Quantum Free Particle question

[Formslip]

Homework Statement

[PLAIN]http://img168.imageshack.us/img168/6892/img002pd.jpg
this is the first half of a test we had last friday.

Homework Equations

[tex]\phi[/tex](k) = [tex]\frac{1}{2\pi}[/tex][tex]\int[/tex][tex]\Psi[/tex](x,0)e[tex]^{-ikx}[/tex]dx
that looks a little funky but i think you understand what I am trying to say.

The Attempt at a Solution

a) so for the first part i know youre supposed to normalize to solve for A so i tried with [tex]\int[/tex][tex]^{\infty}_{0}[/tex]([tex]\Psi[/tex])[tex]^{2}[/tex]=1
[PLAIN]http://img258.imageshack.us/img258/7572/picture001b.jpg

and its the same thing when you do it for x<0. assuming that's right the next part

b) is just a matter of plugging in and chugging.
[PLAIN]http://img259.imageshack.us/img259/7357/picture003ivl.jpg

which I've probly chugged incorrectly, doesn't seem like there should be an i in there..
anyway

c) i can do np once i have the correct [tex]\Psi[/tex]

d) i don't know what this is really asking, but its not as important as figuring out a & b


e) i have no idea how to do this. any help appreciated.



tl;dr basically i kindof understand a) and b) though i think i got em wrong, and i do not understand e) at all
this is my first post here, i hope I've done everything I am supposed to.

 

[diazona]

Both your parts (a) and (b) suffer from the same problem: you can't magically change the limits of integration from (-∞,∞) to (0,∞). Since the wavefunction is defined with two different functions on two different domains, you will need to take the full integral from -∞ to ∞ and break it up into two pieces, which then need to be added together to get the result of the full integral.

 

[Formslip]

do you mean like this
[PLAIN]http://img24.imageshack.us/img24/6184/picture002lpv.jpg
cause then it seems like when you evaluate it, stuff starts going to infinity all over the place.
thats why i broke it into one integral from 0 to +inf and one integral from 0 to -inf. here in these examples i was doing just the 0 to +inf part just to see if i had the method down correctly, but if i can't do that then I am not sure how to approach this problem.

thanks for the response btw.

 

[diazona]

do you mean like this
[PLAIN]http://img24.imageshack.us/img24/6184/picture002lpv.jpg[/QUOTE]
No, not like that. What you are integrating there is not the square of the magnitude of the wavefunction.

thats why i broke it into one integral from 0 to +inf and one integral from 0 to -inf. here in these examples i was doing just the 0 to +inf part just to see if i had the method down correctly

Breaking the integral down into two parts is correct, but what you tried to do before was breaking the whole problem into two parts, which is not correct.

Try starting from scratch. This is the (only) correct equation to determine the normalization:
[tex]\int_{-\infty}^{\infty}\left|\psi(x)\right|^2\mathrm{d}x = 1[/tex]
Plug in your wavefunction and see what you get.

 

[Formslip]

Try starting from scratch. This is the (only) correct equation to determine the normalization:
[tex]\int_{-\infty}^{\infty}\left|\psi(x)\right|^2\mathrm{d}x = 1[/tex]
Plug in your wavefunction and see what you get.

[PLAIN]http://img203.imageshack.us/img203/7758/picture004nn.jpg
like this perhaps?
sorry its like ten feet tall

 

[diazona]

No. That would be true if the wavefunction were
[tex]\psi(x) = A e^{ax} + A e^{-ax}[/tex]
but that is not the wavefunction.

Hint: break up the integral before plugging in [itex]\psi[/itex].

 

[Formslip]

ok I've been really dense, thanks for persisting

hopefully this is what you're talking about:
[PLAIN]http://img545.imageshack.us/img545/7517/picture005x.jpg

 

[diazona]

Yep, assuming that both your constants (A and a) are real, that's the right way to start. Remember what the sum of those two terms has to be equal to and you should be able to find the value of A now.

 

[Formslip]

Aha! awesome, thanks.

ok so what I've done from that:
[PLAIN]http://img833.imageshack.us/img833/7193/picture002sb.jpg

and doing part b) over i get
[PLAIN]http://img688.imageshack.us/img688/944/picture004ya.jpg

i put a little star next to line 6 cause I am not sure if i did this right. my reasoning is that you can only get this value if a is a negative number, so in my evaluation i put a as being negative where it appears under a square root. so i just pulled the negative out of each one and got an i.

if that's correct then the last line here would be what i got for [tex]\phi[/tex](k)

 

[diazona]

That's pretty close, you just have a few algebra errors. Look at the integral you do for part (a), and think about what the allowed range for a is. Specifically, if a < 0, does the integral converge? What about if a > 0? What about the case a = 0? (You may have already thought about this)

Once you've figured out the allowed range for a, look at your answer for the first part. Does it make sense, given the sign of a? You should find that the value of A is real and positive (because technically, what you're calling A is really |A|).

