@ShuklaSannidhya Yeah.... I've noticed that in some expressions... ;-)
@ShuklaSannidhya Isn't that a fraction? o_O
Even in my mechanics, they're using the same expression... They just multiply on both sides -_-
Well, $dv$ and $dt$ aren't just numbers to be multiplied on both sides... I think that's more like an approximation, may work for small order derivatives like this one :)
I installed Windows 8 in my laptop. And, my browser is Firefox 22 (as usual). FF22 didn't have any issues with XP or Windows 7. But in Windows 8, it doesn't respond when some page is loading.
Say, I've opened a link. Now, I click anywhere on the window (it may be menu, another tab, minimize butt...
@hwlau Nope... I don't think Physics even exists in our College...
A phrase you can find in my blog...
> I don’t give much importance to my academics, because I’ve already begun to hate “Engineering” which I believe is due to our educational system, currently (for more than a decade) affected by a pedagogical virus. While my fellow students who’ve gotten used to this scheme, can rote everything mercilessly, I can’t even digest a single formula (due to my low IQ or something)…
The Madras Institute of Technology (MIT) is an engineering institute located in Chromepet, Chennai (Madras), India.It is one of the four autonomous constituent colleges of Anna University It was established in 1949 by C. Rajam as the first self-financing engineering institute in the country, and later it merged with Anna University. The institute was at that time an experiment in technical education, for it introduced to India new areas of specialization: aeronautical engineering automobile engineering, electronics engineering and instrumentation technology.
The MIT is the first i...
@ShuklaSannidhya It's not, it's the operator d/dt of v. However, for first order total derivatives, you can split it. It's an abuse of notation, but it works and it's not too bad to use it
Well, $dv$ and $dt$ aren't just numbers to be multiplied on both sides... I think that's more like an approximation, may work for small order derivatives like this one :)
We seem to be missing an off-topic close reason. I think. Unless I'm missing something.
When I select off-topic, I can choose from:
homework with insufficient research
engineering
fringe rather than mainstream physics
belongs on math.SE or meta.physics
But there are potentially plenty of off...
This is an offshoot of ChatJax, which enables MathJax along with mhchem on chat.
Copy the text below:
javascript:(function(){if(window.MathJax===undefined){var%20script=document.createElement("script");script.type="text/javascript";script.src="https://d3eoax9i5htok0.cloudfront.net/mathjax/lat...
@CrazyBuddy Tried using that but it was broken. Replacing the "https://d3eoax9i5htok0.cloudfront.net/mathjax/latest/MathJax.js" url with "http://cdn.mathjax.org/mathjax/latest/MathJax.js" seems to fix it. Cool script!
@shukla if y(x) is a diffeomorphism, yes. Basically, y(x) must be continuous and have a continuous inverse. Also y'(x) must not be zero in the domain of interest
When one has to normalise $\psi(x)$, you occasionally encounter the integral $\int_{-\infty}^{\infty} \cos(x) dx$. However, this integral's value is not defined (as it takes different values according to how you go about calculating the area). What is done here?
This is time-independent Schroedinger equation, by the way, although I doubt the answer wouldn't generalise.
I'll give a concrete example.
Let there be a constant potential $V(x)=V<E$. Then
I think I see why now. If space is homogeneous, then the probability of a free particle being at a point is independent of the point chosen. Thus if you integrate the probability over all space and expect to yield a finite value, $|\psi(x)|^2$ must be $0$, as for all other functions the integral is infinite.
Strictly speaking the space of legal states in quantum mechanics (the "Hilbert space") only includes wavefunctions which are normalisable. So the plane waves aren't actually valid quantum states, but it is so convenient to use them that many people pretend they are. This can be justified mathematically by going to a "rigged Hilbert space", but really in the end you always have to make wavepackets.
Does that mean that there is a probability of $0$ of finding a free particle at any specified point in space, or does the whole quantum-mechanical probability thing not apply for free particles?
@MichaelBrown So is what is strictly spoken (i.e. the statement that only functions in Hilbert space are valid wavefunctions) incorrect? Sorry for being pedantic but as a novice I can't differentiate rigorous statements from 'oh well, that sort-of works'.
@ManishEarth Normalizability is usually required. Like I said you can relax it by going to a rigged Hilbert space, but the only fundamental reason for doing so is convenience. Unless you are dealing with literally infinite spaces :)
@Alyosha That quantum states belong to a Hilbert space is an axiom of quantum mechanics. Unfortunately quantum mechanics is usually introduced historically so you may not see the axioms for a while...
The reason you can get away with using plane waves is because quantum mechanics is linear. So you can describe any normalizable state by a superposition of plane waves and work out everything one plane wave at a time, so to speak.
Another trick that is commonly used is to put your system in a big imaginary box. Now space is finite and the plane waves are normalizable. If the size of the box is much bigger than your system then the edge effects don't matter to any observables. You then take the limit of volume -> infinity at the end of the day.
@MichaelBrown I assume this is basically the same as in my crude example of specifying exactly how the integral $\int_{-\infty}^{\infty} \cos(x) dx$ is calculated (e.g. $\lim_{n \rightarrow \infty}2n\int_{0}^{2 \pi} \cos(x) dx$), with the way you expand your box equivalent to the form of the limit of the integral?
Well, I'm just trying to get a mathematical way of determining how nice two notes sound together. I know the music theory behind it (overlapping harmonics), but I want a way to plug in two waveforms and get a quantitative estimate of how they sound. Probably a futile goal, but a fun thing to play with in y spare time.
So one parameter I was looking at was the energy overlap.
@michael :) I posted some graphs earlier. The energy integral has some issues (which I sorta expected but I looked into it anyway) but I'm going to see if I can mathematically measure beats from arbitrary superpositions.
I've had the singular bad fortune to naïvely try (and fail) to learn GR on my own through MTW, and try to learn QM through Dirac. Yeah, that went well :P
I think so.
Sakurai isn't bad, but I forgot where it starts. Also there are two Sakurais. Be careful about which one you want.
(the Advanced one is about the Dirac/Klein Gordon eqns and whatnot)
For what it's worth I would recommend reading as many books you can. :) Feynman is pretty good for physical examples and intuition. Dirac is good too, though he takes a very abstract algebraic approach which might be hard to digest (depends on your taste). My uni used Messiah & Griffiths which are fairly low level but explain things with more examples. Landau&Lifshitz is good but heavy... even heavier than Dirac. Never had the pleasure of reading Sakurai, though I hear good things about it...
@ManishEarth He taught it that way though didn't he? The legend goes that at the end of a lecture a student raised something and said "I don't understand this." Dirac sat there silent for a minute until the student said "Well?" And Dirac said "That wasn't a question."...
I won't be able to read much variety over the next few months, though, as I'm applying for university and I need to have read thoroughly through a few books rather than learn different bits from different books.
@Alyosha you're asking pretty good questions for someone applying to undergrad. They only start teaching QM at that sort of level at 2nd year undergrad around here.
To bring you guys back on-topic I've pulled out a (funny) question right out of my NCERT textbook
> A drunkard walking on a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires 1s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13m away from the start.
@ShuklaSannidhya infinity, the drunkard will take 2 steps, stop to urinate on a door and pass out on the floor in his own vomit, never reaching the pit