense in the Hilbert space.

Here we have an example, where some eigenfunctions of the Hamiltonian belong not to the domain of the operators we are interested in. Though $\hat{p}^2$ is well defined applied to these states, the result is not in the domain. That's why another application of $\hat{p}^2$ (making together $\hat{p}^4$) leads to trouble. Again, for a good treatment check the two nice pedagogical papers

https://arxiv.org/abs/quant-ph/9907069
https://arxiv.org/abs/quant-ph/0103153

To make the usual sloppy physicists' math rigorous (implying of course taking the caveats in situations like the here discussed with $\hat{p}^4$ seriously), the most elegant way is to use the "rigged Hilbert space". This you find in the following dissertation

and in the textbook

A. Galindo, P. Pascual, Quantum Mechanics, Springer Verlag, Heidelberg (1990), 2 Vols.

#### Hans de Vries

Gold Member
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 $r=0$, which appears as a boundary term after a partial integration. In Cartesian coordinates there is simply no boundary at $r=0$, so in partial integration one does not need to worry about it.
Note that for any arbitrary real Anti-symmetric matrix $\mathcal{A}$ the following holds:
$$\langle\,\psi\,|\,\mathcal{A}\,\psi\,\rangle = 0 ~~~~~~~~~~ \langle\,\mathcal{A}\,\psi\,|\,\psi\,\rangle = 0 ~~~~~~~~~~ \langle\,\psi^*\,|\,\mathcal{A}\,\psi\,\rangle = 0 ~~~~~~~~~~ \langle\,\mathcal{A}\,\psi^*\,|\,\psi\,\rangle = 0 ~~~~~~~~~~$$
Therefor we need two independent wave functions for the Self-Adjoint test for an operator $\mathcal{L}$ like in:
$$\langle\,\psi^*\,|\,\mathcal{L}\,\varphi\,\rangle~~=~~ \langle\,\mathcal{L}\,\psi^*\,|\,\varphi\,\rangle$$
Any arbitrary real $\mathcal{L=S\!+\!A}$ with a symmetric part and an anti-symmetric part would pass the "test" below for being Self-Adjoint:
$$\langle\,\psi^*\,|\,\mathcal{L}\,\psi\,\rangle~~=~~ \langle\,\mathcal{L}\,\psi^*\,|\,\psi\,\rangle$$
Because the anti-symmetric part is eliminated during the calculation.

#### Hans de Vries

Gold Member
Here we have an example, where some eigenfunctions of the Hamiltonian belong not to the domain of the operators we are interested in.
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.

Last edited:

#### Hans de Vries

Gold Member
Only in the Lebesgue inner product. But the transformation to spherical coordinates changes the inner product.

Thus you are working in the wrong inner product!
According to Arfken & Weber (10.6) the operator $p^2$ is not self-adjoint in a Cartesian inner product but it is self-adjoint in a spherical radial inner product.

See the $r^2$ factor in post #69.

Last edited:

"Why is p^4 not Hermitian?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving