ATAR Notes: Forum

Uni Stuff => Science => Faculties => Physics => Topic started by: /0 on February 28, 2010, 02:54:31 am

Title: quantum mechanics questions
Post by: /0 on February 28, 2010, 02:54:31 am: A particle of mass $m$ is in the state

$\Psi (x,t) = Ae^{-a\left[\frac{mx^2}{\hbar}+it\right]}$

Where $A, a \in \mathbb{R^+}$

a) Find $A$ .

I tried using the normalising condition $\int_{-\infty}^\infty |\Psi (x,t)|^2\,dx=1$ but I keep getting

$\int_{-\infty}^\infty A^2e^{\frac{-2amx^2}{\hbar}}\,dx = 1$

$\therefore A = \frac{1}{\sqrt{\int_{-\infty}^\infty e^{\frac{-2amx^2}{\hbar}}\,dx}}$

Am I missing something, or is the answer really this messy?

b) "Find $\langle x^2\rangle$ "

Well, $\langle x^2\rangle = \int_{-\infty}^\infty x^2|\Psi|^2\,dx = A^2\int_{-\infty}^\infty x^2e^{-\frac{2amx^2}{\hbar}}\,dx$

how do you do that?!

Thanks!
Title: Re: quantum mechanics questions
Post by: QuantumJG on February 28, 2010, 09:24:20 am: Give me until the end of march and I'll probably know how to do this. This maths seems awesome!
Title: Re: quantum mechanics questions
Post by: mark_alec on February 28, 2010, 09:31:40 am: Look up the integral to a Gaussian.
Title: Re: quantum mechanics questions
Post by: humph on February 28, 2010, 11:48:29 am: Quote from: mark_alec on February 28, 2010, 09:31:40 am
Look up the integral to a Gaussian.
+1.

You're right so far for the second, then make the substitution $u=x^2$ and use integration by parts twice.

I seem to remember helping my friends with questions like this two years ago... :P
Title: Re: quantum mechanics questions
Post by: /0 on March 01, 2010, 01:20:46 am: Are you sure the integral is elementary?

$\int_{-\infty}^\infty \sqrt{u}e^{ku}\,du$

I tried integration by parts on it and also tried plugging it into my calculator but it won't work.

Hold on a sec... how do you do substitution with infinities?

$\langle x^2\rangle = A^2\lim_{a \to -\infty}\lim_{b \to \infty}\int_{a}^b x^2e^{-\frac{2amx^2}{\hbar}}\,dx$

$u = x^2$ , $du = 2xdx$

$a \to a^2$ , $b \to b^2$

Is this right?

$\langle x^2\rangle = \frac{A^2}{2} \lim_{a \to -\infty}\lim_{b\to\infty} \int_{a^2}^{b^2} \sqrt{u}e^{-\frac{2amu}{\hbar}}\,du$
Title: Re: quantum mechanics questions
Post by: humph on March 01, 2010, 10:48:49 am: Hmmm, you're right, actually. Mathematica says the solution isn't elementary...

Also, the way to deal with the integral with that bit is to notice that it's even, so split it up around the origin into two integrals which are equal, then make the substitution.

EDIT: Actually this definite integral is still determinable, according to this. Indeed, we have
$\int_{-\infty}^{\infty} x^2 e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \left ( {2\pi \over a } \right ) ^{1\over 2} = \left ( {2\pi \over a } \right ) ^{1\over 2} {1\over a}$
Title: Re: quantum mechanics questions
Post by: /0 on March 01, 2010, 04:31:07 pm: Thanks humph!

(these problems are so mean :( )
Title: Re: quantum mechanics questions
Post by: /0 on April 07, 2010, 01:55:11 am: The Hamiltonian-Jacobi equation

$\frac{\partial W}{\partial t}+\frac{1}{2m}\left[\left(\frac{\partial W}{\partial x}\right)^2+\left(\frac{\partial W}{\partial y}\right)^2+\left(\frac{\partial W}{\partial z}\right)^2\right] + V(x,y,z) = 0$

Can be re-expressed as $|\nabla W| = \sqrt{2m(E-V)}$ by taking $W = -Et+S(x,y,z)$

Schrodinger says that if we think of the level curves of W, and assign an arbitrary curve the value $W_0$ , that we can take a normal to that paticular level curve (spanning $W_0+dW$ ) to be":

