# Recent content by lukka

1. ### Gaussian Integrals for Quantum States of well Defined Momentum

i can see where i going wrong with this now.. thanks Dauto!
2. ### Gaussian Integrals for Quantum States of well Defined Momentum

Hey that's great! i never thought of transforming it that way.. i guess what i was attempting to do at first before was nonsensical although i'm not sure why, it seems they should have revealed the same? i'm sure i'm on the right track with this now. Thanks for going to the time to show me where...
3. ### Gaussian Integrals for Quantum States of well Defined Momentum

Equation 2.64 implies that the normalisation constants are equivalent. that is.. (2σ √∏) / √h(2∏σ^2)^1/4 = (2∏(hbar)^2 / 4σ^2)^-1/4. here's the algebra i did.. (2σ √∏) / √h(2∏σ^2)^1/4 = √∏ / (2∏/16σ^2)^1/4 .. ..=√h (2∏/16σ^2 ^∏^2)^-1/4 = (2∏(hbar)^2 / 16σ^2)^-1/4.
4. ### Gaussian Integrals for Quantum States of well Defined Momentum

jtbell i apologise for any confusion here but I'm not referring to the normalisation process, it's the constant on the exponential. The equation shows a rationalisation of the numerator to give 1 /(2∏(hbar)^2 / 4σ^2)^1/4. From what i understand if one wants to move a quantity from the...
5. ### Gaussian Integrals for Quantum States of well Defined Momentum

Hey SteamKing, thanks for stopping by to help. I'm not referring to the normalisation process here, it's the constant on the exponential i'm having trouble with.. The equation shows a rationalisation of the numerator to give (2∏(hbar)^2 / 4σ^2)^-1/4. From what i understand if one wants to...
6. ### Gaussian Integrals for Quantum States of well Defined Momentum

Consider the Gaussian Integral (eqn 2.64).. is anyone able to explain how the constant of normalization is rationalised?
7. ### Commutator Relations; Conjugate Product of a Dimensionless Operator

Thanks CompuChip
8. ### Commutator Relations; Conjugate Product of a Dimensionless Operator

Consider the following commutator for the product of the creation/annihilation operators; [A*,A] = (2m(h/2∏)ω)^1 [mωx - ip, mωx + ip] = (2m(h/2∏)ω)^1 {m^2ω^2 [x,x] + imω ([x,p] - [p,x]) + [p,p]} Since we have the identity; [x,p] = -[p,x] can one assume that.. [x,p] - [p,x] =...
9. ### On the Expansion of Exponential Function by Integration

I'm beginning to get the hang of this already.. Thanks very much for all your help guys!
10. ### Do you get annoyed about forgetting old curriculum?

There is a great deal of uncertainty when it comes to knowledge. The problem with the 'learning hysteria' arises essentially from a common misunderstanding about how cognition and the brain operates.. The truth is graduates are not really masters of the information they have assimilated, or at...
11. ### How To Find The Integral of Sin x dx?

Yes i understand what you are saying. That's a valid point you have made. So how would one go about that Micro? expand the complex exponential into the trigonometric form, separate the terms and integrate?
12. ### How To Find The Integral of Sin x dx?

I'm not sure you can for anything but the most vigorous proof, but one will often know from experience that the exponential function that which is a weighted sum of the sine and cosine, equals it's own derivative. Maybe someone else could provide an answer to this..
13. ### How To Find The Integral of Sin x dx?

since i^3 (i) = i^4 = 1 dy/dx [i^3(e^ix)] = e^ix
14. ### How To Find The Integral of Sin x dx?

$$\int e^{ix}~dx=i^3e^{ix} + c$$
15. ### On the Expansion of Exponential Function by Integration

Yes okay i understand. That's pretty cool. So the bar is inserted at the end with the underscore to indicate the limits themselves? I have bookmarked the Latex page for future reference. The sandbox is very helpful for getting started. Might take some getting used to but i'm happy to learn it...