Question

Please let me know about Heisenberg uncertainty principle. I am very unclear about its equation? Thank you so much.

Answer 1

There are two formulations, but one is more commonly used.

`DeltaxDeltap_x >= ℏ` `bblarr`This is more commonly evaluated

`sigma_xsigma_(p_x) >= ℏ"/"2`

where `Delta` is the range of the observable, and `sigma` is the standard deviation of the observable.

In general, we can simply say that the minimum product of the associated uncertainties is on the order of Planck's constant.

This means that the uncertainties are significant for quantum particles, but not for regular-sized things like baseballs or human beings.

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The first equation illustrates how when someone sends focused light through a slit and narrows the slit (thereby decreasing `Deltax`), the light that comes out further splits (thereby increasing `Deltav_x` and thus `Deltap_x`).

Just try lowering `Deltax`. Eventually, you will get to the point where `DeltaxDeltap_x` would be `< ℏ`, violating the `>=` sign. So, `Deltap_x` must increase.

What this says is that the more you know about the `x` position of the quantum particle, the less you know about its momentum in the `x` direction (or similarly for the analogous relations in the `y` or `z` directions).

For once, I will refer the reader to a video!

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The second equation is more often used in higher-level chemistry, like Physical Chemistry, and the standard deviations are defined as the square root of the variance:

`sigma_a = sqrt(sigma_a^2)`

`= sqrt(<< a^2 >> - << a >>^2)`

and the averages in the square root are:

`<< a^2 >> = int_(-oo)^(oo) a^2p(x)dx`
`<< a >>^2 = [int_(-oo)^(oo) ap(x)dx]^2`

with `p(x)` as the probability as a function of `x`.

But since standard deviation can be taken as the uncertainty around the average, it's just another perspective to the same general description of the Heisenberg Uncertainty Principle:

The minimum product of the associated uncertainties is on the order of Planck's constant.