A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1360436634776:dw\[φ''=\frac{2mE}{ℏ^2} φ\] \[ \varphi= A \sin(\frac{\sqrt{2mE}}{ℏ}x)\] \[\varphi(a)=\varphi(a)=0\] \[ \pm\frac{\sqrt{2mE}}{ℏ}a=n \pi\] \[ \frac{\sqrt{2mE}}{ℏ}=\pm\frac{n \pi}{a}\] \[\lambda= \frac{2 \pi}{\pm\frac{n \pi}{a}}=\pm\frac{2a}{n}\] Which, for n=1, is shorter than the real groundstate lavelength (4a). What have I done incorrectly?

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1You haven't written down the correct general solution. You can also have terms \[ ... + B \cos(\omega x) \] with \( \omega = \sqrt{2mE}/\hbar \).

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1This will give you the 4a, as cos(omega.x) = 0 => omega.x = pi/2 + n.pi

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0Aren't you assuming A=0 for that solution? \[Asin(\pi/2 \pm n\pi) \ne 0\]

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1The general solution of y '' + w^2.y = 0 is y(x) = A.sin(wx) + B.cos(wx) or y(x) = A.cos(wx + phi0) Either way, there are two parameters. You are applying one of the boundary conditions. You actually are going to apply another boundary condition to y'(x) in order to completely nail down the solution.

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1Alternatively, the two conditions you are using phi(a) = 0, phi(a) = 0. From this, you get two families of solutions. One with sin, the other with cos. This equivalent to using the second general solution with the phi0 alternating between 0 and pi/2

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0Ah, that's it! Thanks. What are the boundary conditions? That \[\varphi'(a)=\varphi(a)=0\] Because \[\varphi(x)=0\] for \[x>a, x<a\]?

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0*\[\varphi'(a)=\varphi'(a)=0\]

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0That guess came from avoiding making the first derivative jump about at + a

JamesJ
 2 years ago
Best ResponseYou've already chosen the best response.1You have two conditions, which solving for the two parameters in the general solution. I recommend you start with A.cos(wx + phi0) and you'll get what you want.

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0\[A \sin( \omega a)+B \cos( \omega a)\] It is harder to make this expression = 0 in this form than yours, as both summands cannot be made = 0 or the negative of the other by eyeballing it, so yours is simpler. Thanks, I think I've got it.

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0@JamesJ , sorry to tag you, but I have a (very) brief question: the general solution is \[ \varphi= A e^{\pm i \omega x}\] What is done with \[\Im(\varphi)\]? Is only the real part considered, as is often done in classical mechanics? Thanks again
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.