How to Use the Heisenberg Uncertainty Principle in Relation to Position and Momentum
How Physicists Solved the Photoelectric Effect of Light
How de Broglie Showed that All Particles Display Wave-Like Properties

How to Find the Heisenberg Uncertainty Relation from Scratch

If you’ve read through the last few sections, you’re now armed with all this new technology: Hermitian operators and commutators. How can you put it to work? You can come up with the Heisenberg uncertainty relation starting virtually from scratch.

Here’s a calculation that takes you from a few basic definitions to the Heisenberg uncertainty relation. This kind of calculation shows how much easier it is to use the basis-less bra and ket notation than the full matrix version of state vectors. This isn’t the kind of calculation that you’ll need to do in class, but follow it through — knowing how to use kets, bras, commutators, and Hermitian operators is vital in the coming chapters.

The uncertainty in a measurement of the Hermitian operator named A is formally given by

image0.png

That is,

image1.png

is equal to the square root of the expectation value of A2 minus the squared expectation value of A. If you’ve taken any math classes that dealt with statistics, this formula may be familiar to you. Similarly, the uncertainty in a measurement using Hermitian operator B is

Now consider the operators

image2.png

(not the uncertainties

image3.png

anymore), and assume that applying

image4.png

as operators gives you measurement values like this:

image5.png

Like any operator, using

image6.png

can result in new kets:

image7.png

Here’s the key: The Schwarz inequality gives you

image8.png

So you can see that the inequality sign,

image9.png

which plays a big part in the Heisenberg uncertainty relation, has already crept into the calculation.

image10.png

(the definition of a Hermitian operator), you can see that

image11.png

This means that

image12.png

So you can rewrite the Schwarz inequality like this:

image13.png

Okay, where has this gotten you? It’s time to be clever. Note that you can write

image14.png

is the anticommutator of the operators

image15.png

(the constants <A> and <B> subtract out), you can rewrite this equation:

image16.png

Here’s where the math gets intense. Take a look at what you know so far:

The commutator of two Hermitian operators, [A, B], is anti-Hermitian.
The expectation value of an anti-Hermitian is purely imaginary.
The
image17.png
is Hermitian.
The expectation value of a Hermitian is real.

All this means that you can view the expectation value of the equation as the sum of real

image18.png

And because the second term on the right is positive or zero, you can say that the following is true:

image19.png

Whew! But now compare this inequality to the relationship from the earlier use of the Schwarz inequality:

image20.png

Combining the two equations gives you this:

image21.png

This has the look of the Heisenberg uncertainty relation, except for the pesky expectation value brackets, < >, and the fact that

image22.png

appear squared here. You want to reproduce the Heisenberg uncertainty relation here, which looks like this:

image23.png

Okay, so how do you get the left side of the equation from

image24.png

Because an earlier equation tells you that

image25.png

you know the following:

image26.png

Taking the expectation value of the last term in this equation, you get this result:

image27.png

Square the earlier equation

image28.png

to get the following:

image29.png

And comparing that equation to the before it, you conclude that

image30.png

Cool. That result means that

image31.png

This inequality at last means that

image32.png

Well, well, well. So the product of two uncertainties is greater than or equal to one-half the absolute value of the commutator of their respective operators? Wow. Is that the Heisenberg uncertainty relation? Well, take a look. In quantum mechanics, the momentum operator looks like this:

image33.png

And the operator for the momentum in the x direction is

image34.png

So what’s the commutator of the X operator (which just returns the x position of a particle) and

image35.png

you get this next inequality(remember,

image36.png

here are the uncertainties in x and

image37.png

not the operators):

image38.png

Hot dog! That is the Heisenberg uncertainty relation. (Notice that by deriving it from scratch, however, you haven’t actually constrained the physical world through the use of abstract mathematics — you’ve merely proved, using a few basic assumptions, that you can’t measure the physical world with perfect accuracy.)

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