Particle in a box wave function

[ProPatto16]

Homework Statement

for a wave function confined between x=0 and x=L find an expression for A in order that the wave function be normalized

The Attempt at a Solution

for a particle in a box between 0 and L the normalized wave function is integral of C²sin²(n\pix/L).dx = 1

using trig identities this results in C²L/2 so then C = sqrt(2/L)

also C = 2iA so equate equations for C for an ewuation to A

2iA = sqrt(2/L)

so A = (sqrt(2/L)) / 2i

squaring both sides gives A² = (2/L) / (-4)

then this gives A to be a sqrt of a negative number. do i just use i?

so then its A= (1/2L)i ??

then i need to find probability that the particle is in the ragion 0<x<L/4 ? with no expression for x?

 

[vela]

Homework Statement

for a wave function confined between x=0 and x=L find an expression for A in order that the wave function be normalized

The Attempt at a Solution

for a particle in a box between 0 and L the normalized wave function is integral of C²sin²(n\pix/L).dx = 1

No. The wave function is \psi(x) = C \sin(n\pi x/L). To be properly normalized, it must satisfy

\int_0^L \psi^*(x)\psi(x)\,dx = \int_0^L C^2 \sin^2\left(\frac{n\pi x}{L}\right)\,dx= 1

using trig identities this results in C²L/2 so then C = sqrt(2/L)

Right, so the normalized wave function is

\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)

also C = 2iA

Where did this come from?

so equate equations for C for an ewuation to A

2iA = sqrt(2/L)

so A = (sqrt(2/L)) / 2i

squaring both sides gives A² = (2/L) / (-4)

then this gives A to be a sqrt of a negative number. do i just use i?

so then its A= (1/2L)i ??

Aside from the fact you don't need to do any of this, you already had an expression (not an equation) for A, so why did you square it only to take the square root (incorrectly) to find what A equaled?

then i need to find probability that the particle is in the ragion 0<x<L/4 ? with no expression for x?

What do you mean "no expression for x"?

 

[ProPatto16]

oh. i see. it looks like i started doing it from one part of the textbook then finished it with parts of another section :/

the second integral you have there. sqrt(2/L)sin(npix/L)... shows that A = sqrt(2/L)

then for the probability thing... take integral of wave function squared between 0 and L/4?

 

[vela]

then for the probability thing... take integral of wave function squared between 0 and L/4?

Yes, that's what you want to do.

 

[ProPatto16]

i got sqrt2/2...

 

[vela]

That's not correct. Show your work.

 

[ProPatto16]

integrating sqrt(2/L)sin(pix/L) between L/4 and 0... i can't use latex so ill do my best. also i have an image here and it shows n=1 that's why i left it out.

so integrating with respect to x

[sqrt(2/L)cos(pix/L) between L/4 and 0

[sqrt(2/L)cos(piL/4/L] - [sqrt(2/L)cos0]

(sqrt(2/L)(cospi/4) - sqrt(2/L)

and cos(pi/4) = sqrt2/2 = 0.707

so 0.707sqrt(2/L) - sqrt(2/L) = ahhhhh 0.293

i see. i minused them the wrong way round.

that better?

probability would be 0.293A?

 

[vela]

Recall what you wrote earlier:

then for the probability thing... take integral of wave function squared between 0 and L/4?

 

[ProPatto16]

oh. amatuer mistake! -.-

1/4 + 1/L ...

 

[vela]

No, that's obviously not correct because the units on the two terms don't match. Keep in mind you're calculating a probability, so your answer should be a unitless number.

 

[ProPatto16]

well see i had that thought too but then i wasnt sure if the probability depended on L..

but i redid the integral and got 1/4 this time. but that doesn't seem right either because the probability isn't evenly distibuted so i would of thought it can't be 1/4 for L/4...

 

[ProPatto16]

1/4 has to be right. I am confident on my integration!

 

[ProPatto16]

also, I've noticed while doing these integrals, that the probability doesn't depend on n for L/4. because no matter what value n takes you still end up taking the sin of a whole integer multiplied by pi... which is always 0. so does this mean the probability is the same for all n between L/4 and 0?

and then that means that all probabilities for all domains between 0 and L must be independant of n?

going on that, if i change n to 2 and then graph between 0 and L, it shows very obviously the area under the curve between 0 and L/4 is indeed 1/4...

 

[vela]

You're not integrating correctly. You were right when you suspect 1/4 wasn't right because the probability density isn't uniform.

If you just post an incorrect answer, all I can tell you if it's right or wrong. I can't see what mistake you're making unless you show your work.

 

[ProPatto16]

Squaring function gives (2/L)sin^2(pix/L) then using trig identity it becomes (2/L)*1/2(1-cos(2pix\L) then 1/L(1-cos(2pix/L) then multiply 1/L through gives (1/L-(cos(2pix/L)/L) then integrating gives x/L-sin( 2pix/L)/L^2 then subbing in L/4 gives L/4/L + 0 and subbing in 0 gives 0. So that simplifies down to 1/4...

 

[vela]

Recheck your integration of the cosine term. You made a couple of mistakes.

 

[ProPatto16]

[-cos(2pix/L)]/L becomes L^-1[-cos(2pix/L)] so then take L^-1 out front cause it's constant, leaving integral of -cos(2pix/L) which gives (2pix^2/2L)sin(2pix/L) which becomes (pix^2/L)sin(2pix/L)... Then bring L^-1 back in giving [(pix^2/L)sin(2pix/L)]/L

 

[ideasrule]

leaving integral of -cos(2pix/L) which gives (2pix^2/2L)sin(2pix/L)

No, that's not the right integral. Do you remember how to integrate functions of the form cos(ax)? The integral should be sin(ax)/a, which you can check by differentiation.

By the way, you can always check integrals using Wolfram Alpha: http://www.wolframalpha.com/input/?i=integrate+1/L*[-cos%282pix%2FL%29]dx

Failing that, you can always check them by differentiating the result and seeing if it's the same as the original value.

 

[ProPatto16]

Gah. I give up :'(

 

[vela]

Use the substitution u=2πx/L.

 

[ideasrule]

Gah. I give up :'(

I think you should review calculus. It's absolutely essential to physics, and quantum mechanics classes usually involve lots of tedious integration.

 

[ProPatto16]

That substitution is what I thought I did. But I did just find a big mistake I've been making. So I'll try redoing everything from the start

 

[ProPatto16]

So I get to this.

1/L*[x-(sin(2pix/L)/2pi/L)

Now subbing in L/4 for x gives sin(pi/2) =1 so

1/L*[L/4-(1/(2pi/L))

1/L*(L/4-L/2pi)

1/4-1/2pi

 

[ideasrule]

That's only correct for n=1, because you used (2/L)sin^2(pix/L) instead of (2/L)sin^2(n*pi*x/L). What's the general solution, for arbitrary n?

BTW sorry for missing that error earlier.

 

[ProPatto16]

I have an image along with the question that shows n=1. a wave function that only has nodes at x=0 and x=L showing n=1. that's why I omitted it.

 

[ideasrule]

Oh, ok. In that case, you got the right answer, but I highly encourage you to review calculus before tackling any more quantum.

 

[ProPatto16]

Okay. Thanks a lot. And yes. I've alreAdy started making a list of tips and hints for differentiation and integration.