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##p^2## is hermitian then. The contradiction still persists.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.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.There is no “error tending to zero”. There are boundary terms that either are or are not exactly 0.
Does this mean that a measurement of ##p^2## is only approximately real valued? What does that even mean?##p^2## is approximately Hermitian.
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}?##Does this mean that a measurement of ##p^2## is only approximately real valued? What does that even mean?
For ##g=\psi_{100}## and ##f=\psi_{200}##,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}?##
Coincidentally or not, the same question popped on the SE website last week. It received a good answer by user Mani Jha.Why is ##p^4## not hermitian for hydrogen states with ##l=0## when ##p^2## is?
Doesn't this contradict the following theorem?
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}##.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.
This is weird because I get the error expression for ##p^4## in post #7 by assuming ##p^4=(p^2)^2##.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}##.
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?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##.
What exactly do you mean by ##(p^2)^2##? I assumed ##p^4 f = p^2 p^2 f = p^2 (p^2 f)##.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.
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##.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?
##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.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.
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.
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.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##?
Yes.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?
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.Everything involved is always differentiable.
$$\frac{1}{r^2}\frac{d}{dr}r^2\frac{df}{dr}=(\frac{r}{a}-\frac{2}{a}+\frac{16}{r})e^{-r/2a}$$
Since when is p or its powers a matrix?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.
Yes, this is the cause of the problem.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.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)?
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.Why is ##p^4## not hermitian for hydrogen states with ##l=0## when ##p^2## is?
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).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.
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 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).