Question

How do you solve the Heisenberg Uncertainty Principle?

Answer 1

A formal definition of the Heisenberg Uncertainty Principle states that the product of the position and momentum uncertainties `sigma_vecx` and `sigma_(vecp_x)` is no less than `ℏ//2`, where `h` is Planck's constant, `6.626xx10^(-34) "J"cdot"s"`, and `ℏ = h//2pi` is the reduced Planck's constant.

That is,

`\mathbf(sigma_(vecx)sigma_(vecp_x) >= ℏ//2)`

(By the way, this happens to hold for any direction, not just the `x` direction, so this extends to our three dimensions.)

What this really says is that the position and the momentum cannot be simultaneously observed at the same level of uncertainty, where an observation implies a specific uncertainty.

Thus, if you know the momentum to a specific uncertainty, you cannot also know the position to that specific uncertainty.

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EXAMPLE

My textbook (Physical Chemistry: A Molecular Approach, McQuarrie & Simon, pg. 23) provides a different version of the Heisenberg Uncertainty Principle for the simpler examples, but rest assured, it's the same person's principle:

`bb(DeltavecxDeltavecp_x >= h)`

where `h` is Planck's constant as before.

[This is adapted from Example 1-11.]

Suppose we have an uncertainty `Deltavecx = "50 pm"` in the position of an electron. Then, what we do first is simply change the `>=` symbol to an equal sign and solve a regular equation.

`color(blue)(Deltavecp_x)`

`= h/(Deltavecx)`

`= (6.626xx10^(-34) "J"cdot"s")/(50xx10^(-12) "m")`

`= (6.626xx10^(-34) "kg"cdot"m"^2"/s")/(50xx10^(-12) "m")`

`= color(blue)(1.3xx10^(-23) "kg"cdot"m/s") > h`

CHALLENGE: If `vecp = mvecv`, and `m = 9.11xx10^(-31) "kg"` for an electron, what is `Deltavecv`, the uncertainty in the speed of the electron?

You should get something on the order of 10 million `"m/s"`! It means if you know the position to such a low uncertainty, you have a high uncertainty in the speed.

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That was admittedly really simple mathematically, so if that's not what you meant by "solve"... well then I've derived the Heisenberg Uncertainty Principle I listed at the top of the answer.

DERIVATION

An uncertainty `sigma` for some observable `vecv` pertaining to an electron, let's say, is calculated from the average squared quantity, `<< vecv^2 >>`, and the average quantity squared, `<< vecv >>^2` as:

`\mathbf(sigma_(vecv) = sqrt(<< vecv^2 >> - << vecv >>^2))`

where for some set of boundary conditions `[0,a]` for the electron, we have:

• `<< vecv >> = int_(0)^(a) f^"*"(vecv)vecvf(vecv)dvecv`
• `<< vecv^2 >> = int_(0)^(a) f^"*"(vecv)vecv^2f(vecv)dvecv`

where `f(vecv)` is some function of `vecv`, and `f^"*"(vecv)` is its complex conjugate, which in the real numbers is equal to `f(vecv)`.

Sadly this is too complicated for the scope of this question, so I'll simply supply you with the averages ahead of time (Physical Chemistry, A Molecular Approach, pp. 86-89).

For a situation where we have a confined space of width `a`, height `oo`, and depth `0`:

• `<< vecx >> = a/2`
• `<< vecx^2 >> = a^3/3 - a^2/(2n^2pi^2)`
• `<< vecp >> = 0` (the momentum is equally likely to be either in the positive or negative direction)
• `<< vecp^2 >> = (n^2h^2)/(4a^2)`
• `n` is the quantum level, and is an integer.

Then, what we have is:

`color(blue)(sigma_(vecx))`

`= (<< vecx^2 >> - << vecx >>^2)^"1/2"`

`= (a^3/3 - a^2/(2n^2pi^2) - a^2/4)^"1/2"`

`= color(blue)(a/(2pin)((pi^2n^2)/3 - 2)^"1/2")`

Next, we have:

`color(blue)(sigma_(vecp_x))`

`= (<< vecp^2 >> - cancel(<< vecp >>^2)^(0))^"1/2"`

`= ((n^2h^2)/(4a^2))^"1/2"`

`= color(blue)((nh)/(2a))`

Lastly, multiply the two to get:

`sigma_(vecx)sigma_(vecp_x)`

`= cancel(a)/(2picancel(n))((pi^2n^2)/3 - 2)^"1/2"*(cancel(n)h)/(2cancel(cancel(a))`

`= h/(4pi)((pi^2n^2)/3 - 2)^"1/2"`

Now let's isolate `((pi^2n^2)/3 - 2)^"1/2"`. Note that

`(pi^2n^2)/3 - 2 > 0,`

and at worst, `n = 1`, which gives a result of about `1.28`. Thus, `((pi^2n^2)/3 - 2)^"1/2" > 1`.

This means:

`bb(sigma_(vecx)sigma_(vecp_x) >= ℏ//2)`

Just like how it was stated at the very beginning.