What is the significance of the Dirac delta function in quantum mechanics?

[Loren Booda]

In your opinion, what physics of the past 100 years most closely approaches the elegance, simplicity and appeal of Einstein's equation E=mc²?

 

[desA]

How about: 0=0

 

[3trQN]

I don't know any physics from the last 100 years :s

 

[selfAdjoint]

In your opinion, what physics of the past 100 years most closely approaches the elegance, simplicity and appeal of Einstein's equation E=mc²?

$$\Delta x \Delta p = ih$$

 

[K.J.Healey]

$$\Delta x \Delta p = ih$$

I agree. This is much more revolutionary than simple relativity.

And to be technical, Einstein first publised his papers in 1905 on SR, which would make E=mc^2 out of the 100 year range given I believe.

 

[Schrodinger's Dog]

$$\Delta x \Delta p = ih$$

I am going to show my supreme ignorance of physics by asking what this means?

The change in x and p = complex number times the plank constant? Wild guesses here? I'm serious I have no idea what this relates too sorry.

For us physics numptys could you let us in on this breakthrough?

EDIT: I can't believe I didn't conjugate the word numpty properly, I apologise to Scotsmen everywhere!

 

[Loren Booda]

The imaninary number i should be left out of that Heisenberg uncertainty principle, but included in such as the Q. M. operator for momentum.

What do you all think of ther contention that relativity is more fundamental than quantum mechanics?

 

[selfAdjoint]

I am going to show my supreme ignorance of physics by asking what this means?

The change in x and p = complex number times the plank constant? Wild guesses here? I'm serious I have no idea what this relates too sorry.

For us physics numptys could you let us in on this breakthrough?

EDIT: I can't believe I didn't conjugate the word numpty properly, I apologise to Scotsmen everywhere!

Yes the imaginary i comes in in quantizing the momentum to an operator. Ignore it for the present purposes. The change( or range of change ~ standard deviation of distribution of possible values) for the position x, multiplied by the same thing for the momentum p, equals a constant. By picking your units carefully you can make this constant be the Planck constant h.

The point is that the more narrowly position is constrained to be, the broader is the constraint on momentum, and vice versa. In the limit where you know one of them exactly, as you know the momentum of the photons in a single-frequency beam of light, then the position of those photons becomes completely uncertain, you can't pick out one or the other in order; any of them could be anywhere along the beam.

 

[Rade]

..Yes the imaginary i comes in quantizing the momentum to an operator...

Question. Why not also quantize the "position" to an operator--why just "momentum" ? What mathematics results if both position and momentum are simultaneously quantized ?

 

[Kurdt]

General relativity for me would fit the bill. Like quantum mechanics the leap in imagination for the concepts of general relativity was rather large. Also being a theory which makes use of tensors the scope of the field equation is enormous and also lends to an elegance that is beyond most other theories I've studied.

 

[Loren Booda]

The displacement is also assigned an operator, usually the Kronecker delta function.

 

[Rade]

The displacement is also assigned an operator, usually the Kronecker delta function.

Thanks. So how does this knowledge modify this ? : $$\Delta x \Delta p = ih$$

 

[Schrodinger's Dog]

Yes the imaginary i comes in in quantizing the momentum to an operator. Ignore it for the present purposes. The change( or range of change ~ standard deviation of distribution of possible values) for the position x, multiplied by the same thing for the momentum p, equals a constant. By picking your units carefully you can make this constant be the Planck constant h.

The point is that the more narrowly position is constrained to be, the broader is the constraint on momentum, and vice versa. In the limit where you know one of them exactly, as you know the momentum of the photons in a single-frequency beam of light, then the position of those photons becomes completely uncertain, you can't pick out one or the other in order; any of them could be anywhere along the beam.

Thanks that's a very complete answer and easy to understand, although I'm glad I just finished studying a statistics block or I may have had more questions.

 

[K.J.Healey]

Thanks. So how does this knowledge modify this ? : $$\Delta x \Delta p = ih$$

idk, maybe more accurate to say something:

$$\Delta x \Delta p > h/(2 \pi)$$

if I did that LaTeX right...

 

[Gokul43201]

The displacement is also assigned an operator, usually the Kronecker delta function.

I'm sure I do not understand. Could you explain?

 

[Loren Booda]

Actually, the position operator in quantum mechanics is usually given by a Dirac delta function. It may be described 1.) as a function equal to zero except for one neighborhood of the domain approaching a singularity, with non-zero range, and 2.) so that the integral of its range over all its domain equals 1.

Take a sinusoidal wavefunction (technically, slowly tapering off as it approaches infinity, which allows that non-zero position probability can be measured at least finitely along its infinite extent, and normalization can be defined). Once measurement is made, the operator applies and the wavefunction "collapses," and a definite probability of 1 applies to the point of measurement.

Mathematically, the Dirac delta function may choose a discrete value of the variable (position) under consideration when integrating its probability over all space, effectively assigning certainty to the point of measurement. Likewise, position's complement, momentum, introduces an operator (the Fourier transform of the Dirac delta) which under integration yields an infinite range. One might say that a discrete frequency corresponds to a totally uncertain (infinite) wavetrain.