This is simply an identity of Taylor expansion and has nothing to do with the fact that you have operators around, if $$f(x)=\sum_n \frac{f^{(n)}(0)}{n!}x^n$$ then $$\frac{df}{dx} = \sum_n \frac{f^{(n+... View answer Accepted answer 3 votes The gist of such self-consistent is the (implied) assumption of a unique solution. If we indeed have a unique solution then the self-consistent solution we find under certain assumptions is indeed the ... View answer Accepted answer 3 votes At low temperatures (\beta\to\infty) the partition function is dominated by the ground state and the average energy should be the ground state energy. This should be correct when taking the limit on ... View answer 3 votes Dot product of u with itself is constant:$$u\cdot u=-c^2$$Taking derivative with respect to proper time on both sides gives$$u\cdot \frac{du}{d\tau}=u \cdot a=0$$View answer 2 votes The short answer for 1: you can't. Long answer: While the NS are exact equations, getting useful information about transport in air strictly from the equations is unfeasible. The reason is that the NS ... View answer 1 votes The PEP is a direct result of the fermionic statistics of the neutrons. The fact that they are fermions mean that if we write a multiparticle wave function it must satisfy$$\psi(x_1,...,x_i,...,x_j,.....

The relevant matrix that you need to diagonalize is not $\delta H$ itself, but its projection onto each one of the degenerate subspaces of $H_0$. Since the $E=1$ subspace is generated by $v_1=(1,0,0,0)... View answer Accepted answer 0 votes This is not a physics application per se, but I remember that as a first year undergrad I was convinced in the usefulness of matrix diagonalization by the application to obtain a closed expression for ... View answer 0 votes For the Cauchy-Schwartz inequality you need to have some inner product that you can use. The inner product you used $$\left\langle f,g\right\rangle=\int f(x)g(x)dx$$ is valid, but the definition $$\... View answer 0 votes My guess is that by "violation of HUP" you mean that$$\left\langle \theta^2\right\rangle\left\langle L_z^2\right\rangle=0$$The reason for that is that while we can define the operator$L_z=...