[Q]eigenfunction of position operator and negative energy

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The eigenfunction of the position operator is the delta function, δ(x-x'), which is non-normalizable and equal to its complex conjugate. The integral of the square of this function over all space diverges, indicating it cannot represent a physical state. Negative energy can exist in quantum mechanics, particularly in terms of interaction potential energy, but total energy remains positive due to the dominance of rest energy. While potential energy can be negative, it does not imply that total energy can also be negative in realistic systems. The discussion highlights the importance of understanding the implications of eigenfunctions and energy in quantum mechanics.
good_phy
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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?
 
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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.
 
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.
 
good_phy said:
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')?

\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.


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?

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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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