$dn = \frac{dW_0}{\sqrt{2m(E-V)}}$

(In other words, $\frac{dW_0}{dn} = |\nabla W|$ )

Where does this come from? How do we know the normal differential has this value?
Title: Re: quantum mechanics questions
Post by: /0 on April 20, 2010, 11:00:14 pm: Can someone please help me solve:

$\frac{r^2}{R}\frac{d^2R}{dr^2}+\frac{r}{R}\frac{dR}{dr} - r^2k = 0$

It comes from separating the schrodinger equation $-\frac{\hbar^2}{2m}\nabla^2\psi = E\psi$ in cylindrical coordinates.
Title: Re: quantum mechanics questions
Post by: mark_alec on April 20, 2010, 11:06:09 pm: From memory, let u(r) = 1/R and convert your DE to one in terms of du/dr. It will work out to be much nicer.
Title: Re: quantum mechanics questions
Post by: humph on April 20, 2010, 11:10:02 pm: You could always just multiply through by $R/r$ , so it becomes a 2nd order linear ODE. Then try a power series solution perhaps, or any other technique to solve 2nd order linear ODEs (I'm not sure what you're up to in Methods 1...).
Title: Re: quantum mechanics questions
Post by: /0 on April 20, 2010, 11:28:57 pm: Quote from: mark_alec on April 20, 2010, 11:06:09 pm
From memory, let u(r) = 1/R and convert your DE to one in terms of du/dr. It will work out to be much nicer.

thanks mark, I'll have a go at that

Quote from: humph on April 20, 2010, 11:10:02 pm
You could always just multiply through by $R/r$ , so it becomes a 2nd order linear ODE. Then try a power series solution perhaps, or any other technique to solve 2nd order linear ODEs (I'm not sure what you're up to in Methods 1...).

yeah I thought that might work but if possible I want to try to get an analytic solution first
Title: Re: quantum mechanics questions
Post by: /0 on April 23, 2010, 07:08:08 pm: Too busy at the moment for stuff that's outside the course... I'll try the DE at some later... indefinite date.

Speaking of indefiniteness, to find the momentum distribution of a particle in an infinite square well, do you:

a) Decompose $\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L} x\right)$ into $\sqrt{\frac{2}{L}}\frac{e^{\frac{i n\pi x}{L}}-e^{\frac{-i n\pi x}{L}}}{2i}$ and by noting that the momentum operator operating on each of the exponentials gives back an eigenvalue for momentum, conclude that the momentum probability distribution is given by two spikes at $p = \pm \frac{n\pi \hbar}{L}$ ?

OR

b) Use the Fourier Transform $\phi (k) = \frac{1}{\sqrt{2\pi}}\int_0^L \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)e^{-ikx}\,dx$ to find the momentum distribution which has two peaks but smooth peaks, rather than the discrete spikes given in the first method?

This is from an assignment I just handed in. I went with the second method since I trust the Fourier Transform and it seems to make more sense to have continuous position/momentum fourier pairs. but a few of my friends argued using the first method that the momentum has discrete values.

Which one is correct? thx
Title: Re: quantum mechanics questions
Post by: mark_alec on May 13, 2010, 08:37:25 pm: http://mathworld.wolfram.com/FourierTransformSine.html

If you do the Fourier transform, the answer looks strangely like what you'd get if you did it the first way. Indeed, you expect the particle is travelling in one direction, or the other, at a speed dependent upon the energy.
Title: Re: quantum mechanics questions
Post by: /0 on May 14, 2010, 04:10:55 pm: But why is it the dirac delta function? When I integrated the exponentials I got cosines and signs.
Title: Re: quantum mechanics questions
Post by: mark_alec on May 18, 2010, 10:23:56 pm: I don't know how to prove it, but I know that you have to take $\int_{-\infty}^{\infty} e^{-2\pi i(k - k_0) x} dx = \delta(k-k_0)$ as given.
Title: Re: quantum mechanics questions
Post by: /0 on May 19, 2010, 01:02:28 am: Hmm perhaps it's because the integrand is complex at all times except when $k = k_0$

Anyway thanks mark
Title: Re: quantum mechanics questions
Post by: QuantumJG on June 03, 2010, 06:18:40 pm: Ok I have a quantum mechanics question and I thought here is probably the best place to ask it.

