Question

what is the complex conjugate of

Answer 1

usuaally

a-i a+i/a+i right?

Answer 2

\[\psi(x)=(2\pi a^2)^{-1/4}e^{ik_ox}e^{-x^2/4a^2}\]

Answer 3

yeah by problem is this seems to all be imaginary. so is the whole thing just negative?

Answer 4

\[\psi(x)=(2\pi a^2)^{-1/4}e^{ik_ox}e^{-x^2/4a^2}\]
\[\psi(x)^*=(2\pi a^2)^{-1/4}e^{-ik_ox}e^{-x^2/4a^2}\]

Answer 5

How I remember this is that the complex conjugate is exactly same except that you switch all the signs for the i

Answer 6

\[z=a+ib\]\[z^*=a-ib\]

Answer 7

\[z=a-ib\]
\[z*=a+ib\]
@UnkleRhaukus ?

Answer 8

same thing, \(b=-b\)

Answer 9

A complex number consists of two parts 1) Real 2) Imaginary Part..

Complex number is generally shown by: \(x + iy\),

The term that does not contain \(i\) is called Real Part which is \(x\) here.

The term that contains \(i\) is called Imaginary Part which is \(y\) here...

Conjugate of a Complex number is the number in which we only reverse the sign of Imaginary Part that is for getting complex conjugate of \(x+iy\) we have to reverse the sign of \(y\) ie Imaginary Part..

\(Conjugate = x - iy \)

Answer 10

which do you use to denote complex conjugate ?\[{\Large\bar z\text { or }z^*}\]

Answer 11

First one..

Answer 12

\[\large \bar{z}\]

Answer 13

in physics we use \(z^*\)

Answer 14

That is really nice...

Answer 15

Both are correct...

Answer 16

\[z^*z=|z|^2=\]

Answer 17

I can now easily see it using eulers formula
\[\psi(x)=(2\pi a^2)^{-1/4}[\cos(k_ox)-i \sin (k_ox)]e^{-x^2/4a^2}\]
this makes it easy to see the real and complex parts. I was struggling before because there was no addition sign anywhere so it was hard to see how to get the complex conjugate. Either way now \[\psi(x)\psi^*(x)\]becomes super easy to integrate since all that trig stuff cancels out. I sure feel stupid now haha

Answer 18

i mean realy and imaginary part...?

Answer 19

yeah,