Recent content by Ritorufon
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Boundary conditions in finite potential well
oh yea that should be a first order differential, not a second order, okay ill take a look at them again thanks a lot! :)- Ritorufon
- Post #3
- Forum: Advanced Physics Homework Help
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Boundary conditions in finite potential well
Homework Statement Hi guys, I'm having trouble understanding the finite potential well, in particular the boundary conditions The well under scrutiny has potential V(x)= 0 for |x|<a and V(x)=V_0 for >a Homework Equations \frac{d^2\psi}{dx^2}=-\sqrt{\frac{2mE}{\hbar^2}}\psi=-\alpha^2\psi...- Ritorufon
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- Boundary Boundary conditions Conditions Finite Potential Potential well
- Replies: 2
- Forum: Advanced Physics Homework Help
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Trigometric identity conversion within an integral
ah! stupidly obvious like I thought! thanks for your help and patience guys! that's been annoying me for weeks!- Ritorufon
- Post #8
- Forum: Calculus and Beyond Homework Help
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Trigometric identity conversion within an integral
I would understand if the equation was multiplied by 2 but its multipied by 1/2 as the fraction in the end equation is \frac{n\pi}{4a}, and the integral inside is multiplied by 1/2 so the identity doesn't apply right? or am i missing something stupidly obvious?- Ritorufon
- Post #5
- Forum: Calculus and Beyond Homework Help
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Trigometric identity conversion within an integral
whoops the last equation should be <p_x> = B^2 (+{i} {\hbar} ) \frac{n\pi}{4a} \int^{a}_{-a} sin \frac{n\pi}{a} x dx- Ritorufon
- Post #2
- Forum: Calculus and Beyond Homework Help
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Trigometric identity conversion within an integral
Homework Statement The problem is finding the average value of momentum in an infinite potential well but the theory I understand, its the mathematical execution I'm having trouble with. Homework Equations The expectation value for the momentum is found using the conjugate formula...- Ritorufon
- Thread
- Identity Integral
- Replies: 7
- Forum: Calculus and Beyond Homework Help