Is the product of two hermitian matrices always hermitian?

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In summary, the conversation discusses the contradiction between the fact that the product of two Hermitian matrices is Hermitian only if they commute, and the case of ##p^4## not being Hermitian for hydrogen states with ##l=0## when ##p^2## is Hermitian. The discussion also touches on the implications of this for the hydrogen radial wave functions and their differentiability. It is concluded that while ##p^2## is approximately Hermitian, it is not exact, and the error function makes the difference in the case of ##p^4##.
  • #36
fresh_42 said:
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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  • #37
fresh_42 said:
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.
 
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  • #38
A. Neumaier said:
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##?
 
  • #39
Dr.AbeNikIanEdL said:
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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  • #40
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?
 
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  • #41
Dr.AbeNikIanEdL said:
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.
 
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  • #42
Happiness said:
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.
 
  • #43
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?
 
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  • #44
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)?
 
  • #45
vanhees71 said:
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...)
 
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  • #46
vanhees71 said:
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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  • #47
Happiness said:
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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  • #48
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.
 
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  • #49
Demystifier said:
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).
 
  • #50
A. Neumaier said:
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##.
 
  • #51
Demystifier said:
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.

Demystifier said:
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).
 
  • #52
Demystifier said:
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.
 
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  • #53
Dr.AbeNikIanEdL said:
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##.
 
  • #54
Demystifier said:
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!
 
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  • #55
Demystifier said:
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##.
 
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  • #56
Demystifier said:
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?
 
  • #57
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
 
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  • #59
vanhees71 said:
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)
 
  • #60
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!
 
  • #61
Hans de Vries said:
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
How do you obtain these matrices?
 
  • #62
Hans de Vries said:
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
That's not a valid argument since you forgot that the scalar product in this case is (omitting the angular piece, i.e., concentrating on wave functions that are only ##r## dependent (particularly of course the ##\ell=0## energy eigenstates of spherical systems fulfill this constraint)
$$\langle \psi |\phi \rangle=\int_{0}^{\infty} \mathrm{d} r r^2 \psi^*(r) \phi(r).$$
Now consider
$$p^2=-\frac{1}{r^2} \partial_r (r^2 \partial_r)$$
Let's check for hermiticity (as a necessary constraint for self-adjointness) [edit: corrected in view of #63]
$$\langle \psi |p^2 \phi \rangle=-\int_0^{\infty} \mathrm{d} r \psi^*(r) \partial_r (r^2 \partial_r \phi(r)) \\
=-r^2 \psi^*(r) \partial_r \phi(r))|_{r=0}^{\infty} + \int_0^{\infty} \mathrm{d} r r^2 \partial_r \psi^*(r) \partial_r \phi(r) \\
=r^2 [\partial_r \psi^*(r) \phi(r)-\psi^*(r) \partial_r \phi(r))]|_{r=0}^{\infty} - \int_0^{\infty} \mathrm{d} r \partial_r[r^2 \partial_r \psi^*(r)] \phi(r) \\
=r^2[\psi^*(r) \partial_r \phi(r)-\psi^*(r) \partial_r \phi(r)]|_{r=0}^{\infty} + \langle p^2 \psi|\phi \rangle.$$
So ##p^2## is Hermitean indeed if the boundary terms vanish.

This holds for the ##\ell=0## wave functions (see also the quotes from Griffiths's book in the first few postings of the thread, though he forgot to take the conjugate complex of the left wave function, which is unimportant for his case though since there you can choose the eigenstates as real functions).
 
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  • #63
Isn’t there a ##\partial_r## too many in your first boundary term (although not changing the conclusion)?
 
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  • #64
Of course you are right. I edited the posting accordingly.
 
  • #65
Dr.AbeNikIanEdL said:
How do you obtain these matrices?
They are just the difference (discrete) versions of the first and second order differentiation. When you consider a function ##f(x)## then its discrete version becomes ##f(n)## where ##n## is an integer. The discrete version of the first and second order derivatives may be given by:

##\Delta(n) = f(n+1)-f(n)~~~~~~\mbox{and}~~~~~~\Delta^2(n) = f(n+1)-2f(n)+f(n-1)##

These two sequences "slide" along the diagonal of the matrices because a differentiation is in principle a convolution.

Thus ##\partial_x## is a convolution with ##\tfrac{\partial \delta(x)}{\partial x}## and ##\partial^2_x## is a convolution with ##\tfrac{\partial^2 \delta(x)}{\partial x^2}## where ##\delta(x)## is the Dirac function.

