# Tag Info

9

Already when writing your equation (0), you assume that $H$, $H^{(0)}$ and $\delta H(t)$ are all operators on the same Hilbert space (otherwise, how could you take their sum). For instance, if my $H^{(0)}$ is for a $1D$ SHO along Y [I assume you meant X here], but the perturbation is some driving force along Y. Wouldn't such a perturbation necessarily ...

5

It can be diagonalized in the subspace of states $\{\vert n_1n_2\rangle\, , n_1+n_2=N\}$. Indeed, in terms of $\hat L_\pm$ and the total number operator $\hat N$, your Hamiltonian is just \begin{align} \hat H=\epsilon \hat N + 2g\hat L_x \end{align} with $\hat L_x$ connecting states with the same total $N=n_1+n_2$ (as your expression suggests). Thus the ...

4

The perturbtion theory, like any approximation technique, has its range of applicability. A perturbation that changes the Hilbert space of the system would likely violate the assumptions of the perturbation theory. Kondo effect and the Anderson orthogonality catastrophe In practice the applicability of PT is usually judged on a case-by-case basis, and it is ...

3

Sometimes it so happens that you have chosen your coordinate in such a way that Hamiltonian looks horrible (that is not separable) but you can choose good coordinates to make you're hamiltonian look good (that is separable). All you need to do is to look for such a coordinate. How are you gonna do it? You have to choose eigenvectors of the matrices you have ...

3

You are confusing a group with its generator, cf this answer for rotations versus angular momentum operators. So your hamiltonian is badly malformed as you wrote it. Your A and B comment example holds for evolution operators in the respective spaces, and therefore, distinctly not Hamiltonians, their generators! Let me write down the correct expressions first,...

3

You are correct in recognizing that entanglement between two systems is different from the existence of an interaction between two systems. However, I think is a two-step argument that makes what Shankar is saying perfectly Kosher: If two systems start out as unentangled then if there is no interaction between them (i.e., if the Hamiltonian of the full ...

3

Technically you'd need to specify the domain on which these operators act and the inner product... Assuming the domain is the real line and the usual inner product $$\langle \phi\vert\psi\rangle = \int_{-\infty}^\infty dx\,\phi^*(x)\psi(x) <\infty$$ then the first clearly isn't because $\hat x$ and $\hat p$ are Hermitian so that \begin{align} \hat H^\...

2

You need to add the other term so that the Hamiltonian is still hermitian. The adjoint of $$\hat{H}\approx\hat{H}(x_c)+(\hat{x}-x_c)\frac{\partial\hat{H}}{\partial \hat{x}}\Bigg|_{x_c}$$ is $$\hat{H}^\dagger=\hat{H}(x_c)^\dagger+\Bigg((\hat{x}-x_c)\frac{\partial\hat{H}}{\partial \hat{x}}\Bigg|_{x_c}\Bigg)^\dagger=\hat{H}(x_c)+\frac{\partial\hat{H}}{\partial \... 2 Exponentials of operators are defined by their power series.$$ \exp(\hat{O}) = \sum_{n=0}^{\infty} \frac{\hat{O}^n}{n!}$$So your question boils down to: what are the exponentials of a projection operator? We immediately see that (i\omega t |e\rangle\langle e|)^n = (i\omega t |e\rangle\langle e|)\cdot(i\omega t |e\rangle\langle e|)\dots = (i\omega t)^n|e\... 2 Some choice of units has been made to simplify the analysis of the problem. You're already (presumably) comfortable with setting \hbar=1 and m=1/2; apparently some other constant in front of the x^3 term has also been set to 1. If it makes you more comfortable, you can use$$H = \frac{p^2}{2m} + i\alpha x^3$$for some constant \alpha which has ... 2 Actually, downfolding (see section 5.2) on its own does not require any approximations. It also does what you are looking for. One defines two projectors P, Q onto sub-Hilbert spaces, where we are interested in the subspace corresponding to P. Together P+Q = 1, i.e., they together project onto the full Hilbert space. One then gets an effective ... 2 I think your theorem is a thoroughgoing misconception of the p.d.e. factorization, of, e.g., a spherically symmetric system. Nondimensionalizing all silly constants by absorbing them in the relevant units, you have something like -\Delta +V(r) -E=0. Its eigenvectors are not the product of the eigenvectors of all symmetry generators (operators commuting ... 1 There is a very useful identity for exponentials of Pauli matrices (see https://math.stackexchange.com/questions/3236998/exponential-of-pauli-matrices/3237834 for a proof): \begin{eqnarray} e^{i\theta \hat{\bf n}\cdot \sigma} = \cos \theta I + i(\hat{\bf n}\cdot \sigma) \sin \theta \end{eqnarray} For your Hamiltonian \theta = -tg and \hat{\bf n}\cdot \... 1 \renewcommand{\ket}[1]{\left \lvert #1 \right\rangle} \renewcommand{\bra}[1]{\left \langle #1 \right\rvert} To my best guess, the problem started with a Hamiltonian from some bases \ket{1} and \ket{2}. Let me neglect the parameter a. It is irrelevant for now.$$\tag{1} H = \left(\ket{1} \bra{1} - \ket{2}\bra{2} -i \ket{1}\bra{2} + i \ket{2} \bra{1}\...

1

OP's 2 different conditions arise from using 2 different definitions of the Legendre transformation. One definition uses supremum while another definition uses substitution, cf. e.g. my Phys.SE answer here.

1

1. Are these state eigenstates? How can i calculate their energy? Act Hamiltonian on the states $\vert 00,00\rangle$ and $\vert 11,11\rangle$. The 4 number denotes occupation for site 1 spin-up, site 1 spin-down, site 2 spin-up, and site 2 spin-down: $$H=\sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma} + \epsilon_2 c_{2\... 1 It is fairly simple to see that they would be same.$$\begin{aligned} A_{s}(t) & = \text{exp}\left(-\imath H_{0} t\right)A_{s}\text{exp}\left(\imath H_{0} t\right) \\ & = \sum_{n,m}\left(-\imath\right)^{n}\left(\imath\right)^{m}\frac{1}{n!m!}H_{0}^{n}A_{s}H_{0}^{m} \\ & = \sum_{n,m}\left(-\imath\right)^{n}\...

1

You recall that $$PpP=-p; ~~~PxP=-x; \\ ~~~TxT= x; ~~~ TpT=-p; ~~~TiT=-i,\\ P^2=T^2=1\!\!1; ~~~[P,T]=0.$$ These two operators preserve the canonical commutation relations, separately. You may then see both the kinetic and potential terms of your hamiltonian are PT symmetric, $$PT~ix~PT= ix.$$ Yes, real for the (divergent) ground state energy (only). Write ...

1

That a system is not in an eigenstate of an observable, and the dispersion turns out to be zero, should always surprise you and make you suspicious enough to go back to your calculations, to see what you got wrong. In your case, the reason is confusing $\langle H^{2}\rangle$ with $\langle H\rangle^{2}$. You will notice that,  \langle H\rangle^{2}\neq\sum_{...

1

Yes, in the simplified model of time-independent Hamiltonian $H$ effecting a quantum gate $U=\exp(-iHt)$ multiplying $H$ by a constant factor $\gamma > 1$ does shorten the gate's duration. In practice, physics constants, material properties, control electronics etc constrain what Hamiltonians can be engineered on any given hardware platform. For example, ...

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