A Non-selfadjointness and solutions

[SemM]

Hi, I have the two operators:

\begin{equation}
Q = i\hbar \frac{d}{dx} - \gamma
\end{equation}\begin{equation}
Q' = -i\hbar \frac{d}{dx} - \gamma
\end{equation}

where $\gamma$ is a constant. Both of these are not self-adjoint, as they do not follow the condition:

\begin{equation}
\langle {Q}x, y \rangle = \langle x, {Q}^{*}y \rangle,
\end{equation}

thus being non-Hermitian.

This gives , for the system $QQ' = 0$ a non-selfadjoint solution, which cannot be normalized, or used to derive expectation values.

What can I do?

Thanks!

 

[DrDu]

Both Q and Q' are hermitian. Whether they are self adjoint or not depends on the domain of definition, but normally, they can be chosen to be self-adjoint.

 

[SemM]

For Hermiticity one has to follow the given condition:

\begin{equation}
\langle \phi \vert Q \vert \psi \rangle = \langle Q \phi \vert \psi \rangle
\end{equation}doing this on $Q$, with $\phi = x$ and $\psi = x^2$ we have:

\begin{equation}
\langle \phi \vert Q \vert \psi \rangle = \int x (i\hbar \frac{d}{dx} - \gamma)x^2 dx = \int i\hbar 2x^2-\gamma x^3 dx
\end{equation}

\begin{equation}
\langle Q \phi \vert \psi \rangle = \int (i\hbar \frac{d}{dx} x - \gamma x) x^2 dx = \int i\hbar x^2 - \gamma x^3 dx
\end{equation}So:

\begin{equation}
\langle \phi \vert Q \vert \psi \rangle \ne \langle Q \phi \vert \psi \rangle
\end{equation}unless, I have done something wrong with prioritizing x^2 over x in the first integral, given that d/dx is to the left of x^2.The same accounts for $Q'$, only with the minus sign . So I cannot see how these two operators are self-adjoint. Indeed, the solution to $QQ'= 0$ gives a non-square-integrable function, which is discontinuous in $\mathscr{H}$ and gives therefore non-trivial expectation values. That should further confirm that these operators $Q\ and\ Q'$ are not hermitian.

 

[DrDu]

Neither x nor $x^2$ are functions in the hilbert space, hence the integrals do not exist. So your statement that the two integrals are different is pointless.
You have to use partial integration on functions which at least vanish at $\pm \infty$ to show the equivalence.

 

[SemM]

Neither x nor $x^2$ are functions in the hilbert space, hence the integrals do not exist. So your statement that the two integrals are different is pointless.
You have to use partial integration on functions which at least vanish at $\pm \infty$ to show the equivalence.

Thanks Dr Du! I can't imagine a reason for why such a critical point has never been given in relation to these points in the textbooks, as an example. I see indeed that they are not infinetly differentiable.

Would $e^{-x}$ and $e^{-3x}$ be sufficient?

Thanks!

 

[DrDu]

Diverge at - infinity, don't they?

 

[SemM]

Diverge at - infinity, don't they?

Sorry, I can't see that. I see that they approach zero in an interval [a, b]. Although, I could always look up in a book, an example would make this discussion come to a completion also for other readers. Do you have an example?

Thanks!

 

[DrDu]

The discussion is most easy for functions defined on the interval $[ -\infty, \infty]$. Then you might consider for example $\exp(x^2), x*\exp(x^2)$ or the like. On the interval $[0, \infty]$, the momentum operator (and hence your Q) are in deed not self-adjoint and cannot be made so. On the finite interval $[a,b]$, the operators are self adjoint when they are defined, e.g., on the periodic functions f(a)=f(b).

 

[SemM]

So, in QM, working in intervals, we can consider them as self-adjoint, only and if, we use an interval. However, if it was an infinite interval , they are not self adjoint. Is this your conclusion basically?

Thanks!

 

[DrDu]

No
I urge you to read this article:
https://arxiv.org/abs/quant-ph/0103153

 

[SemM]

Printing...

thanks Dr Du!

 

[DrDu]

So, in QM, working in intervals, we can consider them as self-adjoint, only and if, we use an interval. However, if it was an infinite interval , they are not self adjoint. Is this your conclusion basically?

