# [Q]eigenfunction of position operator and negative energy

1. Nov 6, 2008

### good_phy

Hi, Everybody know that eigenfunction of position operator x' is $\delta(x-x')$

But i also knew that integral of square of current state over entire space is 1(probability)

Then, $\int_{-\infty}^{\infty}\delta(x-x')\delta(x-x')^{*} dx$ is 1?

What is conjugate of $\delta(x-x')$?

And i wandered whether negative energy exists. In classical mechanics, I know potential

energy can have negative sign such as product of electron(negative charge) and voltage.

If electron is at position which potential is larger and kinetic energy of electron, electron

can have negative energy.

Is it vaild in QM?

2. Nov 6, 2008

### Fredrik

Staff Emeritus
I think the product of two delta functions is undefined. Also, a particle is never actually in a position eigenstate. A position measurement squeezes the wavefunction so that it gets a sharp peak, but it never actually becomes a delta function.

The eigenvalues of the Hamiltonian must be bounded from below, but they can be negative. Of course you can always replace H with H'=H+bI, where b is a number and I is the identity operator. With a suitable choice of b, the eigenvalues are going to be positive.

3. Nov 6, 2008

### cks

From

$$E^2=p^2c^2+m^2c^4$$ then $$E=\pm\sqrt(p^2c^2+m^2c^4)$$. Usually we just take the positive sign of E, but actually we can't neglect the minus sign, there's a physical meaning to it.

In particle physics, a minus sign means its antiparticle.

4. Nov 6, 2008

### meopemuk

$\delta(x-x')$ is a non-normalizable (sometimes called "improper") eigenfunction. This is a real function, so it is equal to its complex conjugate $\delta(x-x')^{*} = \delta(x-x')$, and the integral you wrote is divergent. If you want to work with normalized eigenfunctions of position, you should consider localized functions, like $\sqrt{\delta(x-x')}$, though they are not frequently mentioned in the literature.

Yes, interaction potential energy can be negative in both classical and quantum mechanics. However, this does not mean that the total energy can be negative too. For realistic physical systems the (negative) potential energy is much smaller than the (positive) rest energy of involved particles $$mc^2$$. The sum of all energies (rest+kinetic+potential) is always positive.