Anyway I have a proton in a 1-D infinite potential well and the wavefunction for it is:

$\psi (x) = Asin \left( \dfrac{ 2 \pi x}{L} \right) , x \in \left( - \dfrac{L}{2}, \dfrac{L}{2} \right)$

One part of the question asked me what is the quantum number n, i.e. if:

$\psi _{n} (x) = Asin \left( \dfrac{ n \pi x}{L} \right) , x \in \left( 0, L \right)$

I found n to be 2 (i.e. the proton is in it's first excited state), by shifting the second equation by $\dfrac{L}{2}$ to the left.

Another part of this question is to prove that this wavefunction is a solution to the time independent Shrodinger equation. Is this just another way of asking you to show the second derivative of $\psi$ is simply a constant times $\psi$ .
Title: Re: quantum mechanics questions
Post by: /0 on June 03, 2010, 07:30:46 pm: I guess you could show by differentiation, but you might as well just solve the time-independent schrodinger equation anyway. It wouldn't be a long derivation and it's probably in your book

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$
Title: Re: quantum mechanics questions
Post by: QuantumJG on June 03, 2010, 07:43:18 pm: Thanks.

How is your exam timetable? Mine is pretty good quantum & special on the 9th, thermal and classical on the 11th, Real Analysis on the 17th and Introductory personal finance on the 18th.

This year I have really learnt that I love physics (definately more than maths), but I still enjoy maths. You are lucky to do PDE's next semester I have to wait until next year.
Title: Re: quantum mechanics questions
Post by: QuantumJG on June 03, 2010, 09:38:08 pm: This is another question and I want critique with my answers.

i.)

Normalisation implies that:

$\int_{-\infty}^{\infty} \Psi (x,t)* \Psi (x,t) dx = 1$

$\int_{-\infty}^{\infty} |a_{0}| ^{2} | \psi _{0} (x) | ^ {2} + |a_{1}| ^{2} | \psi _{1} (x) | ^ {2} dx = 1$

(I cut out a lot of crap since the wave-functions being orthonormalised means that after a bit of tedious work you get to that line)

$|a_{0}| ^{2} \int_{-\infty}^{\infty} | \psi _{0} (x) | ^ {2} dx + |a_{1}| ^{2} \int_{-\infty}^{\infty} | \psi _{1} (x) | ^ {2} dx = 1$

$|a_{0}| ^{2} + |a_{1}| ^{2} = 1$

i.e. I found this to be the normalisation condition.

ii.)

First to see if the energy is an eigenvalue

$[E] \Psi (x,t) = i \hbar \dfrac{d}{dt} \left( a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } \right)$

$[E] \Psi (x,t) = E_{0} a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + E_{1} a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } }$

So the energy is not an eigenvalue and hence the energy is not a sharp observable.

$\langle E \rangle = \int_{-\infty}^{\infty} \left( a _{0} * \psi (x) * e^{ \dfrac{i E_{0} t }{ \hbar } } + a _{1} * \psi (x) * e^{ \dfrac{i E_{1} t }{ \hbar } } \right) \left( E_{0} a _{0} \psi (x) e^{ - \dfrac{i E_{0} t }{ \hbar } } + E_{1} a _{1} \psi (x) e^{ - \dfrac{i E_{1} t }{ \hbar } } \right)$

$\langle E \rangle = E_{0} |a_{0}| ^{2} + E_{1} |a_{1}| ^{2}$

$\langle E \rangle = E_{0} |a_{0}| ^{2} + E_{1} (1 - |a_{0}| ^{2} )$

$\langle E \rangle = |a_{0}| ^{2} ( E_{0} - E_{1} ) + E_{1}$

$\langle E \rangle = |a_{0}| ^{2} ( \dfrac{1}{2} \hbar \omega - \dfrac{3}{2} \hbar \omega ) + \dfrac{3}{2} \hbar \omega$

$\langle E \rangle = \hbar \omega \left( \dfrac{3}{2} - |a_{0}| ^{2} \right)$

I'm not 100% sure if this is right.
Title: Re: quantum mechanics questions
Post by: full of electrons on June 03, 2010, 10:51:27 pm: Back to the original question. Isn't the integral of a Gaussian sqrt(pi)? or something similar, I'm afraid I don't actually know, it'll be in some table of integrals.

Gee, your quantum course seems harder than mine (and i reckon mine is hard enough)
Title: Re: quantum mechanics questions
Post by: /0 on June 03, 2010, 11:05:06 pm: QuantumJG the steps you take seem alright to me... although I wouldn't take my own advice at the moment. My quantum exam is in a while so I haven't really thought much about it recently...

