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Is <math> C_0 </math> in equation 4.28 equal to 2?
I think that's the normalization constant (assuming you mean equation 4.82 rather than 4.28?) so in order to find it, you would integrate <math>\int_0^{\infty}|R_{20}®|^2 r^2 dr = 1</math> and solve. I did this with a computer algebra system and got <math> \text{c0}\to \frac{a^4}{-1+a^2}</math>, but I'm not sure if this is right because it seems like <math>a</math> should be able to be 1…
correction to the last part of today lecture.
In a hurry to show how one can get derivatives such as <math>\frac{\partial\phi}{\partial x}</math> without “knowing” how to differentiate arctan, I made a mistake. I said that
<math>\tan\theta=\frac{y}{x}</math> to show you <math>\frac{\partial\theta}{\partial x}</math>, but this should have been for <math>\phi</math>, not <math>\theta</math>, although the main idea is unchanged.
When you differentiate with respect to x, <math>\tan\phi=\frac{y}{x}</math>, the LHS will be <math>\frac{\partial}{\partial x}\tan\phi=\sec^2\phi\frac{\partial\phi}{\partial x}</math> using the chain rule (is this the right terminology?). The RHS will be <math>\frac{\partial}{\partial x}\frac{y}{x} = -\frac{y}{x^2}</math>. From this, one can find out that <math>\sec^2\phi\frac{\partial\phi}{\partial x}=-\frac{y}{x^2}</math>, and subsequently, <math>\frac{\partial\phi}{\partial x}=-\frac{y}{x^2}\cos^2\phi</math>. Usually, in this kind of context, one wants to express everything in the spherical coordinate variables, this will be further modified to give <math>\frac{\partial\phi}{\partial x}=-\frac{r\sin\theta\sin\phi}{(r\sin\theta\cos\phi)^2}\cos^2\phi=-\frac{\sin\phi}{r\sin\theta}</math>.
Using this method, one never needs to know how to differentiate art tangent function!
Sure, “chain rule” is the right terminology. This whole process is known as implicit differentiation.
If its ok with you guys I'd like to start discussing the practice quiz–
I'll start with question 3 (because it seems a little too simple)
Without giving away the method, the answer I got was 5.155*10^-9 m. How does that compare to your answer?
I think the point is we are supposed to get the Bohr radius. ~5 x 10^-11. About half an angstrom.
I have 5.19*10^-11 m as well.
Oy! I also found the Bohr radius. Mate.
Yup, I must've done something wrong, got 5.3e-11 m.
Half an angstrom here as well
Do we need to calculate the value of the quantity or we can just express it using <math>\hbar, e, m_e, \eps_0</math>?
It seems fair to me that we'd be able to leave it in constants form. The rest of the test is pretty calculator independent… so I wouldn't imagine that if we had the constants correct we'd be penalized for not multiplying the actual values.
i dont know what im doing wrong but i get 0 for the stationary states in question 1 :( and of course that cant be right! i have energies to be (5+/-√5)ε/2. Are my energy values wrong??
im still having trouble with this, this is what i have done: <math> E_1= (5+sqrt5)/2 </math>and <math>E_2= (5-sqrt5)/2 </math> and using <math> H\psi=E_n\psi </math> i want to find the stationary states which is described by the vector [x,y]. here is my equations for <math>E_2= (5-sqrt5)/2 </math>
<math/> \large\begin{eqnarray*} 2x+iy=x(5-sqrt5)/2\\-ix+3y=y(5-sqrt5)/2\end{eqnarray*}</math>
and so this system of equation gives x=y=0. and the other case <math>E_1= (5+sqrt5)/2 </math>
<math/> \large\begin{eqnarray*} 2x+iy=x(5+sqrt5)/2\\-ix+3y=y(5+sqrt5)/2\end{eqnarray*}</math>
also gives x=y=0. i dont know if im solving these systems wrong or if my equations are wrong ?!?!
We got the same energies. <math>E = \frac{5 \pm \sqrt{5}}{2} \epsilon</math>. This was confirmed with Mathematica. Mathematica also gave <math>\left( -\frac{1 \pm \sqrt{5}}{2} i, \ 1 \right)</math> as the eigenvectors (states).
We also got 0 and 0 when we tried to solve, so in hindsight, looking at the Mathematica solution, we realized that you could simply take one of the relations in the system, set y to 1, and solve for x. (Twice, once for each energy/eigenvalue. Either that or maybe do it for both relations in the system, but we haven't tried that and don't know if it would work.)
I also got the same energies and eigenvectors. What I did was take one equation (in this case <math/>2x+iy=x(5-sqrt5)/2</math> and solved for y in terms of x.
Yeah, that's what I ended up getting for my energies too
Those equations actually lead to the same relation between “x” and “y.” That's why you can take one or the other equation to get the right relation between them. If they are NOT redundant, then the only answer will be x=y=0. To avoid this result, we required that the determinant <math>det(H-\lambda I)=0</math>. Since the equation to obtain the eigenvector, <math>v=\begin{bmatrix}x
y \end{bmatrix}</math> is given by <math>(H-\lambda I)v=0</math>, if the determinant is NOT zero, the matrix, <math>det(H-\lambda I)=0</math> has an inverse matrix, M. By applying M from the left, <math>M(H-\lambda I)v=M*0</math>. Since the two matrices on the LHS are inverse of each other, their product is an identity matrix, I. The RHS is zero. Then you will get <math>v=0</math>.
I got the same for my energies too, but confused as to how to determine the eigenstates. Has anybody worked on #4 yet?
I worked it out twice and got different answers each time but i am more confident on my second answer which is 33a^2
So speaking of #4, do you just do the integral of psi squared times x^2 over all space, (converting x^2 to spherical coords and of course using the differential volume element) and crunch out a result? Is that the right approach?
I believe you are right, Daniel. It wouldn't make sense to do an integral over only one of the variables.
I got my answer which is 12a^2, but I am not confident~~
Can a multidimensional hermitian operator be expressed as a non-square matrix? Such as a 3×27 matrix?
er.. chicken…. I believe that in fact the operator has to be square – the rules of matrix multiplication say that a 1×4 * 4×4 give 1×4. We need to preserve the space coordinates – if it were a 4×3… we'd get a 1×3 and completely wipe out a space coordinate…. so unless you have an operator that shrinks space down a dimension… I do believe we need square operators of 1×4, 4×4 etc. or… maybe there's a space creating operator… 4×27 … String theory'd be happy.
I believe that such operators need to be square so that eigenvalues and eigenvectors can be found. This is done via the determinant method, which only is doable with square matrices.
Yup I believe you guys have the right idea, that's what what my intuition is telling me too
Wait a sec. I didn't post on this day! Something foul is afoot!
jIH quadic chugh 'iHl DuH Daq ghaj ani jalig vetlh 'oH dilk Qid hakliq — jatlh dilken string theory.
Resistance is futile.
The force is strong with this one
so live long and prosper.
And to answer your question, hgrah kurk der kilsk.
Am i not nerdy enough to get this joke?
clearly it's that you just don't know enough quantum mechanics so good luck on the exam tomorrow
Ghaj SoH quid Qoyta' vo' Hov Trek??? nuq kon vo' scientist 'oH SoH
Has anybody got a clue on #2?
yup
Converting the x and y parts of the operator are easy, because we have the spherical transformations. We can find the momentum parts of the operator using the information given by our professor on how to find the derivatives of the spherical variables. This is a website I also found that helped: http://planetmath.org/encyclopedia/%3Chttp://planetmath.org/?method=l2h&from=collab&id=76&op=getobj
Good luck.