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## New answers tagged hamiltonian

0

In Lagrangian/Hamiltonian the crucial property of the Legendre transform is that it preserves all information. One way to demonstrate that the Legendre transform preserves all information is to show that it is its own inverse. When the Legrendre transform is applied twice you are back to the starting point. The required properties of the transform are very ...

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 ...

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 ...

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}{\left \lvert #1 \right\rangle}\renewcommand{\bra}{\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. $$\tag{1} H = \left(\ket{1} \bra{1} - \ket{2}\bra{2} -i \ket{1}\bra{2} + i \ket{2} \bra{1}\... 2 You need to add the other term so that the Hamiltonian is still hermitian. The adjoint of$$\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 \... 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 ... 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: $$H=\sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma} + \epsilon_2 c_{2\... 0 The proportionality constant follows from the properties of the delta distribution. Step 1: Substitute the expression of the energy in terms of the momentum in the second orthonormalization relation: \begin{equation} \delta(E-E') = \delta\left(\frac{p^2}{2m}- \frac{p'^2}{2m}\right) =\delta\left(g(p)\right) \end{equation} Here I am considering the \delta to ... 1 It is fairly simple to see that they would be same.$$\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}\... 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\...

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 ...

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 ...

0

I am going to try to answer my own question here, since I think I now understand what I was missing before. We want to understand the link between commutation relation and factorizability of the eigenfunction. This link is all about the definition of separable Hamiltonians and about a proper interpretation of the meaning of commutation relations: Think about ...

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 ...

-2

For the wave function to be a product, it is necesary and sufficient to deal with separated variables in the Schroedinger equation. If the operators $\hat{O}_n$ commute with $\hat{H}$, then $\Psi_E$ may be an eigenstate for each operator in question.

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 ...

0

When you have a time dependent hamiltonian you can always define the instantaneaous eigenstates and eigenvalues (in literature sometimes are called adiabatic eigenstates). The question "can we still define eigenvalues and eigenstates" is a bit confused because of course you can but differently from the time-independant hamiltonian those are not ...

0

The sum over the single particle operator $\vec{A}\vec{\nabla_i}$ can be written as follows $$\sum_{k,q} \langle k+q| \vec{A} \vec{\nabla} |k \rangle c_{k+q}^{\dagger}c_{k}.$$ Using the above equation and inserting the completeness relation into the hamilonian yields $$H=\frac{ie\hbar}{m} \sum_{\vec{k},\vec{q}} \sum_{r} \langle k+q| r\rangle \langle r |\... 1 That a system is not in an eigenstate of an observable, and the dispersion turns out to be zero, should always surprise you and make you suspicious enough to go back to your calculations, to see what you got wrong. In your case, the reason is confusing \langle H^{2}\rangle with \langle H\rangle^{2}. You will notice that,$$ \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, ...

0

If you are working on the entire real line, you can use Fourier transforms to get a matrix with a continuous infinity of of rows and columns. If on a finite interval commensurate with the period of the sine, use a Fourier series. In each case the matrix will be infinite.

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The Feynman lectures volume 3 notes on chapter 8 are in the colloquial style of the lectures. In the relevant part of the text, the author is communicating that $U_{ij}(t_2,t_1)$ are quantities about which we don't have much intuition except that $\Delta t \rightarrow 0$ implies that $U_{ij} = \delta_{ij}$. As a matter of definition, $$U_{ij}(t_2+\Delta t,... 3 Technically you'd need to specify the domain on which these operators act and the inner product... Assuming the domain is the real line and the usual inner product$$ \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^\... 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,... 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 useH = \frac{p^2}{2m} + i\alpha x^3 for some constant $\alpha$ which has ...

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 ...

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