# Heisenberg uncertainty principle

**Statement**: According to Heisenberg uncertainty principle, it is impossible to measure the exact position and momentum of a particle simultaneously within the wave packet.

We know, group velocity of the wave packet is given by

v_{g} =∆ω/∆k

Where ω is the angular frequency and k is the propagation constant or wave number

But v_{g} is equal to the particle velocity v

Thus v_{g} = v = ∆ω/∆k (1)

But ω=2пf

Where f is the frequency

Therefore ∆ ω = 2п ∆ f (2)

Also k=2 п/λ

Since de-Broglie wavelength λ=h/p

By putting this value in equation of k, we get

k=2пp/ λ

Therefore ∆k=2п∆p / λ (3)

Put equations (2) and (3) in equation (1), we get

v= 2пh∆f/2п∆p =h∆f / (4)

Let the particle covers distance ∆x in time ∆t, then particle velocity is given by

_{ } v = ∆x/∆t (5)

Compare equations (4) and (5), we get

∆x/∆t=h∆f/∆p

Or ∆x.∆p=h∆f ∆t (6)

The frequency ∆f is related to ∆t by relation

∆t≥ 1/∆f (7)

Hence equations (6) becomes

∆x.∆p≥ h

A more sophisticated derivation of Heisenberg’s uncertainty principle gives

∆x.∆p=h/2п (8)

Which is the expression of the **Heisenberg uncertainty principle**.

As the particle is moving along x-axis. Therefore, the momentum in equation (8) of Heisenberg’s uncertainty principle should be the component of the momentum in the x-direction, thus equation Heisenberg’s uncertainty principle can be written as,

∆x.∆p_{x}=h/2п (9)

**Note**: There can not be any uncertainty if momentum is along y direction.

Q: Why there is uncertainty in position and momentum?

Answer: Because the particle is always in disturbed state during motion. It is not possible to calculate the position and momentum of particle simultaneously.

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