schrodinger equation derivation?
407070
The short answer is that it simply isn't derived. \n\nStandard quantum mechanics is based on a series of postulates which describe something called a wavefunction. One of the postulates states that the wavefunction must obey the Schrodinger Equation.\n\nThe Schrodinger Equation is somewhat analogous to F=MA (Newton's Second Law) from classical mechanics. There really isn't any derivation of F=MA, it just seems to work. If you are doing an experiment with classical mechanics and you plug the right numbers into F=MA, you get the right answer. Likewise with the Schrodinger Equation, except it works on the quantum scale.\n\nThere have been some attempts to derive the Schrodinger Equation from a grand unified field theory, but for all but the most advanced particle physics applications the Schrodinger Equation is simply taken as a postulate. (just like F=MA)
1048691
(r*x)*(-sin(x))+(cos(x))*(r)+\n(0)*(lnrsin(x))+(r)*(cot(x)) \n\nfor the cot(x) you end up for the derivative of lnrsin(x) is (1/sin(x)) *cos(x)=cot(x)\n\nsorry it cut off on the top. hopefully this works
45463
The last answer is correct.  The equation:\n\nA*sin(bx+c)\n\nCan be graphed.  A is the amplitude of the wave (how high along the y axis).  b is the frequency (how fast do you need to complete one wavelength).  c is a wavelength shift, how far away from x=0 do you have to start the wavelength.
72824
Wave length is often represented by the Greek lower case letter lambda (your stick figure).  Replace that in your equation with "wavelength" or W until it makes sense then substitute lambda back into the equation.  It is just a shorthand notation that scientists have agreed upon (and nothing very magic).  See the link for more.  Good luck.
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
277032
I assume you are talking about any wave equation, not just a QM probability density function.\n\nIn any case, what do you mean that you have a "different x for every instant t"?  That is not the case.  I suspect you are thinking of travelling along with the crest or trough of a wave.  As t changes, the location of the crest and trough changes, but that is not the only location on the wave.\n\nWhat the wave equation specifies is the magnitude of the wave at *any* location x, at any time t.  So, if you are at a constant location x', the magnitude of the wave will vary periodically in time at that location. If you were a cork on the water, you would bob up and down at your location as the waves passed you. But similarly, if we were to somehow freeze time, and travel up and down our x axis, the magnitude of the wave would vary periodically in space as well.  So now, rather than a cork bobbing on the water, imagine you're on a jet ski that is going across the water so fast, compared to the speed of the waves, that it's almost like the waves aren't moving.  Yet you bob up and down as you cross the waves that lie in your path in space.\n\nHence, the wave varies both in x and t, independently.
287143
From Rule 1 and 2 we are ready to obtain an expression for one of the atomic properties, in this case is the velocity of the electron. Multiply both sides of Eq.(1) by r/mv, we obtain\n\nv= Z e2/4p eo mvr\n\nbut mvr in the denominator of this equation can be replaced nh/2 pi. we immediately have an explicit expression for the allowed values of velocity of the electron,\n\nv = Z e2/2peoh n ...(3)\n\nEq. (3) tells us that the velocity of the electron is directly proportional to the nuclear charge Z (in case of hydrogen Z=1) and inversely proportional to the principle quantum number n.\n\nvn = 2.188 x 106 m/s x (Z/n)\n\nFor H, n = 1.\n\n2,188,000 m/s = 4,894,000 miles per hour\n= 0.729 %c, c = speed of light in vacuum, 3 x 10^8 m/s
43146
Use Kirchoff's laws.  Read page at link then follow the link "example."\n\nhttp://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/kirchoff.html\n\n--- \nadded:\nwow! pjw, that UTex link is a cool online tutorial!
956102
Here is the equation\n\n(246,92)U ---> (4,2)He + (242,90)Th + 2Gamma
960070
Two dimensions should be explanation enough. Assume the de Broglie wavelength of an electron is some distance in nanometers. If we draw the nucleus as a dot and draw the electron as a continuous wave around the nucleus, the wave has to be the length of some multiple of its wavelength. If the wave were not the length of a discrete multiple of its wavelength it would not be continuous because a full circle's rotation would cause the wave to jump. If we constrain the atomic model to behave like this, then the answer is simple. \n\nConsider the electron wave that is four wavelengths long. This wave is a function which describes electron density. The next energy level has to correspond to a wave that is five wavelengths long. In this way, the wave cannot increase gradually, but rather the levels are split into discrete shells, one for each number of wavelengths in the wave.
505385
Here ya go!\n\nhttp://en.wikipedia.org/wiki/Bose-Einstein_condensate
835410
You need to get your units in a consistent form.  I suggest changing the velocity of the electron to 2.5E6 m/s so when you divide it into c-----------WAIT A MINUTE!-----Yor formula is NOT right, because you're dividing a speed by a speed, which will give a dimensionless answer.  Wavelength is a LENGTH, so the answer must come out that way.  I forget the formula you're refering to, but what you've typed isn't it................
836806
Are you referencing the time dependency of the Schroedinger equation?  Or are you implying quantization of time?
241834
I beleive this is the one:\n\nhttp://www.phy.ohiou.edu/~braslavs/Biophysics/Articles/Wei%20Zhang/Rosensweig2002.pdf
741723
De Broglie wavelength L = (h/mv) sqrt( 1-v^2/c^2).\n\nSquare, solve for v = hc /sqrt(mc^2 L^2 + h^2) \nH = Planck's constant\nc = the speed of light\nm = mass
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
120626
There's a derivatives tutorial on the following page:\n\nhttp://www.karlscalculus.org/calculus.html#s4