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Basic Quantum stuff

Hi all,

Just wondering if someone could help me with this problem out of my textbook: (question is attached as a picture file because i can't get the SR forums to do the latex stuff anymore :frown:

)

Its right at the start of the book which is strange because to me it seems quite difficult (well i just can't work out how to do the integral i think)

any help would be awesome! :smile:

-Sarah

Picture 1.png33.2KB

Reply 1

AlphaNumeric

If you've never seen how to integrate a Gaussian integral then you'll need to read up on that first. It cannot be done analytically except for the limits of 0 and -+ infinity.

The general result is that

Int e^(-K[x^2]) dx = 2 sqrt(K*pi)

when the limits are -infinity and +infinity.

For the other parts, do integration by parts.

Reply 2

sarahisme

AlphaNumeric

For the other parts, do integration by parts.

other parts?

the gaussian integral thing is for part A right? :P

Reply 3

sarahisme

ok so this is what i did for part A :....

what do you guys think?

Reply 4

AlphaNumeric

The lower limit in your integral is wrong, it's -infinity (the text code is \infty by the way). Otherwise it looks like you're on the right track :smile:

Reply 5

sarahisme

k, yyeah i think i got it

A^2 = (lambda/pi)^{1/2}

but i dont know how to do <x> or <x^2>, i cant seem to find an integral to sub. into it... :frown:

Reply 6

sarahisme

ok for the <x> part i am trying to figure out how to do this integral.... (is attached as a picture)

any suggestions? i have tried integration by parts but that doesnt seem to be working (maybe i am doing it wrong?? :s-smilie:

)

Sarah

Reply 7

Mehh

sarahisme

ok for the <x> part i am trying to figure out how to do this integral.... (is attached as a picture)

any suggestions? i have tried integration by parts but that doesnt seem to be working (maybe i am doing it wrong?? :s-smilie:

)

Sarah

Oh gawd no. Integration by substitution. Its the most simple type.
x^2 = u
2xdx = du
Its not even difficult. You should have seen this type of integration before.

Reply 8

sarahisme

i did integration by parts in the end anyways :smile:

now as for the probability distribution sketch thing?? any ideas?

Reply 9

Mehh

sarahisme

i did integration by parts in the end anyways :smile:

now as for the probability distribution sketch thing?? any ideas?

Its a standard Gaussian distro, you should be able to sketch this freehand. I am guessing it should be in your lecture notes. 'a' is simply is the offset from the central point for teh Gaussian.

Reply 10

sarahisme

ok i will give it a go and get back to you :smile:

Sarah

Reply 11

AlphaNumeric

http://mathworld.wolfram.com/NormalDistribution.html

Looks like the left hand graph, but a lot flatter, and the peak being at x=a, not x=0.

Reply 12

sarahisme

ok so if i draw a graph should probably put on the mean (<x>) the standard deviation (<x^2> - <x>^2)^{1/2}, axis labels is there anything else worth putting on?

also

what does <x^2> tell you? or is it just used for working out the standard deviation?

also how do you know that the graph is very flat?

and

do i need to put anything about the value of A on the graph?

oh and here is my sketch : i reckon its pretty good :P

Thanks (sorry bout the thousands of questions! )

Sarah :P

Reply 13

AlphaNumeric

Ignoring the integral sign (which is for findign the area, not plotting the function) when you set x=a (which is the max of the graph) you end up with y = A, so the max of the graph is A.

The graph's flatness is dependent on lambda, I just suggested doing it flatter to illustrate you know what's going on.

Working out <x> and <x²> is to work out the standard derivation yes.