# Dummy variables in Dyson series

In the Dyson series, it is known that: \begin{align} {\cal T}\exp\left[-\frac{i}{\hbar}\int_0^tH(t')dt'\right] &= I - \frac{i}{\hbar} \int_{0}^{t} dt' H(t') + \left(-\frac{i}{\hbar}\right)^2 \frac{1}{2} \mathcal{T}\left(\int_{0}^{t} dt' H(t')\right)^2 \\& \quad + \left(-\frac{i}{\hbar}\right)^3 \mathcal{T}\left(\frac{1}{3!}\left(\int_{0}^{t} dt' H(t')\right)^3\right) \cdots \\ & = I - \frac{i}{\hbar}\int_{0}^{t}dt' H(t') + \left(-\frac{i}{\hbar}\right)^2 \int_{0}^{t}dt' \int_{0}^{t'}dt'' H(t') H(t'') \\ & \quad + \left(-\frac{i}{\hbar}\right)^3 \int_{0}^{t}dt' \int_{0}^{t'}dt'' \int_{0}^{t''}dt''' H(t') H(t'') H(t''') +\cdots \end{align}

I know that by showing the "square" integration is equal to two "triangle" integrals, we have $$\int_{0}^{t}dt'\int_{0}^{t'}dt'' H(t')H(t'') = \frac{1}{2}\mathcal{T}\left\{\left(\int_{0}^{t}dt'H(t')\right)^2\right\}.$$

I know the R.H.S is the integral of square. But why is $\mathcal{T}$ needed in$$\mathcal{T}\left\{\left(\int_{0}^{t}dt'H(t')\right)^2\right\}~?$$

Since $$\left(\int_{0}^{t}dt'H(t')\right)^2 = \left(\int_{0}^{t}dt'H(t')\right)\left(\int_{0}^{t}dt'H(t')\right),$$

even by changing the dummy variable $$\left(\int_{0}^{t}dt'H(t')\right)^2 = \left(\int_{0}^{t}dt'H(t')\right)\left(\int_{0}^{t}dt''H(t'')\right),$$ the two single integral terms in the product mean the same thing. Without messing around this product form into a double integral form, why $\mathcal{T}$ is needed in this particular term even before changing of dummy variable?

Consider

$$\int_0^t dt'\int_0^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right)$$ \begin{align} &= \int_0^t dt'\int_0^{t'}dt'' \mathcal{T}\left(H(t')H(t'')\right) \\ & \qquad +\int_0^t dt'\int_{t'}^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right) \\&= \int_0^t dt'\int_0^{t'}dt'' H(t')H(t'') \\ & \qquad +\int_0^t dt'\int_{t'}^{t}dt''H(t'')H(t') \end{align}

The last line follows from the definition of time ordering: in the first integral $t''<t'$ checking the limits of integration, so $\mathcal{T}\left(H(t')H(t'')\right)=H(t')H(t'')$, similarly for the second integral.

As you mentioned using the trick with triangles and squares, we can change the second integral to $$\int_0^t dt''\int_{0}^{t''}dt'H(t'')H(t')$$

This is the upper triangle in a $(t',t'')$ diagram. Now relabel dummy variables $$\int_0^t dt'\int_{0}^{t'}dt''H(t')H(t'')$$

And so recombining we get $$\int_0^t dt'\int_0^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right)=2\int_0^t dt'\int_{0}^{t'}dt''H(t')H(t'').$$

Let's take a reasonably straightforward example:

$$H(t)=t\sigma_y +(1-t)\sigma_x$$

$H(0)$ doesn't commute with $H(1)$.

Let $$I(t)=\int_0^t dt' H(t')=\frac{t^2}{2}\sigma_y +t(1-\frac{t}{2})\sigma_x$$

Then \begin{align}I(t)^2 &=\int_0^t dt_1 \int_0^t dt_2 H(t_1) H(t_2)\\ &=(\frac{t^4}{4}+t^2(1-t/2)^2)I+\frac{t^3}{2}(1-t/2)\{\sigma_y,\sigma_x\}=t^2(\frac{t^2}{2}-t+1)I \end{align}

Now with time ordering \begin{align} J(t)=\int_0^t dt_1 \int_0^t dt_2 \mathcal{T}\left(H(t_1) H(t_2)\right)&=\int_0^t dt_1\int_0^{t_1}dt_2 H(t_1)H(t_2)+\int_0^t dt_1\int_{t_1}^{t}dt_2 H(t_2)H(t_1)\\ &=\int_0^t dt_1 H(t_1)\int_0^{t_1}dt_2 H(t_2)+\int_0^t dt_1\int_{t_1}^{t}dt_2 H(t_2)H(t_1)\\ &= \int_0^t dt_1 H(t_1)I(t_1)+\int_0^t dt_1 \left(I(t)-I(t_1)\right)H(t_1)\end{align}

