Is the product of two hermitian matrices always hermitian?

[dextercioby]

Why is $p^4$ not hermitian for hydrogen states with $l=0$ when $p^2$ is?

Doesn't this contradict the following theorem?

Coincidentally or not, the same question popped on the SE website last week. It received a good answer by user Mani Jha.

https://physics.stackexchange.com/q...zimuthal-quantum/496546#comment1119041_496546

 

[fresh_42]

For $g=\psi_{100}$ and $f=\psi_{200}$,

$\left.\left( r^2 f\dfrac{dg}{dr} - r^2g\dfrac{df}{dr} \right)\right|_0^\infty=\left.\left(\frac{1}{8\sqrt{2}\pi a^5}r^3e^{-3r/2a}\right)\right|_0^\infty=0$, where $a$ is a constant.

And if the expression in post #7 is correct, then we have $\langle \psi_{100}|p^4\psi_{200}\rangle = c + \langle p^4\psi_{100}|\psi_{200}\rangle$ with $c\neq 0$. Hence $p^4$ isn't Hermitian, but $\langle f|p^4g \rangle \neq \langle f|(p^2)^2g\rangle$, so $p^4\neq (p^2)^2$ on $\psi_{n00}$.

And as in post #31, the domain matters!

 

[Happiness]

And if the expression in post #7 is correct, then we have $\langle \psi_{100}|p^4\psi_{200}\rangle = c + \langle p^4\psi_{100}|\psi_{200}\rangle$ with $c\neq 0$. Hence $p^4$ isn't Hermitian, but $\langle f|p^4g \rangle \neq \langle f|(p^2)^2g\rangle$, so $p^4\neq (p^2)^2$ on $\psi_{n00}$.

This is weird because I get the error expression for $p^4$ in post #7 by assuming $p^4=(p^2)^2$.

Ignoring the constant factor $-4\pi\hbar$,
$\langle f|p^2(p^2g) \rangle=\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} - r^2\hat{p}^2g\dfrac{df}{dr} \right)\right|_0^\infty+\langle p^2f|p^2g \rangle=\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} - r^2\hat{p}^2g\dfrac{df}{dr} \right)\right|_0^\infty+\left.\left( r^2 \hat{p}^2f\dfrac{dg}{dr} - r^2g\dfrac{d(\hat{p}^2f)}{dr} \right)\right|_0^\infty+\langle p^2p^2f|g \rangle$

The terms that don't vanish are $\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} \right)\right|_0^\infty$ and $\left.\left(r^2g\dfrac{d(\hat{p}^2f)}{dr} \right)\right|_0^\infty$.

 

[fresh_42]

This is weird because I get the error expression for $p^4$ in post #7 by assuming $p^4=(p^2)^2$.

Ignoring the constant factor $-4\pi\hbar$,
$\langle f|p^2(p^2g) \rangle=\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} - r^2\hat{p}^2g\dfrac{df}{dr} \right)\right|_0^\infty+\langle p^2f|p^2g \rangle=\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} - r^2\hat{p}^2g\dfrac{df}{dr} \right)\right|_0^\infty+\left.\left( r^2 \hat{p}^2f\dfrac{dg}{dr} - r^2g\dfrac{d(\hat{p}^2f)}{dr} \right)\right|_0^\infty+\langle p^2p^2f|g \rangle$

The terms that don't vanish are $\left.\left( r^2 f\dfrac{d(\hat{p}^2g)}{dr} \right)\right|_0^\infty$ and $\left.\left(r^2g\dfrac{d(\hat{p}^2f)}{dr} \right)\right|_0^\infty$.

I see the first two equalities, but not your conclusion. Remember that $\hat{p}^2f$ means $\hat{p}^2(f)$. So why should the two terms in the middle be zero?

Did you work with the approximations for $\psi_{n00}$ or do you have an exact expression?

 

[Dr.AbeNikIanEdL]

If it equals zero for p2p2p^2 and does not for p4p4p^4, then p4≠(p2)2p4≠(p2)2p^4\neq (p^2)^2, simple as that.

What exactly do you mean by $(p^2)^2$? I assumed $p^4 f = p^2 p^2 f = p^2 (p^2 f)$.

Of course one would conclude from this that $p^2$ is not hermitian either (or only on some set that is not closed under its application, if that makes sense). So I am a bit confused myself.

Btw. I arrive at the same conclusion as @Happiness, just with the caveat that the notation is not really clear, the $p^2$ in the terms you complain about is only applied to the immediately following g/f.

 

[Happiness]

I see the first two equalities, but not your conclusion. Remember that $\hat{p}^2f$ means $\hat{p}^2(f)$. So why should the two terms in the middle be zero?

Did you work with the approximations for $\psi_{n00}$ or do you have an exact expression?

My conclusion is because I use $p^4=p^2p^2$ to arrive at the error expression for $p^4$, I cannot later use that expression to argue or conclude that $p^4\neq p^2p^2$.

The two terms in the middle simplifies to $r^2(2-\frac{r}{a})e^{-3r/2a}$, ignoring any constant factor.

I used $g=\frac{1}{\sqrt{\pi}}(\frac{1}{a})^{3/2}e^{-r/a}$ and $f=\frac{1}{4\sqrt{2}\pi}(\frac{1}{a})^{3/2}(2-\frac{r}{a})e^{-r/2a}$.

 

[A. Neumaier]

Well, the domain, i.e. where the operators are defined play an important role here. Only if they are all the same, then does the implication $\overline{P}^\tau = P \Longrightarrow \overline{P^2}^\tau = P^2$ hold.