My exam time-table is quite spread out,
8th June: Take-home ODEs and Vector Calc exam, worth 20%
11th June: Thermal Exam, worth 35%
21st June: Analysis Exam, worth 70% (and wow I would be stoked with anything near a 70% on the exam)
23rd June: Quantum Exam, worth 40%
Title: Re: quantum mechanics questions
Post by: Cthulhu on June 03, 2010, 11:12:01 pm: wtf TAKE HOME EXAMS? Do want.

I had so much trouble with my last QM assignment. It involved Bessel Functions and Circular Infinite Wells and it made me sad :(
Title: Re: quantum mechanics questions
Post by: QuantumJG on June 08, 2010, 02:32:20 pm: I just want my answer to this question to be criticised.

i.)

Let $\Psi(x,t) = \psi(x) \phi(t)$

Therefore looking at the LHS of the Schrödinger equation:

$i \hbar \dfrac{ \partial \Psi(x,t) }{ \partial t} = i \hbar \dfrac{ \partial }{ \partial t} ( \psi(x) \phi(t)$
$= i \hbar \psi(x) \dfrac{ d \phi(t) }{dt}$

And now the RHS:

$- \dfrac{ \hbar ^{2} }{2m} \dfrac{ \partial ^{2} \Psi(x,t) }{ \partial x ^{2} } + V(x) \Psi(x,t) = - \dfrac{ \hbar ^{2} }{2m} \dfrac{ d ^{2} \psi(x) }{ dx ^{2} } \phi(t) + V(x) \psi(x) \phi(t)$

$\therefore i \hbar \psi(x) \dfrac{ d \phi(t) }{dt} = - \dfrac{ \hbar ^{2} }{2m} \dfrac{ d ^{2} \psi(x) }{ dx ^{2} } \phi(t) + V(x) \psi(x) \phi(t)$

$\therefore i \hbar \dfrac{1}{ \phi(t) } \dfrac{ d \phi(t) }{dt} = - \dfrac{ \hbar ^{2} }{2m} \dfrac{1}{ \psi(x) } \dfrac{ d ^{2} \psi(x) }{ dx ^{2} } + V(x) =$ Constant (Call it E)

$\therefore i \hbar \dfrac{1}{ \phi(t) } \dfrac{ d \phi(t) }{dt} = E$

$\therefore i \hbar \dfrac{ d \phi(t) }{dt} = \phi(t) E$ Equation to find $\phi(t)$

$- \dfrac{ \hbar ^{2} }{2m} \dfrac{1}{ \psi(x) } \dfrac{ d ^{2} \psi(x) }{ dx ^{2} } + V(x) = E$

$- \dfrac{ \hbar ^{2} }{2m} \dfrac{ d ^{2} \psi(x) }{ dx ^{2} } + \psi(x) V(x) = \psi(x) E$ Equation to find $\psi(x)$

ii.)

$\phi(t) = e ^{ - \dfrac{i E t}{ \hbar} }$
Title: Re: quantum mechanics questions
Post by: /0 on July 24, 2011, 06:03:15 pm: In Zetilli it gives a completeness relation for the basis of spin 1/2 states:

If $|\frac{1}{2},-\frac{1}{2}\rangle =\left(\begin{matrix} 0 & 1 \end{matrix}\right)$ and $|\frac{1}{2},\frac{1}{2}\rangle=\left(\begin{matrix} 1 & 0 \end{matrix}\right)$ , then

$\sum_{m_s=-\frac{1}{2}}^{\frac{1}{2}} |\frac{1}{2},m_s\rangle\langle \frac{1}{2},m_s| = \left(\begin{matrix}0 \\ 1 \end{matrix}\right)(\begin{array}{cc} 0 & 1 \end{array})+\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)(\begin{array} 1 & 0 \end{array})=\left(\begin{array} 1 & 0 \\ 0 & 1 \end{array}\right)$

But how does this matrix multiplication make sense?
Title: Re: quantum mechanics questions
Post by: mark_alec on July 24, 2011, 06:25:38 pm: The matrix multiplication makes sense if you treat it as a direct (or Kronecker) product.
Title: Re: quantum mechanics questions
Post by: /0 on July 26, 2011, 01:16:11 am: Thanks mark, I didn't realise it was a direct product