I Why is p^4 not Hermitian?

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Happiness

There is no “error tending to zero”. There are boundary terms that either are or are not exactly 0.
$p^2$ is hermitian then. The contradiction still persists.

fresh_42

Mentor
2018 Award
There is no “error tending to zero”. There are boundary terms that either are or are not exactly 0.
If it equals zero for $p^2$ and does not for $p^4$, then $p^4\neq (p^2)^2$, simple as that.

• Klystron and jim mcnamara

Happiness

$p^2$ is approximately Hermitian.
Does this mean that a measurement of $p^2$ is only approximately real valued? What does that even mean?

fresh_42

Mentor
2018 Award
Does this mean that a measurement of $p^2$ is only approximately real valued? What does that even mean?
The question is: What is the value of $\left.\left( r^2 f\dfrac{dg}{dr} - r^2g\dfrac{df}{dr} \right)\right|_0^\infty$ for $f=\psi_{n00}$ and $g=\psi_{m00}?$

Happiness

The question is: What is the value of $\left.\left( r^2 f\dfrac{dg}{dr} - r^2g\dfrac{df}{dr} \right)\right|_0^\infty$ for $f=\psi_{n00}$ and $g=\psi_{m00}?$
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.

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dextercioby

Homework Helper
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.

• • Heikki Tuuri, aaroman, vanhees71 and 2 others

fresh_42

Mentor
2018 Award
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

Mentor
2018 Award
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 $f=\frac{1}{4\sqrt{2}\pi}(\frac{1}{a})^{3/2}(2-\frac{r}{a})e^{-r/2a}$.

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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.

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• vanhees71

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?

• dextercioby

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.

• Dr.AbeNikIanEdL

Demystifier

2018 Award
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?

• dextercioby

vanhees71

Gold Member
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....)

• vanhees71

Demystifier

2018 Award
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.

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• vanhees71

Demystifier

2018 Award
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.

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Demystifier

2018 Award
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.

• Klystron

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

2018 Award
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 $r\rightarrow 0$.

"Why is p^4 not Hermitian?"

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