 playing the role of Doctor. This talented music director is surely going to add his jingles and theme music to make the show even more interesting. Gold even spoke about his plans to create new theme music for Doctor Who. He has already got over the discussion phase and has even started to work on the project. Viewers of Doctor Who are in for a treat with Gold trying his best to recreate the old magic!
Source: Music From The Movies
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This entry was posted on Tuesday, January 5th, 2010 at 2:06 am and is filed under All News, Doctor Who, TV Shows. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.
January 5th, 2010 at 3:07 am
I just saw Doctor Who: The End of Time Part 2 and it was amazing! It is a little sad to see David Tennant go but I did some research on Matt Smith and I have high hopes for him.
I’m like a little kid in a candy shop every time I hear there is a new episode of Doctor Who out.
February 9th, 2010 at 5:14 am
It always amazes me when folks talk about how ‘amazing’ a series fanale is when talking about DW. My own opinion is that these storylines are the most contrived, masturbatory plotlines the series writers ever seem to come up with.
Strangely, as far as writing goes, I’ve always found the ‘Doctor-lite’ stories to be amazingly well-written (more so than some other series stories)–no doubt because they must eliminate silly plot devices like sonic screwdrivers that solve any dilemma. Don’t get me wrong: I love the Doctor, with or without all his gadgets. But I just don’t see a diagetic reason to include Sarah-Jane, Luke, Mickey, Martha, or any of the characters “resurrected” in the final 15 minutes of “End of Time”.
Murray Gold, however, has been doing a great job with incidental music and variations on the DW theme. I am glad to hear he is back on board for series five!With the Democrats need to ram their filthy legislation down America's throat they continually dust off Ted Kennedy from his deathbed and shuttle him into Washington for his vote (they will probably continue this after his passing because we all know they love the "dead vote") like the movie character Bernie.
With the democracts nearing the "filabuster" proof Senate (except for Minnesota and the pesky illness of Kennedy) they are desperate in their attempts to keep the "bullet train moving forward at breakneck speed" of liberal socialist change.
The House and Senate just sailed through Obama's $3.55 trillion budget (once again shuttling in the ill Kennedy for his much needed "hanging chad") without a single republican vote in both chambers.
The democrats own this legislation lock-stock-and-barrell. In case you missed it, the legislation also includes a federal hate crime provision that now protects necrofiliacs, pedophiles and the like. That's right, protects their rights.
I guess the socialists in charge need their massive overhaul of the health care industry so they can get their hearing checked because it is obvious they didn't hear a single thing we screamed at them from the tea parties.Q: Do periodic boundary conditions affect the k-space Hamiltonian for the Hubbard model? In real space, the Fermi-Hubbard model can be written as:
$$ H = -t \sum_{\langle i,j \rangle\sigma} (c^\dagger_{i\sigma} c_{j\sigma} + c.c.) + U \sum_in_{i\uparrow}n_{i\downarrow}$$
The only difference between having periodic boundary conditions and not having them is that the nearest-neighbor pair, $\langle i, j \rangle$ will also include the pair $(N, 1)$ for periodic boundary conditions.
If we apply a transform to move into k-space, the real space summation really comes into effect when simplifying the exponential (in the form: $ e^{i(k-k')\cdot r_i}$). (I'm assuming that the sites are equally spaced.) I'm wondering if the k-space Hamiltonian is the same, whether or not we have periodic boundary conditions in real space. It seems that the only difference is that our sum over all real sites, $i$, is restricted from 1 to $N$ in the periodic boundary condition case (in order to include $(N, 1)$ at the end, but goes from 1 to $N-1$ in the non-periodic boundary condition case.
The delta function comes (for example in the kinetic term) from:
$$ -\frac{t}{N^2}\sum_{i \sigma}\sum_{kk'}e^{-i(k-k')r_i}e^{ik'\delta}a^\dagger_{k\sigma}a_{k'\sigma} +...$$
where the $a$'s are my k-space creation/annihilation operators. I'm assuming $\sum_ie^{-i(k-k')r_i} = N\delta_{kk'}$ regardless of whether we sum $i$ to $N$, or $i$ to $N-1$.
Is this true? Is there also a way to physically think about this, to justify whether the k-space Hamiltonian should change?

A: Short answer: yes and no.
Please keep in mind that $\sum_i e^{-i (\vec{k}-\vec{k}\prime) \cdot \vec{r}_i} = N \delta_{\vec{k} \vec{k}\prime}$ is not always valid in periodic boundary conditions, i.e., it implies that $\langle \vec{k} | i \rangle = e^{i \vec{k} \cdot \vec{r}_i}$, which is not true, since the electrons described by the Hubbard model are definitely not free electrons that can be described by plane waves!
Instead, you need to involve the Bloch theorem $\psi_i(\vec{r}+\vec{G})=e^{i \vec{k} \cdot \vec{G}} \psi_i(\vec{r})$ where $\vec{G}$ is an integer multiple of the lattice vectors, and $\psi_i(\vec{r}) = \langle \vec{r} | i \rangle$ is the electron wavefunction in real space for an electron occupying site $i$.
$\langle \vec{k} | i \rangle$ actually represents a basis transformation from $\vec{k}$-space to the local basis {i} space. The annihilation and creation operators are also transformed according to $\hat{a}_{\vec{k}\sigma}=\sum_i \langle \vec{k} | i \rangle \hat{c}_{i \sigma}$ and $\hat{a}^\dagger_{\vec{k}\sigma}=\sum_i \langle i | \vec{k} \rangle \hat{c}^\dagger_{i \sigma}$. As expected, these operators anticommute in both bases, $[\hat{c}_{i\sigma},\hat{c}^\dagger_{j\sigma}]_{+} = \delta_{ij}$ and $[\hat{a}_{\vec{k}\sigma},\hat{a}^\dagger_{\vec{k}\prime\sigma}]_{+} = \delta_{\vec{k}\vec{k}\prime}$
In fact, in $\vec{k}$-space, the kinetic term of the Hubbard model takes the form of $\sum_\sigma \epsilon(\vec{k}) \hat{a}^\dagger_{\vec{k}\sigma} \hat{a}_{\vec{k}\sigma}$, and $\epsilon(\vec{k})$ depends on the geometry of the system. For instance, with the 1-D Hubbard model (assume monoatomic chain), $\epsilon(\vec{k})=-2t \cos(|\vec{k}|a)$, where $a$ is the lattice constant. Thus, it indeed changes going from the local basis {i} to momentum basis $\vec{k}$.
(Side note: I wish I can send you my professor's lecture notes on this topic, where he lays out the algebra for the basis transformation.)
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