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 Quantum Physics Quantum Physics Help Forum Nov 2nd 2018, 09:15 AM   #11
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Ok, thanks.

Do you maybe know in an example of a particle in a box, https://www.conservapedia.com/Schrodinger_equation how do we get to this:

 The continuous constraint is only satisfied when $\displaystyle \omega a = n \pi$ where $\displaystyle n$ is an integer.
It's a statement at the end of the page.   Nov 2nd 2018, 09:59 AM   #12

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 Originally Posted by Nforce Ok, thanks. Do you maybe know in an example of a particle in a box, https://www.conservapedia.com/Schrodinger_equation how do we get to this: It's a statement at the end of the page.
A wavefunction is (almost) always a continuous function, at least when the potential energy is given by a continuous function. The result $\displaystyle \omega a = n \pi$ comes from substitution of the trig function solution at the endpoints of the box. ( $\displaystyle \psi (x) = 0$ at x = 0 and x = a.)

As to the form of the momentum operator I don't know who came up with it, but there is something called Ehrefest's theorem that gives a link to Classical mechanics. The pertinent equation here is $\displaystyle m \dfrac{d}{dt} < x > = <p>$ where $\displaystyle < A > = \dfrac{d}{dt} \int \psi ^* A \psi ~ dx$ for an operator A. The momentum operator can be inferred from this equation.

-Dan
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See the forum rules here.   Nov 2nd 2018, 10:25 AM #13 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 481 The boundary conditions are set to be $\displaystyle \psi(0) = 0$ $\displaystyle \psi(a) = 0$ Therefore, following substitution for $\displaystyle \psi$, you get $\displaystyle A=0$ and $\displaystyle B\sin \omega a = 0$ You could just set B = 0 as well, but that doesn't yield a very interesting solution. That just states that when the wave function is 0, you satisfy the potentials. Therefore, the more interesting solution is when you look at the fluctuating sine wave and compare the results with 0. It turns out that the LHS is equal to 0 when $\displaystyle \omega a = \pi n$ where n is any integer. If the above is true, B can be anything and it will still be valid. You can verify relationship by picking an integer (say, n=2) and then plot the sine wave. You'll see that no matter what you choose, the curve will drops down to x=0 at the boundary. The only exception is if you pick n=0 solution, which gives the same result as the B=0 solution. Another constraint is required to pin down what B is. In this case, it is found using normalisation (since the area under the probability density function, which is the square of the wave function, must be 1). The relationship $\displaystyle \omega a = \pi n$ is useful when characterising the energy, E, which must have certain discrete values. This is the key feature of quantum mechanics; a lot of the solutions to problems are discrete (i.e. have an integer "n" in it somewhere) topsquark and Nforce like this.  Tags equation, schroedinger Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post VonDamian General Physics 0 Jan 18th 2017 11:40 AM Torgny Quantum Physics 1 Apr 4th 2016 03:39 AM alexandros87 Thermodynamics and Fluid Mechanics 10 Nov 22nd 2015 03:56 PM pinkprincess08 Thermodynamics and Fluid Mechanics 2 Feb 19th 2014 12:15 PM petdem Thermodynamics and Fluid Mechanics 8 Feb 15th 2010 05:51 AM