Question

(Schrodinger's Eqn.) - sign switching and I'm not sure why.

Answer 1

Hey, folks. I'm watching a lecture where somebody is showing the normalization of the waveform and they're messing around with i/some complex conjugates in the derivation/"proof" of this, but a sign switch occurs and I don't understand why. Here's the specific math below:

Answer 2

\[i \hbar \frac{\partial \psi(x,t)}{\partial t} = \bigg( - \frac{\hbar ^2}{2m} \frac{\partial^2 \psi(x,t)}{\partial x^2}+ V(x,t)\bigg) \psi (x,t)\]
And then, assume the next pat has already been arrived at/achieved:

Answer 3

(Sorry, give me a minute. I'm getting confused.)

Answer 4

They divide through by \[i \hbar \]on the right hand side, but the sign changes for the first term from negative to positive, and the second from positive to negative. Why does this happen? @agent0smith

Answer 5

e.g.\[\frac{\partial \psi (x,t)}{\partial t} = \bigg(i\frac{\hbar}{2m}\frac{\partial^2 \psi(x,t)}{\partial x^2} -i\frac{V(x,t)}{\hbar}\bigg)\psi(x,t)\]

Answer 6

I'm assuming it has to do with the i, but I'm not certain at all.

Answer 7

@dan815

Answer 8

They multiplied both sides by i, and divided both sides by h. Then multiplied both sides by -1.

Answer 9

Just multiple both sides by -i/h, you get it

Answer 10

They explicitly said that they divided through by i, I don't really get why they said that, but this derivation makes sense. Thank you, guys.

Answer 11

You're allowed to divide through by i, right? Is that not allowed? I'm not big on complex/imaginary numbers.

Answer 12

If they divided by i, then they had to also multiply both sides by i^2. Kinda pointless.

Answer 13

Why would they have to multiply both sides through by i^2? Why wouldn't you be allowed to keep it as is?

Answer 14

Again, to be clear here, is i just not allowed to be in your denominator or something because of it equaling sqrt(-1)?

Answer 15

I just meant if they did what they said they did. I think they just misspoke.

Yeah it's just one of those things like square roots being in denominators. I don't think they ever divided by i, anyway.

Answer 16

Alright, cool. Thank you!