# How did Heisenberg derive his famous principle?

1. ### Michio Cuckoo

84
Where

How did he come up with that?

2. ### genericusrnme

615
If I'm not mistaken Heisenberg himself didn't 'derive' the principle, he thought experimented it and got $\Delta x \ \Delta p \approx \hbar$
It was later on that other people found general formulas for uncertanties related with any operators;
$\sigma _A \sigma _B \ge \frac{[A,B]}{2i}$
And since the commutator (the [A,B] part) for position and momentum is simply $i\ \hbar$ we get
$\sigma_x \sigma_p \ge \frac{i\ \hbar}{2 i} = \frac{\hbar}{2}$

3. ### bapowell

1,833
The mathematical inequality actually follows from Fourier analysis. It's application as a physical result -- to be interpreted as physical uncertainties of measurable quantities -- is generally taken to be an axiom of the theory.

4. ### ThomasT

Check out "The Physical Principles of the Quantum Theory". First published around 1935 I think.

### Staff: Mentor

This means that a similar principle applies in other areas where we use Fourier analysis. For example, an electrical signal pulse contains a range of frequencies (or wavelengths) that is inversely proportional to the time duration or spatial width of the pulse. Likewise for pulses of electromagnetic radiation (e.g. light).

This has been known since probably sometime in the 1800s, so when physicists started to think of describing particles using waves, and superpositions of them, it was only a matter of time before someone applied this general uncertainty principle to QM.

6. ### ThomasT

This is an enlightening observation, imho. The Heisenberg uncertainty relations weren't just plucked from nothing. They have historical, mathematical basis. Heisenberg, like Einstein, Bohr, de Broglie, Jordan, Dirac, et al., was a genius of sorts, but not a magician. Everything to do with the development of the quantum theory has some basis in prior mathematics and classical conceptualization. Or so I like to believe.

I don't really understand the basis for Heisenberg's matrix mechanics, and I don't really understand the basis for Schroedinger's wave equatiion. But those are topics for other threads.

7. ### bapowell

1,833
Well, everyone "knew" that QM was about matter waves. Heisenberg was fiddling around with algebra of Hermitian operators that acted on physical states, but what was missing was a wave equation that actually described the evolution of these states. The impetus to develop a wave equation was largely driven by de Broglie's result and the experimental indications of the wave nature of matter.

8. ### ThomasT

Thanks bapowell. Every bit of input from certain perspectives helps. But I fear that I might be too old (and too lazy) to ever really understand this stuff.

### Staff: Mentor

Schrödinger's invention of his famous equation (1925-26) actually preceded the Davisson-Germer experiment (1927) which as far as I know was the first direct verification of de Broglie's hypothesis (1924) that electrons have wavelike properties.

Schrödinger came up with his equation by making an analogy between mechanics and optics, which has two distinct "pictures"; geometrical or ray optics on one hand, and wave optics on the other. He reasoned that classical mechanics of particles might be analogous to ray optics, and invented a new mechanics which is analogous to wave optics:

The early development of modern quantum mechanics is based on guesses like this. Schrödinger, Heisenberg and others were groping around, trying to come up with something that works. Only gradually did these bits and pieces come together into what you see in today's textbooks.

10. ### bapowell

1,833
Huh. I need to refresh my history. I recall reading that Schroedinger and Co. had some inklings that matter had wave-like properties, and that's what directed him to "look" for a wave equation. Were there any other experiments being done at the time that might have provided hints of wave-particle duality?

11. ### emailanmol

297
Hey, a quick question.

Isn't the relation dependent on the method we are using to find the variables (like x and v of electron?
Like on the microscope?

Like for eg if I had superman eyes (i hope this assumption is not an infraction.I am trying to make a point :-) )then I would be able to see the electron move,see exactly where it is and also calculate its exact velocity.No problem would occur.

Last edited: Mar 19, 2012
12. ### genericusrnme

615
Nope, it is completely independant of any measuring aparatus although this is commonly used as a way to explain it for the case of the x-p uncertainty before we learn that they are a fourier transform away from eachother.
No matter what your measuring device you will never be able to tell its position and momentum exactly.

Take a dirac delta function (or a wave with one really big spike somewhere), you can tell where it is but you cannot tell it's frequency since it's just one spike. Similarly, take a uniform sine wave, you can tell it's frequency but you cannot tell where it is.
This is more the nature of the x-p uncertainty principle.

If, in the position basis we have a delta function (this just means we know its position exactly) we cannot gain any useful information about its momentum (we cannot tell it's frequency). If we have a delta function in the momentum basis (we know it's momentum exactly) we have a specific frequency (just like the uniform sine wave) and we cannot gain any information about its position.
We can get interesting little 'wave packets' where we can have some information about the position and some about the momentum, the gaussian wavepacket gives us the minimum uncertainty $\sigma_x \sigma_p = \frac{\hbar}{2}$

Also, going back to your superhuman eyes example, for these super human eyes to be able to give precice information about where the electron is, you need to bombard it with short and shorter wavelength photons which have higher and higher energies. If you bombard it with long wavelengthed photons, you won't disturb it much but you can essentially only know if the electron is in an area of size proportional to the wavelength of the photon.

13. ### emailanmol

297

Oh.Now i get it.
The way they usually talk about in books, led me to think it was an apparatus dependent error.
Thanks a lot for clearing that up :-)

Last edited: Mar 20, 2012
14. ### genericusrnme

615
No problem buddy