Question

Given that ψ(x) = Axe^–ibx for 0 < x < 10 and ψ(x) = 0 for all other values of x:
i) What is the value of the normalization constant?
ii) What is the expectation value, 〈x〉?

Answer 1

\[\psi(x) = Axe ^{-ibx}\] in case that wasn't clear

Answer 2

So to normalize a wave function you need to use the normalization condition which sums all possible states to 1.
\[P = \int\limits_{-\infty}^{\infty} [\psi (x)]^{2}dx=1\]

Answer 3

\[1= \int\limits_{-\infty}^{\infty}\psi(x)\psi(x)^{*}dx\]

Answer 4

Can you type the complex conjugate of your wavefunction psi^*

Answer 5

Complex conjugate would be \[\psi(x)^{*} = Axe ^{ibx}\] then?

Answer 6

Yep, just get rid of the negative. Nice work.

Now to accommodate your problem. We see that the only non-zero wave function given goes from 0<x<10, so we can ignore all the other values of the integral. This gives us finite limits which is a wonderful thing.

So we need to solve the following:\[1=A^{2} \int\limits_{0}^{10} x^{2}e^{-ibx}e^{ibx}dx=A^{2} \int\limits_{0}^{10}x^{2}dx\]

Answer 7

So I get \[A ^{2} \frac{ 1 }{3} x ^{3} = 1\]from x = 0 to 10. Which gives \[A ^{2} \frac{ 1 }{ 3 } 10^{3} = 1\]
\[A = \sqrt{\frac{ 3}{ 1000}} = 0.055\]

Answer 8

Okay. So for the second part, the way that you calculate expectation value is the same type of integral where you have the wave function squared. But now, you add in the operator associated with position, which is just x.

\[<x> = \int\limits _{-\infty}^{\infty}\psi(x)^{*}x\psi(x)dx\] again, we only consider the limits necessary here.

Answer 9

So we can take x out and get \[x \int\limits_{0}^{10} 0.055^{2} x ^{2} dx\]then this gives\[0.055^{2}\frac{ 1 }{ 3 } x ^{4}\]from 0 to 10 which gives\[<x> = 10.083\]

Answer 10

Well the 'x' is still the variable you're integrating over. It should remain in the integral. In fact you now have the same integral, just x^3 instead of x^2. And you have a solve A^2 to add in which you did very nicely.

Answer 11

Ahhhh okay >_< so I get 7.5625 instead then?

Answer 12

I'll trust your calculator haha. If you got an integral:

0.055^2 * 1/4 * x^4, then yes haha

Answer 13

That's the one! You are a true scholar, I'd give you ten medals if I could!

Answer 14

Not a problem. Best of luck with everything. Quantum Mechanics is an amazing subject to study.