# Interaction (Dirac) picture and time dependent perturbation theory

Why in many, if not all, references that discuss the time dependent perturbation theory, they start the discussion with the interaction (Dirac) picture, although, what we need is only solving the time dependent Schrodinger equation?

In another way:

Why these references do not start with the time dependent Schrodinger equation? why we need to discuss the interaction (Dirac) picture to explain the time dependent perturbation theory?

Do I need to explain the interaction (Dirac) picture in order to explain the time dependent perturbation theory, or I can start with time dependent Schrodinger equation?

I hope I am clear in conveying my question.

• But... the Dirac representation is the best way of solving the TDSE! – Cosmas Zachos Dec 6 '19 at 20:06
• I did not get it, any detailed explaination will be appreciated. – Logic Dec 6 '19 at 20:08
• Your text should explain that, if it were any good. Solve a simple problem in all three pictures, and compare. – Cosmas Zachos Dec 6 '19 at 20:14

Start with the time-dependent Schrodinger equation $$i\hbar\frac{d}{dt}\vert\psi(t)\rangle=H\vert \psi(t)\rangle\, , \tag{1}$$ and assume $$U(t)$$ so that $$\vert \psi(t)\rangle =U(t)\vert\psi(0)\rangle \tag{2}$$ Insert (2) in (1) to get \begin{align} i\hbar \frac{d}{dt}U(t) \vert\psi(0)\rangle&=H U(t)\vert\psi(0)\rangle\, ,\\ \frac{dU}{dt}&=-\frac{i}{\hbar} HU(t) \tag{3} \end{align} If $$H$$ does not depend on time then by inspection $$U(t)=e^{-i Ht/\hbar}$$ satisfies (3). However, if $$H(t)$$ does depend on time, it is NOT possible to directly integrate the right and side of (3), i.e. write the evolution operator as $$U(t)=e^{-i \hat H(t)/\hbar}$$ and one must instead solve (3) as an integral equation: \begin{align} U(t)=-\frac{i}{\hbar}\int_0^t dt’ H(t’)U(t’) \tag{4} \end{align} where $$U(0)=\hat 1$$ has been used. This is difficult to bring to a series solution because there is no natural small expansion parameter: $$H(t)$$ is the full Hamiltonian so the matrix elements are not expected to necessarily be small.

In essence the interaction picture looks for an evolution in the form $$U=e^{-i H_0 t/\hbar}U_I(t) \tag{5}$$ where $$H(t)=H_0+\epsilon V(t)$$, with $$\epsilon$$ small. If we insert this into the Schrodinger equation we get \begin{align} i\hbar e^{-i H_0 t/\hbar} \left(-\frac{i}{\hbar} H_0 U_I(t) + \frac{dU_I}{dt}\right)&=\left(H_0+\epsilon V(t))\right)e^{-iH_0t/\hbar}U_I(t)\, ,\\ i\hbar e^{-iH_0t/\hbar}\frac{dU_I}{dt}&=\epsilon V(t)e^{-iH_0t/\hbar}U_I(t)\, ,\\ i\hbar \frac{dU_I}{dt}&=\epsilon e^{iH_0t/\hbar} V(t) e^{-i H_0t/\hbar}U_I(t)\, ,\\ &=\epsilon V_I(t)U_I(t) \tag{6} \end{align} The ansatz (5) has eliminated $$H_0$$, assumed to be the dominant part of $$H$$: the right hand side of (6) now depends on the small parameter $$\epsilon$$ - unlike the RHS of (3) - so it is possible to start an expansion for $$U_I(t)$$ in powers of $$\epsilon$$ and solve $$U_I$$ iteratively order by order in $$\epsilon$$. The “cost” is the transformation $$V_I(t)=e^{iH_0t/\hbar}V(t)e^{-i H_0 t/\hbar}$$ which may not be trivial to evaluate and indeed might have to be evaluated using the usual expansion in nested commutators $$e^A B e^{-A}= B+[A,B]+\frac{1}{2!}[A,[A,B]]+\ldots$$

Disclaimer: I don't know any of the proper functional analysis to make these arguments rigorous. This is going to be very "physicists attempting math" so follow at your own risk. In fact, this is an argument I've sort of made up myself so there might be some glaring issue with it and I would be happy to be corrected if that is the case.

I follow the arguments in wikipedia for Dyson Series a bit so there may be more/better explained detail there.

You are correct. There is no need whatsoever to go into the interaction picture. In particular, for typical situations there is no actual need for "small expansion" parameters.

Consider briefly the expansion

\begin{align} e^x = \sum_{n=0}^{\infty} \frac{1}{n!} x^n \end{align}

You might naively think that for the sum to converge it is necessary for $$|x|<1$$. However, we know that this Taylor series converges for any value of $$x$$. This is because $$n!$$ grows faster than $$x^n$$ for any $$x$$. The argument for the Dyson series will follow similarly.