 

[Formslip]

for the case where a>0 the integral diverges, for a=0 the whole thing becomes 0, so the only value a can have where A has a finite value (other than 0) is when a<0.
so a is negative.
i tried doing the first one over again with replacing all the a's with (-a)'s but i got the same answer. maybe A = sqrt(abs(a)) since a is always a negative number, but here were just talking about the magnitude of a i guess you could say.
that would take care of keeping A real.

 

[diazona]

for a=0 the whole thing becomes 0

No, that's not correct. (What's the value of [itex]e^{2ax}[/itex] when [itex]a=0[/itex], for any number [itex]x[/itex]?) You're right that [itex]a=0[/itex] is not an allowed case, but the reason is slightly different.

so a is negative.

Right.

i tried doing the first one over again with replacing all the a's with (-a)'s but i got the same answer. maybe A = sqrt(abs(a)) since a is always a negative number, but here were just talking about the magnitude of a i guess you could say.
that would take care of keeping A real.

Nope, you can't just arbitrarily insert absolute value signs when you feel like it.

Check your math on the first part, specifically where you plug in values for [itex]x[/itex] after doing the integration.

 

[Formslip]

i asked my teacher today after class about this part, he showed me a similar way of doing it where he took the wave function and just used the abs(x) so that

[PLAIN]http://img213.imageshack.us/img213/504/picture001tq.jpg

if you do it this way, and solve still assuming a<0, then you end up getting the answer where

[PLAIN]http://img180.imageshack.us/img180/1930/picture007ib.jpg

which is the same answer i was getting. so is my answer right and my method wrong? or is there something here were looking over altogether?

 

[diazona]

No, your method is right but your answer of [itex]A = \sqrt{a}[/itex] is wrong. It's a simple math mistake.

By the way, using the absolute value of x as you did is a shortcut that is valid when the wavefunction is an even function. It follows from splitting the integral up.

 

[Formslip]

[PLAIN]http://img815.imageshack.us/img815/7193/picture002sb.jpg

ok this is my last shot
i realized i was doing what i like to call "crazy math", here's my attempt at some 'sane math'.

"A" has to be real idk how to handle the positive part, but here "a" is going to be a negative number so the negative in front will make it positive, thus making [tex]A[/tex] Real.

edit: also i asked about the a=0 thing, he said its not allowed cause its not normalizable, the graph is like a straight line or something. does that sound about right?

 

[diazona]

Yep, that's correct, [itex]A=\sqrt{a}[/itex]. And for the a=0 case, it's not normalizable for the reason I was hinting at previously, because the integral diverges (i.e. is not finite).

Now try fixing up your calculation for part (b). Specifically, look at how you got from line (5) to line (6).

 

[Formslip]

I really appreciate you helping me out with this btw. I've tried many forums/many googles trying to find help on this sort of thing. this is the first place I've gotten real answers i can use.

[PLAIN]http://img44.imageshack.us/img44/1109/picture004so.jpg

thats my second attempt on part b)
hopefully a bit cleaner than the other one.

 

[diazona]

Where did the factor of 2 in the first part of line 7 come from? Other than that, I think it looks okay.

 

[Formslip]

ah yea i thought that looked weird nvm. scratch the two i see my mistake, it was in the subtracting, i basically subtracted the same thing twice and got an extra factor of 2.


the only thing left i did not understand on the test is part e)
[PLAIN]http://img513.imageshack.us/img513/218/51156239.jpg

i don't even know what this is asking me to do.
any help appreciated.

 

[Saxonphone]

ah yea i thought that looked weird nvm. scratch the two i see my mistake, it was in the subtracting, i basically subtracted the same thing twice and got an extra factor of 2.


the only thing left i did not understand on the test is part e)
[PLAIN]http://img513.imageshack.us/img513/218/51156239.jpg

i don't even know what this is asking me to do.
any help appreciated.

He's asking you to show that the inner product of the function is zero for n=1 and m=2.
[tex]\langle \psi_1 | \psi_2 \rangle = 0 [/tex]