I Is the product of two hermitian matrices always hermitian?

[Happiness]

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

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.

 

[fresh_42]

Where have you read this? If $p^2$ is Hermitian, so is $p^4$. I suspect that $p$ is a different one in the two cases.

 

[Happiness]

Where have you read this? If $p^2$ is Hermitian, so is $p^4$. I suspect that $p$ is a different one in the two cases.

Intro to QM, David Griffiths, p269

 

[fresh_42]

This is not very much context you give us. I read it as:
"We use Hermiticity of $p^2$, although $p^4$ is not Hermitian ...",
i.e. he knowingly accepts an error in the specific $l=0$ case.

 

[Happiness]

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.

 

[fresh_42]

Yes, this is more or less an obvious fact. You could as well simply write:
$$
\left(\overline{P}^\tau \right)^4=\left( \overline{P}^\tau \right)^2 \cdot\left( \overline{P}^\tau \right)^2 = \left( P^2 \right)^2 =P^4
$$
$P^4$ is Hermitian if $P^2$ is. The Hermiticity of $P^2$ has some implications which Griffith explains or wants you to solve. This does not include the $l=0$ case, since then $P^4$ is not Hermitian and neither can $P^2$ be in these specific states.

 

[Happiness]

Yes, this is more or less an obvious fact. You could as well simply write:
$$
\left(\overline{P}^\tau \right)^4=\left( \overline{P}^\tau \right)^2 \cdot\left( \overline{P}^\tau \right)^2 = \left( P^2 \right)^2 =P^4
$$
$P^4$ is Hermitian if $P^2$ is. The Hermiticity of $P^2$ has some implications which Griffith explains or wants you to solve. This does not include the $l=0$ case, since then $P^4$ is not Hermitian and neither can $P^2$ be in these specific states.

He is saying $p^2$ is hermitian but $p^4$ is not for the same pair of states, f and g, both with $l=0$.

 

[fresh_42]

Mathematically we have: $P^2$ Hermitian $\Longrightarrow \; P^4$ Hermitian. So either he made a mistake, or $P^2$ isn't Hermitian in $l=0$ states, or $P^4 \neq (P^2)^2$. I don't know what $P^4$ means, as $P^2$ doesn't look like a square either. What would be $P$ to get $P^2= -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right)$?

 

[Happiness]

Mathematically we have: $P^2$ Hermitian $\Longrightarrow \; P^4$ Hermitian. So either he made a mistake, or $P^2$ isn't Hermitian in $l=0$ states, or $P^4 \neq (P^2)^2$. I don't know what $P^4$ means, as $P^2$ doesn't look like a square either. What would be $P$ to get $P^2= -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right)$?

$p$ is the momentum operator as seen from [6.52].
$p^4=(p^2)^2$ and $p^2$ is hermitian as seen from [6.51].

 

[Dr.AbeNikIanEdL]

I actually find it weird to talk about “an operator is hermitian for these states”. I assumes it means $\langle x|p y\rangle = \langle p x| y\rangle$ for these states, but is that a standard nomenclature? In this case I think

Mathematically we have: P2P2P^2 Hermitian ⟹P4⟹P4\Longrightarrow \; P^4 Hermitian.

is not obvious, since it applies to hermitian operators in the sense of the relation above holding for all states.

 

[fresh_42]

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.

In other words: Are the kets still differentiable such that $P^4$ can be applied?

 

[Happiness]

is not obvious, since it applies to hermitian operators in the sense of the relation above holding for all states.

$p^2$ is hermitian does indeed hold for all states, including $l=0$, but not $p^4$.

 

[Happiness]

Are the kets still differentiable such that $P^4$ can be applied?

Yes they are since hydrogen radial wave functions all have the $e^{-r}$ term.

 

[Dr.AbeNikIanEdL]

Hm, it is anyway only proven for $\psi_n$ scaling like $\exp(-r)$, my point was that $p^2 \psi_n$ is not scaling like that anymore, so in $\langle x,p^2 p^2 y\rangle$ you can not even make use of this hermiticity for $p^2$. Can you show your work for the boundary terms for $p^4$? I do agree with Griffith they are not vanishing.

 

[Happiness]

Hm, it is anyway only proven for $\psi_n$ scaling like $\exp(-r)$, my point was that $p^2 \psi_n$ is not scaling like that anymore

It is, because no matter how many times you differentiate $e^{-r}$, multiplying and dividing by $r^2$, in whatever order of these 3 operations, the result is still proportional to $e^{-r}$.

 

[Dr.AbeNikIanEdL]

It will always stay proportional to $e^{-r}$, but you can have other terms depending on $r$ multiply that. Just do compute all terms in $p^2 \psi_n$ explicitly (you don’t really need all in the $r\to 0$ limit, but you must make sure to get the most important one).

 

[Happiness]

It will always stay proportional to $e^{-r}$, but you can have other terms depending on $r$ multiply that. Just do compute all terms in $p^2 \psi_n$ explicitly (you don’t really need all in the $r\to 0$ limit, but you must make sure to get the most important one).

Yes the boundary term does not vanish, indeed. But how can this be? All the hydrogen radial wave functions are infinitely differentiable. Doesn't it contradict the theorem?

 

[Dr.AbeNikIanEdL]

Where did anything become not differentiable?

 

[fresh_42]

Post #7 explains it. We do not have exact Hermiticity, only approximately:
$$
\langle f|p^2g \rangle = \underbrace{-c(r,n)}_{\approx 0} + \langle p^2f|g \rangle
$$
whereas
$$
\langle \psi_n|p^4\psi_m \rangle = \underbrace{d(n,m)}_{\not\approx 0} + \langle p^4\psi_n|\psi_m \rangle
$$
So the error function makes the difference. Neither is Hermitian, but $p^2$ is approximately Hermitian.

 

[Dr.AbeNikIanEdL]

I don’t understand this comment. What is your $c(r)$?

Edit: If it is supposed to be the boundary term in #7: That is actually exactly 0. You have to evaluate the term at the boundaries, e.g. at 0 and infinity (or in the respective limits). There is no approximation involved.

 

[Happiness]

Where did anything become not differentiable?

Everything involved is always differentiable.

Taking $f=\psi_{200}=(2-\frac{r}{a})e^{-r/2a}$ and ignoring all constant factors,

$$r^2\frac{df}{dr}=(4r^2-\frac{r^3}{a})e^{-r/2a}$$

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

 

[Happiness]

but $p^2$ is approximately ate $r=0$.

c(r) has the factor $r^2$, so it is exactly 0 at r=0.

 

[fresh_42]

I don’t understand this comment. What is your $c(r)$?

Yes, I made a mistake and corrected the error terms, resp. the dependent parameters $r,n,m$.
I'm not sure how to interpret $r,n,m$, i.e. the integral since it isn't mentioned and I'm only a mathematician, who stumbled upon a seemingly contradiction in a standard textbook. However, with exact statements, this contradiction isn't one any longer. One error tends to or is zero, the other does not.

 

[Happiness]

One error tends to zero, the other does not.

All operators for observables must be hermitian. If $\hat{p}^4$ is not hermitian, then what would you obtain when you measure $p^4$ or $E^2$? Would you get complex-valued measurements? What would it mean?

 

[Dr.AbeNikIanEdL]

One error tends to or is zero, the other does not.

There is no “error tending to zero”. There are boundary terms that either are or are not exactly 0.

 

[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]

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.

 

[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]

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.

 

[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 $r\rightarrow 0$.