Recent content by barefeet

Homework Statement In a youtube video() it is explained how gaussian beams propagate through an optical lens. Using the complex parameter q \frac{1}{q} = \frac{1}{R} - \frac{j\lambda}{\pi n w^2} (with R the radius of curvature), one can use the ABCD matrix to calculate the effect of an optical...

Homework Statement In a book I find the following derivation: \int (J \cdot \nabla ) \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'}= -\sum_{i=1}^3 \int J_i \frac{\partial}{\partial r_i'} \frac{\bf{r} - \bf{r}'}{|\bf{r} - \bf{r}'|^3} d^3\mathbf{r'} \\ = -\sum_{i=1}^3 \int J_i...

I guess: S_{x,L}S_{x,R} + S_{y,L}S_{y,R} = \frac{1}{2}(S_+ + S_-)_L\frac{1}{2}(S_+ + S_-)_R + \frac{1}{2i}(S_+ - S_-)_L\frac{1}{2i}(S_+ - S_-)_R \\ = \frac{1}{4} ( S_+S_+ + S_+S_- + S_-S_+ + S_-S_- - S_+S_+ + S_+S_- + S_+S_- - S_-S_- ) =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R}) Or do you...

Ok, S_{x,L}S_{x,R} + S_{y,L}S_{y,R} = \frac{1}{2}(S_+ + S_-)_L\frac{1}{2}(S_+ + S_-)_R + \frac{1}{2i}(S_+ - S_-)_L\frac{1}{2i}(S_+ - S_-)_R \\ =\frac{1}{2}(S_{+,L}S_{-,R} + S_{-,L}S_{+,R} ) But I still get the same answer. Whiech term exactly should I recheck?

Ah I see, I am missing a factor \frac{1}{2} in the S_x term and a \frac{1}{2i} in the S_y term So it will be: H \mid \uparrow \uparrow \rangle = g\mu_BhBh \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \uparrow \uparrow \rangle + \frac{Jh^2}{4} \mid \downarrow \downarrow \rangle -...

Ah ok so I have to construct H as: H = \left( \begin{array}{ccc} \langle \uparrow \uparrow \mid H \mid \uparrow \uparrow \rangle & \langle \uparrow \uparrow \mid H \mid \uparrow \downarrow \rangle & \langle \uparrow \uparrow \mid H \mid \downarrow \uparrow \rangle & \langle \uparrow \uparrow...

Where do I go wrong? Is my expression for the Hamiltonian(first line) at least correct? And am I missing an identity operator in the expressions and is B S_{L,z} actually B S_{L,z} 1\!\!1_R . But that wouldn't change much. Then: H \mid \uparrow_L \uparrow_R \rangle = \left(\begin{array}{c}...

Ok, but then would it still be the same except that the hamiltonian is not a vector so just a summation of all the terms? Like: H = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} + J\mathbf{S}_{L,y} \mathbf{S}_{R,y} + B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z} Or is this...

You mean like H_x = J\mathbf{S}_{L,x} \mathbf{S}_{R,x} \\ H_y = J\mathbf{S}_{L,y} \mathbf{S}_{R,y} \\ H_z = B(\mathbf{S}_{L,z} + \mathbf{S}_{R,z}) + J\mathbf{S}_{L,z} \mathbf{S}_{R,z} ?

Homework Statement Consider two spins, L and R, in a magnetic field along the z-axis, i.e. B = (0, 0, B) . The magnetic moments of the two spins are coupled to each other so that the total Hamiltonian reads H = g\mu_B\mathbf{B}\cdot(\mathbf{S}_L + \mathbf{S}_R) + J \mathbf{S}_L\cdot...

Homework Statement Using J^2 \mid j,m_z \rangle = h^2 j(j+1) \mid j,m_z \rangle J_z \mid j,m_z \rangle = hm_z \mid j,m_z \rangle Derive that : \langle j,m_z \mid J_-J_+ \mid j,m_z \rangle = h^2[ j(j+1) - m_z(m_z+1)] Homework Equations J_- = J_x - iJ_y J_+ = J_x + iJ_y The...

Homework Statement Consider three electrons in three different orbital levels n, m and p. We assume that there is one electron in each orbital level. How many states are then possible? Homework Equations Equations for constructing symmetric and asymmetric wavefunctions: Symmetric under pair...

13.6 eV is just the result of all the constants in the E_n term. Off the top of my head, I think you can write : E_n = \frac{-13.6}{n^2} But do check this

Yes, I could have guessed that, but I wasn't trying to prove QM wrong. That QM is wrong in relativistic situations, doesn't mean that all the principles of QM are incompatible with SR/GR. It is interesting to see what exactly is incompatible with SR/GR and in this case I was asking about the...

This is the answer to (a part of) my question. Nevertheless there are still a few cases where my question can be relevant. If there is no upper limit for the relation between measurement time and technique, I still see a problem. Even though you can't measure infinitely precise within a finite...