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View Full Version : The probability as an absolut value of the square of the amplitude


TubbaBlubba
Aug17-10, 09:58 PM
All right, I don't have a problme with the concept, just a specific question.

Is the absolute value of the amplitude abs(r^2 + (xi)^2) or abs(r^2) + abs((xi)^2)

Or, to put it in a simpler way - Do you absolute the value of the square of the imaginary part?

The difference would be, say

2^2 + (5i)^2 = 4 + (-25) = (-21)
abs(-21) = 21

and

2^2 + (5i)^2 = 4 + (-25)
abs(4) + abs(-25) = 29

The latter seems more physically sound to me, but the former seems more mathemathically sound. Can anyone clear this up for me?

betel
Aug18-10, 06:50 AM
The imaginary part is the part standing next to the "i", so in your example the 5 itself. Then you get the absolute value of the complex number by
|z|^2 = \Re(z)^2+\Im(z)^2
Or by using the complex conjugate
|z|^2 = z \cdot \bar z

For the first part I suspect you write your complex number in polar coordinates
z = r \exp{i \xi}
In this case the absolute value would be just |z|=r

TubbaBlubba
Aug18-10, 07:04 AM
Ah, I think I see, the absolute value is the distance from the origin? Thanks for clearing it up.

betel
Aug18-10, 07:41 AM
Ah, I think I see, the absolute value is the distance from the origin? Thanks for clearing it up.

Yes. If you draw the complex number in the complex plane you can recognize the above formula as an application of Pythagoras' theorem.