Thanks!

First, I want to note that this discussion is specific to the momentum operator. Other operators require a discussion on it's own.
I forgot to say that the momentum operator is or can be made self-adjoint for functions defined on the range $[-\infinity, \infinity]$, which is the most important case.

 

[SemM]

Thanks, indeed the main three operators are:

position, x,

Momentum:
\begin{equation}
i\hbar \frac{d}{dx}
\end{equation}

Kinetic energy:
\begin{equation}
-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}
\end{equation}

The operators Q and Q' give the solution (in $QQ'\Psi=0$):

\begin{equation}
\psi(x)= \frac{1} {4} e^{-2i\gamma x/\hbar}+\frac{3}{4}+\frac{i \gamma x }{2 \hbar}
\end{equation}with I.C. $\psi(0)=1$ and $\psi'(0)=0$This solution gives non-trivial values for the kinetic operator, infinite non-existing value for the momentum operators, and for the position operator, a very large value.

I am basically stuck with this state, and can't really do anything with it to give sound physical answers, or I can try to alter the initial conditions, but that does not change this state much, and the problematic terms remain.

 

[SemM]

First, I want to note that this discussion is specific to the momentum operator. Other operators require a discussion on it's own.
I forgot to say that the momentum operator is or can be made self-adjoint for functions defined on the range ##[-\infinity, \infinity], which is the most important case.

Dr Du, should this formality criterion for adjointness:

\begin{equation}
\langle \mathcal{T}\phi, \psi \rangle = \langle \phi, \mathcal{T}^{*}\psi \rangle,
\end{equation}be ONLY applied to admissible functions in Hilbert space in order to tell if T is adjoint of not?

Thanks!

 

[DrDu]

When do you call a function admissible?

 

[SemM]

When do you call a function admissible?

When it is :

1. Square integrable
2. Continuous everywhere
3. Single-valued
4. Infinitely differentiable

 

[DrDu]

This is not sufficient. Confer to the section 3 of the article by Bonnet for a counterexample and to the beginning of chapter 4 for a general definition of a self-adjoint operator.

 

[SemM]

This is not sufficient. Confer to the section 3 of the article by Bonnet for a counterexample and to the beginning of chapter 4 for a general definition of a self-adjoint operator.

Thanks. In QM, these are known as the four main properties of admissible of a wavefunction. I will look at this chapter you mention, and get back to you!

Thanks!

 

[DrDu]

In QM, these are known as the fundamental prescriptions of a wavefunction.

Who says so? Even the eigenfunctions of the hydrogen atom are not even differentiable at the origin. Even discontinuous wavefunctions appear in physical applications.

 

[SemM]

Who says so? Even the eigenfunctions of the hydrogen atom are not even differentiable at the origin. Even discontinuous wavefunctions appear in physical applications.

It is given in the book Molecular Quantum Mechanics, Friedman Atkins. But let me scan that page and make sure I am not saying anything wrong. http://www.fjordforsk.no/scan_MWM.pdf.

In any case, your point, which is intriguing, makes my question even more difficult to answer (and to pose). I will go through your article suggestion tomorrow.Thanks

 

[DrDu]

Atkins, ok...

 

[SemM]

Atkins, ok...

Dr Du,on p 5 in the arxiv article you posted, the authors seem to cancel out the momentum operator by some non-shown operation:

\begin{equation}
-i\hbar \int_0^L dx\frac{d}{dx}[\overline {\psi(x)} \phi(x)] = -i\hbar [\overline {\psi(L)} \phi(L) - \hat {\psi(0)} \phi(0)
\end{equation}

Is that correct? Where does the d/dx term go?

Thanks

(why are some putting the integral dx in front of the expression, and not as correctly done, at the end (far right)?)

 

[DrDu]

C'mon:
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

 

[SemM]

I am not sure here this is so c'mon. I would have written first as:

\begin{equation}
-i\hbar \int_0^L \frac{d}{dx}[\overline {\psi(x)} \phi(x)] dx
\end{equation}

the use the product rule:

\begin{equation}
-i\hbar \int_0^L u'v dx
\end{equation}

where $u'$ is $d/dx \psi$ and v is $\phi$