Recent content by Stan12

Now with this expression ψ(x) I can find the probability density (ψ*(x)ψ(x)) = √(km/π) 2x2/hbar2 exp[-(km)x4/4hbar2

I will try this from scratch

ψ(x) = C2xe-x2/2 ψ*ψ = |C2xe-x2/2|2 ∫|C2xe-x2/2|2 dx = C2∫4x2ex4/4 is this correct, all terms in x so only when an exponential expression is canceled out using pd is when it function is imaginary :ex: e-ix Now by doing what I did above is the pdf corrected? I squared the function and set it...

-So we have ψ1 = C1 2se-s2/2 as our wave function of this specific harmonic oscillator n = 1 - Probability density is given as ψ*ψ and within the intervals of -d to d giving us ∫d-d ψ*(x)ψ(x)dx ψ*ψ = (C1(2s)es2/2)(C1(2s)e-s2/2) = C2(2s)2 I'm still not getting this problem..and how the...

∫-ss ψ*(x)ψ(x) dx ? P = C24s2e-s2/2 + eis2/2 x = [√h-bar/mk1/4] * 1/√2 s = [(km)1/4/√h-bar] x s = 1/√2 so... this doesn't seem right, am I suppose to get a function out of probability density?

the intervals would be from ∫-dd. Is there a way to simplify down the distance given? d = (mk)-1/4√h-bar/2 s = [(km)1/4/√h-bar] x d = x/s√2 ; -d = -x/s√2 this doesn't seem right. the integral, α = ∫-dd x2e-x2/2

Homework Statement For the n = 1 harmonic oscillator wave function, find the probability p that, in an experiment which measures position, the particle will be found within a distance d = (mk)-1/4√ħ/2 of the origin. (Hint: Assume that the value of the integral α = ∫01/2 x2e-x2/2 dx is known...

z = √7 - r2 s(x,y) = < x, y, (7 - x2 - y2)1/2 s(r,θ) = < rcosθ, rsinθ, (7 - (rcosθ)2 - (rsinθ)2)1/2

So the limits to r is from 0 to r = √x2 + y2 or r = √z and limit of z is from 0 to 9 ? since the paraboloid begins at 0 and enclosed by the plane at z = 9 ∫∫∫ r2cos2θ rdrdzdθ

∫∫z r3 cos2θ drdθ 9 ∫ r4/4 cos2θ dθ (9 * 34)/4 ∫cos2θ dθ (9 * 34)/4 ∫(cos2θ + 1)/2 dθ = (9 * 34)/4 *1/2 [ (sin2θ)/2 + θ ] 0<θ<2∏ I got (9 * 34 * 2∏)/8 = 729∏/4

z = x2 + y2 and plane z = 9 I set x2 + y2 = r2 and found that 0<r<3 and 0<z<9 since it's given and 0<θ<2∏ x=rcosθ in cylindrical coord. f(x,y,z) = x2 --> x2 in cylindrical coord. = (rcosθ)2 I set the function under a triple integral with restriction above so I got ∫∫∫ (rcos)2...

Ok. let me start from scratch, I'm going to set x = rcosθ , y = rsinθ and z = √(7 - (rcosθ)2 - (rsinθ)2), cylindrical coordinates. I found the restriction to r by setting x2 + y2 = r2 in portion of sphere given. which gives us r2 + z2 = 7 Now, it says that portion lies above a plane z =...

I apologize for the bad notation, I found that 0 < r < 2 and 0 < θ < 2∏ as restriction

z = √(7 - r2cos2θ - r2sin2θ) s(r,θ) = (2cosθ, 2sinθ, √(7 - r2 (cos2θ - sin2θ)) ) 0 < r < 2, 0 < θ < 2∏

Hmm. I guess my attempt to parametrize was totally incorrect, my question now is how do I parametrize this problem?