# Show that the position operator does not preserve H

• Moolisa
In summary: So I think it's not necessary to worry about the fact that the ##\psi## is not differentiable at the origin.In summary, the conversation discusses finding the integral of a function using the position operator and the definition of a stationary state. It is suggested to use a clever function that is not differentiable at the origin, but can be approximated to any accuracy and has the same asymptotic behavior.

#### Moolisa

Homework Statement
Consider a Hilbert space H that consists of all functions ψ(x) such that

##\int_{-\infty}^{\infty} |\psi(x)|^2, dx##

is finite. Show that there are functions in H for which ˆxψ(x) ≡ xψ(x) is not in H.
That is to say, there are functions in H that are taken out of H when acted upon by
the position operator (or equivalently, the position operator does not preserve H)
Relevant Equations
##\int_-\infty^\infty |ψ(x)|^2\, dx##
ˆxψ(x) ≡ xψ(x)

The attempt

##\int_{-\infty}^{\infty} |ψ^*(x)\, \hat x\,\psi(x)|\, dxˆ##

Using ˆxψ(x) ≡ xψ(x)

=##\int_{-\infty}^{\infty} |ψ^*(x)\,x\,\psi(x)|\, dxˆ##

=##\int_{-\infty}^{\infty} |ψ^*(x)\,\psi(x)\,x|\, dxˆ##

=##\int_{-\infty}^{\infty} |x\,ψ^2(x)|\, dxˆ##

I'm pretty sure this is not the correct approach. ψ(x) is a stationary right? Is it safe to attempt this using the definition of a stationary state? I apologize if I'm completely wrong, I've read the text and lecture notes but am still having a lot of trouble understanding this

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If I parsed your question correctly,then think of this function:
##\psi=1/x##

Does this meets hypothesis?

Abhishek11235 said:
If I parsed your question correctly,then think of this function:
##\psi=1/x##

Does this meets hypothesis?
I think so? If acted on by the position operator it would no longer be finite would it?

Abhishek11235
Abhishek11235 said:
If I parsed your question correctly,then think of this function:
##\psi=1/x##

Does this meets hypothesis?

That function is not defined at ##x = 0##. And, the integral of ##\psi^2## does not converge in any case.

Abhishek11235
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Moolisa
Wolfram integrator is your friend. You can use integral test to have an idea what will converge. Consider e.g. Sum ## 1/ n^s ##. Edit: My idea is to use the comparison criteria: We know ##\Sigma \frac {1}{n^s}## converges for ##s>2##. Using the integral test ( and doing something to avoid the 0 in the denominator), create an integral that converges, barely, and then multiply by ##x## to tweak it out of converging.

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Moolisa
For some problem where the potential energy is singular at the origin, a wavefunction like

##\psi (x) = \frac{1}{1+\left|x\right|}##

could be a meaningful state. And even if the ##\psi## is required to be differentiable everywhere, it's possible to approximate that function with a differentiable function to any accuracy you want while having the same asymptotic behavior in the limit ##\left|x\right|\rightarrow\infty##.

DEvens and Moolisa

## 1. What is the position operator?

The position operator, denoted as x, is a linear operator in quantum mechanics that represents the position of a particle in space.

## 2. What does it mean for the position operator to not preserve H?

In quantum mechanics, the Hamiltonian (H) represents the total energy of a system. If the position operator does not preserve H, it means that the position of a particle can change in time, resulting in a change in the energy of the system.

## 3. How can it be shown that the position operator does not preserve H?

To show that the position operator does not preserve H, we can use the Heisenberg uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we know its momentum. This implies that the position and momentum operators, x and p, do not commute, and therefore, do not preserve H.

## 4. What implications does this have in quantum mechanics?

The non-preservation of H by the position operator has significant implications in quantum mechanics. It means that the position of a particle cannot be precisely measured without affecting its energy, making it impossible to know both the position and energy of a particle at the same time. This is a fundamental principle in quantum mechanics and has led to many interesting phenomena, such as wave-particle duality.

## 5. How does this relate to the uncertainty principle?

The non-preservation of H by the position operator is directly related to the Heisenberg uncertainty principle. This principle states that there is a fundamental limit to the precision with which certain pairs of physical properties, such as position and momentum, can be known simultaneously. Therefore, the non-commutativity of the position and momentum operators is a manifestation of the uncertainty principle in quantum mechanics.