Lizwi 4 Posted July 1 Please show me how to move from equation 1.30 to 1.31 I tried using integration by parts 1 Share this post Link to post Share on other sites

joigus 188 Posted July 1 (edited) Consider: \[\int dx\left(-\frac{\partial\psi^{*}}{\partial x}\psi\right)=\int dx\psi^{*}\frac{\partial\psi}{\partial x}\] and what I told you in the other post about fields vanishing fast enough at infinity. You get twice the first integral in 1.30. Edited July 1 by joigus edit formula 1 Share this post Link to post Share on other sites

Lizwi 4 Posted July 1 Ohhhhh! I can see 2 is cancelled as coefficient of m You equated the first integral to its complex conjugate?? 0 Share this post Link to post Share on other sites

joigus 188 Posted July 1 (edited) 47 minutes ago, Lizwi said: Ohhhhh! I can see 2 is cancelled as coefficient of m You equated the first integral to its complex conjugate?? No, no. Careful. That's not the point. The point is that the integrals, \[\int dx-\left(\frac{\partial\psi^{*}}{\partial x}\psi\right)\] and, \[\int dx\psi^{*}\frac{\partial\psi}{\partial x}\] differ in what is called "a surface term" or "a boundary term". Because in quantum mechanics the boundary is at infinity, they can be identified for all intents and purposes. If you equate one of these integrals to its complex conjugate, what you're saying is that the integral is real. That's not quite so correct. The integrals are equal except terms that vanish at infinity. The point is a bit subtle, but that's the way to read its meaning. Edit: In this case, the surface term is, \[\left.\left(\psi^{*}\psi\right)\right|_{\textrm{infinity}}\] Edited July 1 by joigus addition 1 Share this post Link to post Share on other sites

Lizwi 4 Posted July 1 Mmmh okay. I see. Now for all computations in QM, shall I assume this? 0 Share this post Link to post Share on other sites

joigus 188 Posted July 1 6 minutes ago, Lizwi said: Mmmh okay. I see. Now for all computations in QM, shall I assume this? Exactly. Under the integral sign, yes. Actually, it's used as a matter of course in all of field theory. Field variables at infinity always go to zero "fast enough", so you can shift the derivative from one factor to the other factor (under the integral sign) by just changing a sign. Sorry. I made a mistake before. The surface term should not be the derivative, but the term that is derived. I've corrected the formula. This is what I wrote: \[\left.\frac{d}{dx}\left(\psi^{*}\psi\right)\right|_{\textrm{infinity}}\] This is what is should be (already corrected in the original post): \[\left.\psi^{*}\psi\right|_{\textrm{infinity}}\] 0 Share this post Link to post Share on other sites

Lizwi 4 Posted July 1 Thanks for your time now lastly, can you show me a small example where you shift the derivative from one factor to another changing the sign? or it’s just as in your first post? 0 Share this post Link to post Share on other sites

joigus 188 Posted July 1 (edited) I think I can do a little bit more than that. Most, if not all, interesting wave functions in QM have a behaviour that goes to zero as a Gaussian at infinity. If you take a look at most eigenfunctions of "realistic"^{*} Hamiltonians, for example, the harmonic oscillator, hydrogen atom, etc. The all are dominated by exponential damping at infinity. Example: \[\psi\left(x,0\right)=\frac{e^{-x^{2}/2-if\left(x\right)}}{x^{n}}\] Now it's very easy to see that no matter what power of x is integrated against the exponential, the idea works. \[\int_{\mathbb{R}}dx\frac{e^{-x^{2}/2+if\left(x\right)}}{x^{n}}\frac{d}{dx}\left[\frac{e^{-x^{2}/2-if\left(x\right)}}{x^{n}}\right]=\left.\frac{e^{-x^{2}}}{x^{2n}}\right|_{-\infty}^{+\infty}-\int_{\mathbb{R}}dx\frac{d}{dx}\left[\frac{e^{-x^{2}/2+if\left(x\right)}}{x^{n}}\right]\frac{e^{-x^{2}/2-if\left(x\right)}}{x^{n}}\] Watch out for silly mistakes. * Meaning nothing pathological, like Airy functions, or something like that. Edited July 1 by joigus 1 Share this post Link to post Share on other sites

joigus 188 Posted July 1 9 minutes ago, Lizwi said: Okay thanks a lot Glad to be helpful. 0 Share this post Link to post Share on other sites

Mordred 1370 Posted July 1 +1 to both of you. It's nice to see this caliber of thread on this forum. @Lizwi have you looked at Dirac notation yet ? You will find this notation helpful to understand https://www.google.com/url?sa=t&source=web&rct=j&url=https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_04.pdf&ved=2ahUKEwj16sf4sazqAhVTvZ4KHc8WCzsQFjAaegQIAxAB&usg=AOvVaw28oEJ_F36vW2fsOmfOJg2B This will help you in your studies. 0 Share this post Link to post Share on other sites

Lizwi 4 Posted July 1 Thanks I will have a look on it, I am going through Griffiths QM textbook. I am at the beginning of it. 0 Share this post Link to post Share on other sites

Mordred 1370 Posted July 1 (edited) I have a copy of his second edition. It's a decent textbook. Griffith has a section on Dirac notation. Edited July 1 by Mordred 1 Share this post Link to post Share on other sites

joigus 188 Posted July 1 2 hours ago, Lizwi said: I am going through Griffiths QM textbook. Ah. It did ring a bell. 1 hour ago, Mordred said: I have a copy of his second edition. It's a decent textbook. Griffith has a section on Dirac notation. +1. I agree. It's a bit outdated maybe, but good stuff. 0 Share this post Link to post Share on other sites

Lizwi 4 Posted July 2 I am looking for good statistical physics and classical mechanics notes ,graduate level for self study. 0 Share this post Link to post Share on other sites

studiot 2018 Posted July 2 17 minutes ago, Lizwi said: I am looking for good statistical physics and classical mechanics notes ,graduate level for self study. You should be able to get standard student note/handbooks very cheaply second hand. Look for Classical Mechanics B P Cowan Classical Mechanics J W Leech Statistical Physics F Mandl Statistical Thermodynamics Andrew Maczek All good intro notes for university 1 Share this post Link to post Share on other sites