This is where we get the two sequences from:

##\begin{array}{lrrrrrrrrr}
\mbox{The discrete delta function:} & ~~0&~~0&~~0&~~0&~~1&~~0&~~0&~~0&~~0 \\
1^{st}~\mbox{difference of delta function:} & 0&0&0&1&-1&0&0&0&0 \\
2^{nd}~\mbox{difference of delta function:} & 0&0&0&1&-2&1&0&0&0 \\
\end{array}##

Note we used f(n+1)-f(n-1) for the first order difference in the matrices. Technically this avoids the extra unwanted shift of an 1/2 lattice step
 
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  • #66
ok, I see what you did but not really how it applies here. I would have assumed that, to get a matrix representation, I would have to calculate something like

## (p^2)_{mn} = \langle\Psi_m|p^2|\Psi_n\rangle##

and this would be symmetric ##(p^2)_{mn} = (p^2)_{nm}##?
 
  • #67
As for any self-adjoint operator it should read
$$(p^2)_{mn}=(p^2)_{nm}^*.$$
What's to prove is that this is not the case for the ##\ell=0## bound states of the hydrogen atom for the operator ##p^4##, which is shown in #62.

The conclusion is that the ##\ell=0## bound states of the hydrogen atom are not in the domain of the operator ##p^4##.
 
  • #68
vanhees71 said:
As for any self-adjoint operator it should read

(p2)mn=(p2)∗nm.(p2)mn=(p2)nm∗.​

(p^2)_{mn}=(p^2)_{nm}^*.

Sure, but in this specific case everything is real anyway.

vanhees71 said:
What's to prove is that this is not the case for the ℓ=0ℓ=0\ell=0 bound states of the hydrogen atom for the operator p4p4p^4, which is shown in #62.

The conclusion is that the ℓ=0ℓ=0\ell=0 bound states of the hydrogen atom are not in the domain of the operator p4p4p^4.

I know, and I think we have arrived at these conclusions already multiple times during this thread. With my latest posts I was trying to understand @Hans de Vries comments about ##p^2##, since I am not familiar with this “getting matrices from discretised versions of functions and operators”.
 
  • #69
vanhees71 said:
... the scalar product in this case is (omitting the angular piece, i.e., concentrating on wave functions that are only ##r## dependent ...

1) The trivial Self-Adjoint, Hermitian Operators.

The Laplacian ##\nabla^2## is symmetric while ##p^2## is the part that acts on the radial part of the eigenstates. We have in a trivial way the following (integral) equations:

##\langle\,\bar{g}\,|\,\nabla^2 f\,\rangle ~~=~~ \langle\,\overline{\nabla^2 g}\,|\,f\,\rangle~~=~~ \langle\,\nabla^2 \bar{g}\,|\,f\,\rangle##

because for each individual point of the fields ##f## and ##g## the following holds.

##g(\nabla^2 f)~~=~~ ((\nabla^2)^\intercal g) f ~~=~~ (\nabla^2 g) f##

The integral expressions are trivial in this case because the expressions above hold for every point so their integral over any bounded area holds as well. An operator here is classified as Self-Adjoined / Hermitian based on the integral equations. This is one step beyond qualifying them based on their (infinite-dimensional) matrix properties. This is a key difference.


2) The original question in the OP

Now if ##p^2## was real-symmetric then ##p^4## would be as well. They both would classify as (infinite dimensional) Self-Adjoined Hermitian operators in a trivial way just like ##\nabla^2##. This seems to be the expectation in the OP. He expects ##p^4## to be Hermitian operator but based on the wrong assumption that ##p^2## is an (infinite dimensional) Self-Adjoint, Hermitian matrix, but this is not the case.

##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)##

The first order derivative term is real anti-symmetric. Therefore the transpose of ##p^2## is:

##\Big(p^2\Big)^\intercal~~=~~ \hbar^2\left(\dfrac{2}{r}\dfrac{\partial}{\partial r}-\dfrac{\partial^2}{\partial r^2}\right)##

Nevertheless ##p^2## becomes a Self-Adjoint, Hermitian operator according to the integral definition, not in the trivial way as ##\nabla^2## does.

3) Non trivial Self-Adjoint, Hermitian Operators.

We consult the excelent Arfken & Weber to see under which conditions a second order differential operator ##\mathcal{L}## becomes a Self-Adjoined operator. In other words when ##\langle\,v\,|\,\mathcal{L}\,u\,\rangle = \langle\,\mathcal{L}\,v\,|\,u\,\rangle##, starting with the simpler case ##\langle\,u\,|\,\mathcal{L}\,u\,\rangle = \langle\,\mathcal{L}\,u\,|\,u\,\rangle##

AaW_ODE1.JPG
AaW_ODE2.JPG


It is Arfken & Weber (10.6) which tells us if ##\mathcal{L}## is self-adjoint.4) Operator ##p^2## as a Self-Adjoint Operator.

At first it seems that acording to (10.6) that ##p^2## is not a Self Adjoined operator.

