# I Why is p^4 not Hermitian?

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#### 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.

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#### fresh_42

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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

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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

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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 $l=0$.  Last edited:

#### fresh_42

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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]. Last edited:

#### 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

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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?

• Demystifier

#### 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.

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#### 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

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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.

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#### 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

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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.

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