$p^4$ has a smaller domain than $p^2$. Thus boundary conditions (at infinity and at poles of the interaction, i.e. zero radius) matter.

 

[Dr.AbeNikIanEdL]

p4p4p^4 has a smaller domain than p2p2p^2. Thus boundary conditions (at infinity and at poles of the interaction, i.e. zero radius) matter.

I would like to understand this bettet (sorry @Happiness for the slight highjack). So is the domain of $p^4$ smaller because it should only be defined on the set that is mapped onto itself by $p^2$? And the boundary conditions enter by restricting what functions are potentially in the domain of $p^2$ and hence of $p^4$?

 

[A. Neumaier]

I would like to understand this bettet (sorry @Happiness for the slight highjack). So is the domain of $p^4$ smaller because it should only be defined on the set that is mapped onto itself by $p^2$? And the boundary conditions enter by restricting what functions are potentially in the domain of $p^2$ and hence of $p^4$?

The domain is the set of vectors in the Hilbert space that map into the Hilbert space. This requires more (weak) differentiability for $p^4$ than for $p^2$. For $l=0$, the differentiability is not enough.

 

[Dr.AbeNikIanEdL]

Ok thanks, I think I understand.

So is it correct to say that the actual problem is that we naively apply $p^4$ to the $l=0$ states in the first place (where it is not defined as an operation on the hilbert space), and the apparent non-hermitian behavior is rather a symptom of this incorrect(?) application?

 

[A. Neumaier]

Ok thanks, I think I understand.

So is it correct to say that the actual problem is that we naively apply $p^4$ to the $l=0$ states in the first place (where it is not defined as an operation on the hilbert space), and the apparent non-hermitian behavior is rather a symptom of this incorrect(?) application?

Yes.

 

[Demystifier]

Everything involved is always differentiable.

$$\frac{1}{r^2}\frac{d}{dr}r^2\frac{df}{dr}=(\frac{r}{a}-\frac{2}{a}+\frac{16}{r})e^{-r/2a}$$

Note the term $16/r$ which makes it non-differentiable at $r=0$. Since the boundary term in partial integration involves quantities at $r=0$, this explains why the boundary term may be non-zero. That's the origin of the result that $A^2$ may be non-self-adjoint even when $A$ is self-adjoint.

 

[DrDu]

Doesn't this contradict the following theorem?

The product of two hermitian matrices, A and B, is hermitian if and only if they commute, AB=BA.

Since when is p or its powers a matrix?

 

[vanhees71]

I've not thought about it very hard, but isn't this a problem due to the short-distance singularity of the $1/r$-Coulomb potential? What happens, if one regularizes it by taking the nucleus as an extended object with a smooth charge distribution (or put more precisely take into account a form factor)?

 

[A. Neumaier]

I've not thought about it very hard, but isn't this a problem due to the short-distance singularity of the $1/r$-Coulomb potential? What happens, if one regularizes it by taking the nucleus as an extended object with a smooth charge distribution (or put more precisely take into account a form factor)?

Yes, this is the cause of the problem.

The phenomenon is nonetheless interesting as there are many singular Hamiltonians of interest in physics. It shows the importance of boundary conditions in arguments about self-adjointness. (You cold add the example to your article on sins...)

 

[Demystifier]

I've not thought about it very hard, but isn't this a problem due to the short-distance singularity of the $1/r$-Coulomb potential? What happens, if one regularizes it by taking the nucleus as an extended object with a smooth charge distribution (or put more precisely take into account a form factor)?

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.

 

[Demystifier]

Why is $p^4$ not hermitian for hydrogen states with $l=0$ when $p^2$ is?

In view of my post above, now I think I have a simple answer. The $p^4$ looks non-hermitian when it is expressed in "bad" coordinates (spherical coordinates) that do not have a good behavior at $r=0$. When $p$ is expressed in the good old Cartesian coordinates, then $p^4$ is hermitian.

 

[Demystifier]

Those familiar with differential geometry or general relativity may also find illuminating that the flat metric
$$dl^2=dx^2+dy^2+dz^2=dr^2+r^2d\theta^2+r^2 {\rm sin}^2\theta d\varphi^2$$
is also singular at $r=0$ in spherical coordinates.

 

[A. Neumaier]

In view of my post above, now I think I have a simple answer. The $p^4$ looks non-hermitian when it is expressed in "bad" coordinates (spherical coordinates) that do not have a good behavior at $r=0$. When $p$ is expressed in the good old Cartesian coordinates, then $p^4$ is hermitian.

For the free particle you refer to, $p^4$ is Hermitian in any coordinates. When transforming to spherical coordinates you need to transform the inner product as well, and Hermiticity is with respect to this transformed inner product (which is different from the inner product for the Coulomb setting).

 

[Demystifier]

For the free particle you refer to, $p^4$ is Hermitian in any coordinates. When transforming to spherical coordinates you need to transform the inner product as well, and Hermiticity is with respect to this transformed inner product (which is different from the inner product for the Coulomb setting).

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 $r\rightarrow 0$.

 

[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]

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]

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]

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]

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

 

[vanhees71]

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

https://aapt.scitation.org/doi/abs/10.1119/1.1328351https://arxiv.org/pdf/quant-ph/0103153
and another one

https://arxiv.org/ct?url=https://dx.doi.org/10.1088/0034-4885/63/12/201&v=daefa909https://arxiv.org/abs/quant-ph/9907069

 

[Hans de Vries]

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.

(Thanks for the other links)

 

[vanhees71]

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!