##p^2~~=~~ -\dfrac{\hbar^2}{r^2}\dfrac{d}{dr}\left( r^2\dfrac{d}{dr} \right) ##

but if we multiply it by a factor ##r^2## then ##r^2 p^2## becomes a Self Adjoined operator. This factor ##r^2## is readily explained when we look at the calculations of Hendrik van Hees a few post back. What he does is identical to the steps in Arfken & Weber but now for the specific case of ##p^2##.

vanhees71 said:
Now consider
$$p^2=-\frac{1}{r^2} \partial_r (r^2 \partial_r)$$
Let's check for hermiticity (as a necessary constraint for self-adjointness) [edit: corrected in view of #63]
$$\langle \psi |p^2 \phi \rangle=-\int_0^{\infty} \mathrm{d} r \psi^*(r) \partial_r (r^2 \partial_r \phi(r)) \\
=-r^2 \psi^*(r) \partial_r \phi(r))|_{r=0}^{\infty} + \int_0^{\infty} \mathrm{d} r r^2 \partial_r \psi^*(r) \partial_r \phi(r) \\
=r^2 [\partial_r \psi^*(r) \phi(r)-\psi^*(r) \partial_r \phi(r))]|_{r=0}^{\infty} - \int_0^{\infty} \mathrm{d} r \partial_r[r^2 \partial_r \psi^*(r)] \phi(r) \\
=r^2[\partial_r \psi^*(r) \phi(r)-\psi^*(r) \partial_r \phi(r)]|_{r=0}^{\infty} + \langle p^2 \psi|\phi \rangle.$$
So ##p^2## is Hermitean indeed if the boundary terms vanish.

The factor ##r^2## is explained by the integration. We integrate over spheres so we need a factor of ##r^2## to account for the surface of the spheres depending on ##r##. The multiplication of ##p^2## by ##r^2## thus makes it into a self-adjoint operator according to (10.6)

There seems to be no reason that a combination of ##p^4## and ##r^2## would also make a valid self-adjoint operator as in the trivial cases.
 
Last edited:
  • #70
Dr.AbeNikIanEdL said:
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?
You are misunderstanding. The problem is not the value of the integral over a function. The problem is the value of a function itself at ##r=0##, which appears as a boundary term after a partial integration. In Cartesian coordinates there is simply no boundary at ##r=0##, so in partial integration one does not need to worry about it.
 
<h2>1. What does it mean for a matrix to be hermitian?</h2><p>A hermitian matrix is a square matrix that is equal to its own conjugate transpose. This means that the elements on the main diagonal are real numbers, and the elements below the main diagonal are the complex conjugates of the elements above the main diagonal.</p><h2>2. Can a non-square matrix be hermitian?</h2><p>No, a matrix must be square in order to be considered hermitian. This is because the conjugate transpose operation only applies to square matrices.</p><h2>3. Is the product of two hermitian matrices always hermitian?</h2><p>Yes, the product of two hermitian matrices is always hermitian. This is because the product of two matrices is equal to the conjugate transpose of the product of the conjugate transposes of the individual matrices. Since the conjugate transpose of a hermitian matrix is itself, the product of two hermitian matrices will also be hermitian.</p><h2>4. What are some real-world applications of hermitian matrices?</h2><p>Hermitian matrices are commonly used in physics, particularly in quantum mechanics, to represent observables and operators. They are also used in signal processing and control theory.</p><h2>5. Are all eigenvalues of a hermitian matrix real numbers?</h2><p>Yes, all eigenvalues of a hermitian matrix are real numbers. This is because the eigenvalues of a hermitian matrix are equal to its diagonal elements, which are always real numbers.</p>

1. What does it mean for a matrix to be hermitian?

A hermitian matrix is a square matrix that is equal to its own conjugate transpose. This means that the elements on the main diagonal are real numbers, and the elements below the main diagonal are the complex conjugates of the elements above the main diagonal.

2. Can a non-square matrix be hermitian?

No, a matrix must be square in order to be considered hermitian. This is because the conjugate transpose operation only applies to square matrices.

3. Is the product of two hermitian matrices always hermitian?

Yes, the product of two hermitian matrices is always hermitian. This is because the product of two matrices is equal to the conjugate transpose of the product of the conjugate transposes of the individual matrices. Since the conjugate transpose of a hermitian matrix is itself, the product of two hermitian matrices will also be hermitian.

4. What are some real-world applications of hermitian matrices?

Hermitian matrices are commonly used in physics, particularly in quantum mechanics, to represent observables and operators. They are also used in signal processing and control theory.

5. Are all eigenvalues of a hermitian matrix real numbers?

Yes, all eigenvalues of a hermitian matrix are real numbers. This is because the eigenvalues of a hermitian matrix are equal to its diagonal elements, which are always real numbers.

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