We have \begin{align} \frac{d}{dt}U(t) = \left(-\frac{i}{\hbar}\right) H(t)U(t) \end{align}

We integrate and iterate

\begin{align} U(t_0) =& U(0) + \left(-\frac{i}{\hbar}\right)\int_{t_1=0}^{t_0}dt_1 H(t_1)U(t_1)\\ =& I + \left(-\frac{i}{\hbar}\right)\int_{t_1=0}^{t_0} dt_1H(t_1) + \left(-\frac{i}{\hbar}\right)^2\int_{t_1=0}^{t_0}\int_{t_2=0}^{t_1} dt_1 dt_2 H(t_1)H(t_2)U(t_2)\\ \end{align}

Each term of the continued series can be written as \begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n\int_{t_1=0}^{t_0}\ldots\int_{t_n=0}^{t_{n-1}}dt_1\ldots dt_n H(t_1)\ldots H(t_n) \end{align}

We can define a remainder term

\begin{align} R_n = \left(-\frac{i}{\hbar}\right)^{n+1}\int_{t_1=0}^{t_0}\ldots\int_{t_n=0}^{t_{n-1}}\int_{t_{n+1}=0}^{t_n}dt_1\ldots dt_n dt_{n+1} H(t_1)\ldots H(t_n) H(t_{n+1}) U(t_{n+1}) \end{align}

We then have

\begin{align} U(t) = \sum_{n=0}^N U_n(t) + R_N(t) \end{align}

Note that $$t_n \le t_{n-1} \le \ldots \le t_2 \le t_1$$

This means

\begin{align} H(t_1)\ldots H(t_n) = \mathcal{T}(H(t_1)\ldots H(t_n)) \end{align}

So we have that

\begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n\int_{t_1=0}^{t_0}\ldots\int_{t_n=0}^{t_{n-1}}dt_1\ldots dt_n \mathcal{T}(H(t_1)\ldots H(t_n)) \end{align}

Note now that the integrand is symmetric in the time argument.

Consider now the related but different integral

\begin{align} K_n = \left(-\frac{i}{\hbar}\right)^n\int_{t_1=0}^{t_0}\ldots\int_{t_n=0}^{t_0}dt_1\ldots dt_n \mathcal{T}(H(t_1)\ldots H(t_n)) \end{align}

This is an integral over a hypercubic region with one corner at $$t=0$$ and one at $$t=t_0$$. This region can be broken up into $$n!$$ regions which have the same size but different time orderings as the integrals in $$U_n(t)$$. Because the integrand is symmetric the value is the same in all of these different regions. we thus have

\begin{align} U_n(t) = \frac{1}{n!} K_n(t) \end{align}

This is beginning to look a bit like the exponential series I introduced initially. We now suppose the operator $$H(t)$$ is a bounded operator in some sense. Roughly this could mean its largest eigenvalue is finite. We notate this by

$$|H(t)| \le M$$

Where $$M$$ is a positive real number (with dimensions of energy). It then follows that

\begin{align} |K_n(t)| \le \frac{1}{n!}\frac{M^n t_0^n}{\hbar^n} \end{align}

This follows because the integrand includes $$n$$ factors of $$H(t)$$ and the volume of the integration region is $$t_0^n$$. Similarly the remainder term

\begin{align} |R_n(t)| \le \frac{1}{(n+1)!}\left(\frac{Mt_0}{\hbar}\right)^{n+1} \rightarrow 0 \end{align}

as $$n\rightarrow \infty$$ no matter the value of $$t_0$$.

This means the series

\begin{align} |U(t_0)| = \bigg|\sum_{n=0}^{\infty} U_n(t_0)\bigg| \le \sum_{n=0}^{\infty} \frac{1}{n!} \left(\frac{M t_0}{\hbar}\right)^n = e^{\frac{Mt_0}{\hbar}} \le \infty \end{align}

so the series converges*

Ok, this is possibly very crude and handwaivey but I think the jist of the argument holds. That is, the Dyson series converges nicely even if the Hamiltonian which we are expanding in is not small.

Again.. this argument may not be correct so I'd wait to hear from those better versed in these matters before ruling on this answer.

*very (physicists attempting math-ey)

edit: And to directly answer your question as to why references always do include the interaction picture stuff? I think it is because in practice the sorts of time-dependent Hamiltonians which arise in, for example, atomic physics, it is simply the case that there is a time-independent and large static Hamiltonian $$H_0$$ and a small-time dependent Hamiltonian $$V(t)$$. In that case the calculations are simplified by first moving into the interaction picture. However, I do think it is correct that one could teach time-dependent perturbation theory as a general mathematical method for solving a general time-dependent Schrodinger equation.