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The Schrodinger and Heisenberg pictures make sense to me. But the interaction picture which is a hybrid of the two does not.

Author of this text first splits the Hamiltonian up as

$$H=H_0+H_{int}$$

The time dependence of operators is governed by $H_0$ while the time dependence of states is governed by $H_{int}$.

(Question, does this say that the Hamiltonian is broken into a time dependent $H_0$ and a time independent $H_{int}$ parts?)

The states and operators in the interaction picture is denoted by a subscript $I$:

$$|\psi(t) \rangle_{I}=e^{iH_{0}t}|\psi(t) \rangle_{S}$$

$$\mathcal{O}_{I}(t)=e^{iH_{0}t}\mathcal{O}_{S}e^{-iH_{0}t}$$

The last equation applies to $H_{int}$, which is time dependent

(Question, is this a typo? should it be time independent?)

The interaction Hamiltonian in the interaction picture is then

$$(H_{int})_{I}=e^{iH_{0}t}(H_{int})_{S}e^{-iH_{0}t}$$

$$H_{I}=(H_{int})_{I}$$

(Question, this implies that $(H_0)_I=0{}$, what sense this notation makes? is it that time dependent part of the Hamilotinain in the interaction picture vanishes? but why)

This result is used to find the Schrodinger equation for states in the interaction picture

$$i\frac {d|\psi\rangle_{I}}{dt}=H_{I}(t)|\psi\rangle_{I}$$.

(Question, this also assumes that $(H_0)_{S}=H_0$, what sense this notation makes?) Perhaps an overall question that relates all confusion to each other. When **we do not use any of the subscripts, we are in which pictures? $S, H, I$?

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Author of this text first splits the Hamiltonian up as

$$H=H_0+H_{int}$$

The time dependence of operators is governed by $H_0$ while the time dependence of states is governed by $H_{int}$.

(Question, does this say that the Hamiltonian is broken into a time dependent $H_0$ and a time independent $H_{int}$ parts?)

No, not necessarily, the full Hamiltonian is broken into two parts arbitrarily (i.e., in a way that you get to choose).

The states and operators in the interaction picture is denoted by a subscript $I$:

$$|\psi(t) \rangle_{I}=e^{iH_{0}t}|\psi(t) \rangle_{S}$$

$$\mathcal{O}_{I}(t)=e^{iH_{0}t}\mathcal{O}_{S}e^{-iH_{0}t}$$

The last equation applies to $H_{int}$, which is time dependent

(Question, is this a typo? should it be time independent?)

No, not necessarily. It will only be time-independent if it it: 1) Commutes with the unperturbed Hamiltonian; and 2) has no explicit time-dependence. Typically the first condition is not satisfied, so the interaction Hamiltonian in the interaction picture does have time dependence.

The interaction Hamiltonian in the interaction picture is then

$$(H_{int})_{I}=e^{iH_{0}t}(H_{int})_{S}e^{-iH_{0}t}$$

$$H_{I}=(H_{int})_{I}$$

**(Question, this implies that $(H_0)_I=0{}$,

No, it is not zero. This should say $(H_0)_I=H_0$. Because $H_0$ commutes with $e^{iH_0t}$.

what sense this notation makes? is it that time dependent part of the Hamilotinain in the interaction picture vanishes? but why)**

As explained above, what you wrote above is incorrect.

This result is used to find the Schrodinger equation for states in the interaction picture

$$i\frac {d|\psi\rangle_{I}}{dt}=H_{I}(t)|\psi\rangle_{I}$$.

**(Question, this also assumes that $(H_0)_{S}=H_0$, what sense this notation makes?)

$(H_0)_H=e^{iAt}H_0e^{-iAt}$, where $A=H$. So, typically $(H_0)_H\neq H_0$ because $H_0$ doesn't typically commute with $H$.

$(H_0)_I=e^{iAt}H_0e^{-iAt}$, where $A=H_0$. So, $(H_0)_I=H_0$ because $H_0$ commutes with itself.

$(H_0)_S=e^{iAt}H_0e^{-iAt}$, where $A=0$. So, $(H_0)_S=H_0$ because $H_0$ commutes with $1$ (i.e., $e^0$).

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  • $\begingroup$ Thanks all clear now. Also it seems that this equation is indeed a convention on notation, right? $$H_{I}=(H_{int})_{I}$$ $\endgroup$ – user56963 Mar 12 '15 at 5:49
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$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\timex}[1]{\mathrm{e}^{\mathrm{i} #1 t}}\newcommand{\ntimex}[1]{\mathrm{e}^{-\mathrm{i} #1 t}}$The interaction picture split of the Schrödinger picture Hamiltonian

$$ H_S = H_{0,S} + H_{\text{int}_S} $$

has nothing to do with the Hamiltonians themselves being time (in)dependent. In fact, the choice of $H_{0,S}$ and $H_{\text{int},S}$ as summands of the total Hamiltonian $H_S$ is totally arbitrary, and every choice yields a "different" interaction picture. It is common, however, to choose $H_{0,S}$ as a free or unperturbed Hamiltonian and $H_{\text{int},S}$ as the terms that contain the interactions or perturbations from a simpler system.

Now, in the interaction picture, a bit confusingly, all the states and operators are time-dependent, in contrast to the Heisenberg and Schrödinger pictures. Starting from the Schrödinger picture with states $\ket{\psi_S(t)}$ and time-independent operators $A_S$, we define

$$ A_I(t) = \timex{H_{0,S}}A_S\ntimex{H_{0,S}}$$

$$ \ket{\psi_I(t)} = \timex{H_{0,S}}\ket{\psi_S(t)}$$

Where the first definition makes it so operators now evolve by $H_0$ and the second "removes" the evolution by $H_0$ from the evolution of the Schrödinger state, leaving the evolution by $H_\text{int}$. The equations of motion reflect that - they are Schrödinger for the states with Hamiltonian $H_\text{int}$ and Heisenberg for the operators with Hamiltoian $H_0$.

Furthermore, $H_{0,I}(t) = H_{0,I}(0) = H_{0,S}$ at all times, since $H_0$ commutes with its own exponential, so we need not distinguish between $H_{0,S}$ and $H_{0,I}$.

When we don't use subscripts, we are being sloppy and assume the picture is understood from context.

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  • $\begingroup$ Right. What I'd highlight is that the eigenfunctions of an operator are generally complete which means you can take any wavefunction $|\Psi\rangle$ and express it in terms of the basis functions $|\chi_i\rangle$ of a totally unrelated Hamiltonian $\hat h$ -- e.g. you can write particle-in-a-box states as a superposition of quantum harmonic oscillator states. The "interaction picture" for $\hat h$ is, as you say, just transforming into the Heisenberg coordinates for $\hat h$. $\endgroup$ – CR Drost Mar 11 '15 at 19:48
  • $\begingroup$ In the beginning, can we just split as well the Heisenberg picture? and have the same result? $\endgroup$ – user56963 Mar 12 '15 at 5:54

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