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Infinite potential well 
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#1
Feb113, 05:22 PM

P: 192

In one dimensional problem of infinite square potential well wave function is ##\phi_n(x)=\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}## and energy is ##E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}##. Questions: What condition implies that motion is one dimensional. Did wave function describes motion of particle? When we say particle did we mean electron? What kind of particle mass are good enough to be treated in this model? And also why we don't draw ##\phi_n(x)##.



#2
Feb213, 09:42 AM

P: 5,632

one distance variable x, implies one dimension.
be sure to see the dynamic illustration near the beginning of the article here....http://en.wikipedia.org/wiki/Infinite_square_well 


#3
Feb213, 10:30 AM

P: 104




#4
Feb213, 08:48 PM

P: 192

Infinite potential well
I didn't get the answer that I'm looking. My questions:
Did wave function describes motion of particle? What kind of particle mass are good enough to be treated in this model? And And also why we don't draw ##\phi_{n}(x)##? 


#5
Feb313, 12:14 AM

P: 229




#6
Feb313, 10:27 AM

P: 5,632

Did you read the links I posted.....they will help a lot. " Interpretations of quantum mechanics address questions such as what the relation is between the wavefunction, the underlying reality, and the results of experimental measurements." There are different interpretations about exactly what the wavefunctions means...there are different ways to think about it. 


#7
Feb313, 03:45 PM

P: 351

Usually in one dimensional problem you can constrain a particle between two infinitely large parallel plates, that makes the particle only subject to the boundary conditions in one dimension.
Also, I don't think that particles are described by equations of motion in QM. In QM we describe particles in Hilbert space, in classical mechanics we use phase space which describe the equations of motion of a particle. Basically any particles follows QM, but when n is large, by Correspondence Principle, classical mechanics must be a good approximation of QM 


#8
Feb513, 02:59 PM

P: 192

One more question. By solving Sroedinger eq we get
##\varphi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}## ##E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}## But for example solution of Sroedinger eq is also ##C_1sin\frac {\pi x}{a}+C_2\sin\frac{2\pi x}{a}## in which state is particle? What is the energy level? How could I know in which state is particle? 


#9
Feb513, 03:10 PM

Mentor
P: 11,779

Before the measurement happens, the answer to the question, "which of the two energy states is the particle really in?" depends on which interpretation of QM you subscribe to. 


#10
Feb513, 03:27 PM

P: 192

Well from
[tex]\int^{a}_0 (C_1\sin \frac {\pi x}{a}+C_2 \sin \frac{2\pi x}{a})^2=1[/tex] I get ##C_1^2+C_2^2=\frac{2}{a}## what next? 


#11
Feb513, 08:42 PM

Mentor
P: 11,779

It's better to start out by normalizing each individual function:
$$\psi(x) = C_1 \psi_1(x) + C_2 \psi_2(x)\\ \psi(x) = C_1 \sqrt{\frac{2}{a}} \sin {\left(\frac{\pi x}{a}\right)} + C_2 \sqrt{\frac{2}{a}} \sin {\left(\frac{2 \pi x}{a}\right)}$$ Then when you normalize ##\psi## as a whole by doing an integral like yours, you end up with $$C_1^2 + C_2^2 = 1$$ We interpret ##C_1^2## as the probability that the particle will be measured to have energy E_{1}, and ##C_2^2## as the probability that the particle will be measured to have energy E_{2}. The two probabilities add to 1 because they are the only two possibilities for this particle as far as energy is concerned. 


#12
Feb613, 01:25 AM

P: 192

Ok but from ##C_1^2+C_2^2=1## and some other physical behavior could I say something about ##C_1^2## and ##C_2^2##?



#13
Feb613, 09:35 PM

Mentor
P: 11,779

Obviously you need additional information about your specific situation in order to find C_{1} and C_{2}. It's impossible to say more unless you have a specific situation in mind.



#14
Feb713, 02:01 AM

P: 192

I understand that. But could you give me example of some specific situation.



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