Recent content by CMJ96

Hi, a field operator has been used and split up into two parts, the condensate part and the non-condensate part ## \hat{\psi}= \hat{\phi} + \hat{\delta} ## , ##\hat{ \phi}## is the condensate and ##\hat{\delta} ## is the non-condensate. A Bogoliubov approximation was made so ##\hat{\phi}##...

Homework Statement I'd like to diagonalise the following Hamiltonian for quasiparticle excitations in a Bose Einstein Condensate $$H= K_0 + \hat{K}_1 + \hat{K}_2 $$ where $$K_0 = \int d^3 r \left[ \phi_0 ^* (\hat{h}_0- \mu) \phi_0 + \frac{g}{2} |\phi_0| ^4 \right]$$ $$\hat{K}_1= \int d^3 r...

Ahhh yes, I understand now, thank you very much, you have been very helpful

This is a more concise way of doing it, how exactly does it immediately reduce?

Thank you for the help and I apologise for taking so long to reply Your help was very useful, I used the commutation relation $$\left[a_k,a_i ^{\dagger} \right] = a_k a_i ^{\dagger} - a_i ^{\dagger} a_k = \delta_{ki} $$ Rearranging this and replacing ##a_k a_i ^{\dagger} ## (and using the...

Hi, to help further my understanding of the second quantization for one of my modules I would like to show that the following expressions $$ \hat{H} = \Sigma_{ij} \langle i| \hat{T} | j \rangle \hat{a_i }^{\dagger} \hat{a_j} $$ $$\hat{\psi}(r,t)= \Sigma_k \psi_k(r) \hat{a}_k(t)$$ Obey the...

Hi I've gone away and had a think about this, and now I feel I understand what is happening pretty well, however I'm still struggling with applying it. $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{\sqrt{3}}{2} & \frac{-1}{2} \\ 0 & 0 & \frac{1}{2} & \frac{\sqrt{3}}{2}...

So when A is introduced, would it be another 4x4 unitary matrix (assuming the control U gate is a 4x4 matrix)?

Hello I have the following quantum circuit identity for converting a controlled U gate (4x4 matrix) into a series of CNOT gates and single qubit gates $$ U= AXA^{\dagger}X$$ where A is a unitary matrix. Here is a picture of the mentioned identity. Can someone help me understand conceptually...

Ahhh yes, I have applied my method to the 3x3 block in the ##Q## matrix, the ##U_1##, ##U_2## and ##U_3## when multiplied together give ##I_n## which looks good, as it happens ##U_1## and ##U_2## don't change when taking ##U_i^{\dagger}## so this has simplified things a bit. I probably should...

I see, should I only apply my 3x3 method of decomposition to the ##Q_s## matrix?

Homework Statement I want to decompose the following matrix into a product of two level matrices ##V_i## $$ \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & \frac{-\sqrt{3}}{2} & 0 & \frac{-1}{2} \\ \frac{\sqrt{3}}{2} & \frac{-1}{4} & 0 & \frac{\sqrt{3}}{4} \\ \frac{1}{2} & \frac{\sqrt{3}}{4} & 0 &...

I've got it now, thank you very much for the assistance!

Ah yes, sorry, I forgot to include the 4th term in my reply, I have added it now, the denominator should be ##(1-N^2)^{\frac{1}{2}} ## but for some reason latex won't display it

Ahhhh yes, silly mistake... now I'm getting the correct second term (I just have to use the trig identity to tidy it up) , the only issue is that the first term is slightly wrong, I can't see how I can get the ## (1-N^2)^{\frac{1}{2}} ## as a denominator $$\frac{d^2 \theta}{dt^2} = -E_c E_J...