# Uncertainty Principle & Ground State of Harmonic Oscillator

• alejandrito29
In summary, the ground state of a harmonic oscillator has:- Delta p = Delta x = 1/2- Delta t = Delta E = 1/2- Delta E = hbar/2
alejandrito29
For the ground state of harmonic oscillator i have

$$\Delta p \Delta x =\frac{\hbar}{2}$$

why if i do

$$\frac{1}{F} \Delta p \Delta x \cdot F =\frac{\hbar}{2}$$

$$\Delta t \Delta E = \frac{\hbar}{2}$$

$$\Delta E = \frac{\hbar}{2} \frac{1}{\Delta t}$$

$$\Delta E = \frac{\hbar}{2} f$$

but my answer should be $$\Delta E = \frac{\hbar}{2} \omega$$ ?

The delta-terms are measurements of uncertainty.
i.e. ##\Delta E## represents the precision to which energy is measured.

Simon Bridge said:
The delta-terms are measurements of uncertainty.
i.e. ##\Delta E## represents the precision to which energy is measured.

ok, but ##<\Delta E^2>= <E^2>-<E>^2= (<p^2>/(2m) + 0.5 m \omega^2 <x^2> )^2- ( <p>^2/(2m)+ 0.5 m \omega^2 <x>^2 ) ^2##

##=E_0^2-0^2=E_0 ^2 \to \Delta E = E_0##

then, my aswer should be ##\Delta E= \hbar \omega /2##, and not ##\Delta E = \hbar f/2##

Last edited:

This is a very significant physical result because it tells us that the energy of a system described by a harmonic oscillator potential cannot have zero energy. Physical systems such as atoms in a solid lattice or in polyatomic molecules in a gas cannot have zero energy even at absolute zero temperature. The energy of the ground vibrational state is often referred to as "zero point vibration". The zero point energy is sufficient to prevent liquid helium-4 from freezing at atmospheric pressure, no matter how low the temperature.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc4.html

##\Delta E## is ##\hbar \omega /2## or ##\hbar f /2## ?

The mean deviation of energy in an energy eigenstate is obviously zero.
The manipulations you made in the original post are not correct, the deltas are not differences or infinitesimals

alejandrito29 said:
ok, but ##<\Delta E^2>= <E^2>-<E>^2= (<p^2>/(2m) + 0.5 m \omega^2 <x^2> )^2- ( <p>^2/(2m)+ 0.5 m \omega^2 <x>^2 ) ^2##
I think you are still very confused.

##\langle (\Delta E)^2\rangle## reads: "expectation of the variance of E", and you put it equal to the variance, which is an odd thing to say ... but I think you meant to write:

##(\Delta E)^2 = \langle E^2\rangle-\langle E\rangle^2## would be the definition of variance.

For an energy eigenstate ##\langle H\rangle = E_n##
You seem to want to work on the ground state:
##\cdots =E_0^2-0^2=E_0 ^2 \to \Delta E = E_0##

then, my aswer should be ##\Delta E= \hbar \omega /2##, and not ##\Delta E = \hbar f/2##
Who says that ##\Delta E = \hbar f/2## is correct?
Where are you getting these ideas from?

Checking:
For the ground state:

##(\Delta E)^2 = \langle E^2\rangle-\langle E\rangle^2 = \langle E^2\rangle-\langle E\rangle^2 = E_0^2-E_0^2=0\implies \Delta E = 0##

... which means that the energy eigenstates of the harmonic oscillator are infinitely thin
... the energy of the lowest eigenstate is exactly ##E_0## and so on.

(I think you missed out a term in your subtraction.)

Since ##\Delta t \propto 1/\Delta E##, this means that the lifetime of the state is infinitely uncertain. i.e. the state lasts forever.

It may be easier of you write the relations out as: $$\sigma_x\sigma_p\geq \frac{\hbar}{2},\; \sigma_E\sigma_t\geq \frac{\hbar}{2}$$

Or it may be that you are referring to something else - I can't really tell.
Please tell us where these ideas are coming from.

Meantime, the statistics of the harmonic oscillator are detailed in:

## What is the Uncertainty Principle?

The Uncertainty Principle, also known as Heisenberg's Uncertainty Principle, states that it is impossible to simultaneously know the exact position and momentum of a subatomic particle. This principle is a fundamental concept in quantum mechanics and is based on the idea that the act of measuring one of these properties will inevitably disturb the other.

## How does the Uncertainty Principle relate to the Ground State of the Harmonic Oscillator?

The Ground State of the Harmonic Oscillator is the lowest possible energy state of a quantum mechanical system. The Uncertainty Principle is applicable to the Ground State because the more precisely the position of the particle is known, the less certain we can be about its momentum and vice versa. This means that the Ground State of the Harmonic Oscillator cannot have a definite position and momentum at the same time.

## What is the significance of the Ground State of the Harmonic Oscillator?

The Ground State of the Harmonic Oscillator is significant because it represents the most stable state of a quantum mechanical system. This means that any perturbations or disturbances to the system will push it back towards the Ground State. Additionally, the energy of the Ground State of the Harmonic Oscillator is the lowest possible energy for that system, making it a reference point for measuring the energy of other states.

## How does the Uncertainty Principle impact our understanding of the physical world?

The Uncertainty Principle challenges our traditional understanding of the physical world by showing that there are inherent limitations to our ability to measure and predict the behavior of subatomic particles. It highlights the probabilistic nature of quantum mechanics and the idea that there are fundamental limits to our knowledge about the universe.

## Can the Uncertainty Principle be violated?

No, the Uncertainty Principle cannot be violated. It is a fundamental principle of quantum mechanics that has been extensively tested and confirmed through experiments. Any apparent violations of the Uncertainty Principle can be explained by the limitations of our measurement tools and techniques.

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