I Do I need energy to measure position without uncertainty?

[gmastrogiovanni]

I state the following study and then expose my doubt at the end.

MY SYSTEM
A free particle (absence of forces) on one-dimensional space (X axis).

INITIAL STATE
At t=0 the wave function is a Gaussian wave packet NOT normalizet centered in $x_{0}$ and with standard deviation $\sigma$:
$$\psi_{\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}\thinspace\,e^{-\frac{(x-x_{0})^{2}}{2\sigma^{2}}}$$ In a non-strict way, I can write: $$\underset{\sigma\rightarrow0}{lim}\,\psi_{\sigma}(x)=\delta(x-x_{0})\,\,\,\,\,\,\,\,\,\,(1)$$ The genaralized function Dirac delta $\delta(x-x_{0})$ is the "improper" eigenvector of operator $\hat{X}$ with eigenvalue $x_{0}$.
The normalized wave function wave is: $$\psi_{N\sigma}(x)=\frac{1}{\sqrt{\sigma\sqrt{\pi}}}\thinspace\,e^{-\frac{(x-x_{0})^{2}}{2\sigma^{2}}}$$
CALCULATION OF ENERGY
The hamiltonian operator is: $$\hat{H}=-\,\frac{\,\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}$$ From my calculations it results: $$<E_{\sigma}>=<\psi|\hat{H}\psi>=\int_{-\infty}^{+\infty}\psi_{N\sigma}^{*}\hat{H}\,\psi_{N\sigma}\,\,dx=\frac{\hbar^{2}}{4m\sigma^{2}}$$
From which we deduce that: $$\underset{\sigma\rightarrow0}{lim}\,<E_{\sigma}>=\infty\,\,\,\,\,\,\,\,\,\,(2)$$
THEREFORE
From (1) and (2) if $\sigma\rightarrow0$: $$\psi_{\sigma}(x)\rightarrow\delta(x-x_{0})$$ $$<E_{\sigma}>\rightarrow\infty$$

1 QUESTION
Is the reasoning wrong?

2 QUESTION
Is $\delta(x-x_{0})$ a "not real" state with infinity energy?

3 QUESTION
At t=0 I measure the position x and I get the value $x_ {0}$ without error and,
immediately after the measurement, because of wave function collapse, the wave function is $\delta(x-x_{0})$.
To measure the position of a particle without error, do I have to supply energy $E\rightarrow\infty$?

 

[king vitamin]

1 QUESTION
Is the reasoning wrong?

There's nothing mathematically wrong with your conclusions (barring some technical issues about \psi_{\sigma} as a nascent delta function), but you should be careful in interpreting them. The object \psi_{\sigma} is not normalized, and similarly, neither is \delta(x - x_0), so these objects are not giving you intuition about actual quantum states.

2 QUESTION
Is \delta(x - x_0) a "not real" state with infinity energy?

It is not a normalizable state, so talking about its energy (or physical properties of it in general) does not make much sense.

3 QUESTION

First of all, as I've already stressed, \delta(x - x_0) is not a valid quantum state (it is not normalizable), so there's no sense in which it is ever the wave function of a particle. In practice, we do not measure particle positions to infinite precision, so the nonexistence of such localized states isn't too much of a problem physically.

Your computations do show an important physical property of the free particle: it takes a large amount of energy to try to localize it. This is because localized position leads to large fluctuations in momentum, and therefore kinetic energy. If your measurement does lead to a wave function which is much more localized in space than the initial wave function, the average energy has increased. This energy must have been inputted by you during your measurement.

 

[Simon Bridge]

The reasoning states up front that the particle has no forces acting on it, and then asserts that it has a stationary wavefunction at t=0.
How do you get that state without "forces"?

Are we to understand that the wavefunction is that of the position of the particle after measurement?

It looks like all you are showing is that it takes lots of energy to squeeze a particle into a small space. You start out with a HO potential, and investigate what happens to the average energy as the potential gets narrower.

The particle energy is still uncertain, and becomes more uncertain the more confined the position. You are adding energy to the system by squeezing it.
The average energy (the expectation value calculated) can be expected to become arbitrarily high for arbitrary precision in the position measurement... which is the usual was HUP is discussed. To get high precision in position you have to hit it harder... in your case, the way you are getting the position precision looks like squeezing.

Have you tried writing out the wavefunction explicitly as a superposition of energy eigenstates of the system?

The top question is: "Do I need energy to measure position without uncertainty?"
The answer is that you need energy to measure position ... period.

If you mean, "Do I need to know the energy to know the position?"
No, you don't.

However, your maths does not show this.

I think the question needs to be more carefully phrased: what is it you are trying to show... exactly?

 

[gmastrogiovanni]

Your computations do show an important physical property of the free particle: it takes a large amount of energy to try to localize it. This is because localized position leads to large fluctuations in momentum, and therefore kinetic energy. If your measurement does lead to a wave function which is much more localized in space than the initial wave function, the average energy has increased. This energy must have been inputted by you during your measurement.

OK. So if the particle is free (with low energy) to locate it in a small region of space, my measurement must necessarily provide large energy to the particle. I understood. Very clear!

Instead we now assume that the particle is already confined in a small region of space (forced by a force field) and therefore is already in a high-energy state.
In such case my measurement does not necessarily need to provide energy to the particle... (It's right?)
... but I need a large amount of energy to explore the small area, ie to explore the small area I have to shoot another particle with short wavelength ie high energy. (It's right?).

Thanks for your help!

 

[gmastrogiovanni]

The reasoning states up front that the particle has no forces acting on it, and then asserts that it has a stationary wavefunction at t=0.
How do you get that state without "forces"?

I do not make any further hypothesis.

Are we to understand that the wavefunction is that of the position of the particle after measurement?

No. The wavefunction is before measurement.

what is it you are trying to show... exactly?

I tried to understand what the physical state of Dirac delta represents, but as King Vitamin said, Dirac delta is not a physical state.
Furthermore, I am trying to understand better what happens after the wave function collapse.

Thanks for your help!

 

[king vitamin]

Instead we now assume that the particle is already confined in a small region of space (forced by a force field) and therefore is already in a high-energy state.
In such case my measurement does not necessarily need to provide energy to the particle... (It's right?)

If we assume that the total energy of the particle and the apparatus doing the measurement is conserved, then if the average energy of the particle did not change much before and after the measurement, then the average energy of the apparatus must not have changed much either.

(I am using the phrase "average energy" because that is all that can be defined for normalizable wave functions evolving under the free particle Hamiltonian.)

... but I need a large amount of energy to explore the small area, ie to explore the small area I have to shoot another particle with short wavelength ie high energy. (It's right?).

It is true that you need a high energy photon to probe a localized state, but if the measurement did not change the energy of the particle appreciably, then the energy of the rest of the system must not have changed. This energy may still be in the photon, which can scatter off of the particle without losing much energy.