Author Topic: quantum mechanics questions (Read 11027 times) Tweet Share

mark_alec · « **Reply #15 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.

/0 · « **Reply #16 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

QuantumJG · « **Reply #17 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 $\psi$ .

/0 · « **Reply #18 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

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$

QuantumJG · « **Reply #19 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.

QuantumJG · « **Reply #20 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.

full of electrons · « **Reply #21 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)

/0 · « **Reply #22 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%

Cthulhu · « **Reply #23 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

QuantumJG · « **Reply #24 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.)

$\phi(t) = e ^{ - \dfrac{i E t}{ \hbar} }$

/0 · « **Reply #25 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?

mark_alec · « **Reply #26 on:** July 24, 2011, 06:25:38 pm »

The matrix multiplication makes sense if you treat it as a direct (or Kronecker) product.

/0 · « **Reply #27 on:** July 26, 2011, 01:16:11 am »

Thanks mark, I didn't realise it was a direct product

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