It's useful to commute some commutators:

\begin{align}\left[H(t_1),H(t_2)\right]&=\left[ t_1\sigma_y +(1-t_1)\sigma_x , t_2\sigma_y +(1-t_2)\sigma_x \right]\\ &=t_1(1-t_2)[\sigma_y,\sigma_x]+t_2(1-t_1)[\sigma_x,\sigma_y]\\&=2i(t_2-t_1)\sigma_z\end{align} And

\begin{align} \left[H(t_1),I(t_1)\right] &=\left[ H(t_1) \,, \int_0^{t_1} dt_2 H(t_2)\right]\\&= \int_0^{t_1} dt_2 \left[ H(t_1) \,,H(t_2)\right]\\&=-it_1^2\sigma_z \end{align}

So \begin{align} J(t)&=\int_0^t dt_1 \left[H(t_1),I(t_1)\right] +I(t)^2\\&=-i\frac{t^3}{3}\sigma_z+I(t)^2 \end{align}

Finally lets consider

$$K(t)=2\int_0^t dt_1 \int_0^{t_1} dt_2 H(t_1)H(t_2) \overset{?}{=} J(t)$$

Well

\begin{align} K(t)=2\int_0^t dt_1 H(t_1)I(t_1)&=\int_0^t dt_1 \left[H(t_1),I(t_1)\right]+\{H(t_1),I(t_1)\}\\&=-i\frac{t^3}{3}\sigma_z + \int_0^t dt_1 \{H(t_1),I(t_1)\} \end{align}

Now

\begin{align} \{H(t_1),I(t_1)\}=\int_0^{t_1} dt_2 \{H(t_1),H(t_2)\}&=\int_0^{t_1} dt_2 \left(2t_1 t_2 +2(1-t_1)(1-t_2)\right)I\\&=I\left(t_1^3 +2(1-t_1)(1-\frac{t_1}{2})t_1\right) \end{align}

Finally if you multiply out and integrate you will find that

$$K(t)=-i\frac{t^3}{6}\sigma_z +t^2(\frac{t^2}{2}-t+1)I=-i\frac{t^3}{6}\sigma_z+I(t)^2=J(t)$$

As an extra note, the definition I've seen for the time ordered exponential is (with the convention that the first term is the identity operator):

$$\mathcal{T}\exp\left(-i/\hbar \int_0^t h(t')dt'\right):=\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{-i}{\hbar}\right)^n \int_0^t dt_1 \ldots\int_0^t dt_n \mathcal{T}\left(H(t_1)\ldots H(t_n)\right)$$

And for the Dyson series:

$$U^\dagger(t,0)=I+\frac{-i}{\hbar}\int_0^t dt' H(t')+\left(\frac{-i}{\hbar}\right)^2\int_0^t dt'\int_0^{t'}dt'' H(t')H(t'')+\ldots$$

The above reasoning at least formally suggests them to be equivalent.

• But I starts from $\int_0^t dt'\int_0^{t}dt'\mathcal{T}\left(H(t')H(t')\right)$ where the time index is the same. Why is time ordering still needed there? – diff Jul 21 '16 at 11:36
• @diff You won't get the same answer is the main reason. Here's an example to show why you need separate dummy variables: Let $I=\int_0^t dx \,x=t^2/2$, then $I^2=t^4/4$, however $\int_0^t dx \int_0^t dx [\,x\cdot x]= \int_0^t dx \,t^3/3=xt^3/3\big|_0^t=t^4/3$. You'll get the correct answer if you evaluate $\int_0^t dx \int_0^t dx' \,x\,x'$ – snulty Jul 21 '16 at 11:45

Time-ordering is needed if the Hamiltonians $H(t^{\prime})$ and $H(t^{\prime\prime})$ at different times do not commute.

Example: If the Hamiltonian is $$H(t) ~=~ \left\{\begin{array}{ccl} \color{Red}{\diamondsuit}&\text{if}& t<0, \cr \clubsuit &\text{if}& t>0, \end{array} \right.$$ then the non-time-ordered product $(\int\! dt ~H(t))^2$ has all $2\times2=4$ possible terms: $$\clubsuit\clubsuit , \quad \color{Red}{\diamondsuit}\color{Red}{\diamondsuit}, \quad \clubsuit\color{Red}{\diamondsuit}, \quad \text{and} \quad \color{Red}{\diamondsuit}\clubsuit.$$ In contrast, the time-ordered product $T\left[(\int\! dt~ H(t))^2\right]$ does not contain terms with anti-chronological ordering $\color{Red}{\diamondsuit}\clubsuit$.

See also this related Phys.SE post for the necessity to introduce time-ordering.

• Specifically in that order $\color{Red}{\diamondsuit}\clubsuit$ right? – snulty Jul 21 '16 at 14:01
• $\uparrow$ Yes. – Qmechanic Jul 21 '16 at 18:09

You have to keep in mind that an integral is (more or less) a sum, and therefore in a double or multiple integral, you have cross terms that have to be correctly time-ordered. The usual Riemann integral does not take in account the non-commuting nature of its integrand, so you have to make it manifest with the time-ordering symbol $\mathcal{T}$.

Imagine for a moment that you discretize your integral. Without the time-ordering symbol $\mathcal{T}$, you would get $$\left(\int_0^t dt' H(t') \right)^2 = \int_0^t dt' \int_0^t dt'' H(t') H(t'') \to \sum_{j=0}^{t/\Delta t} \sum_{k=0}^{t/\Delta t} (\Delta t)^2 H(j\Delta t) H(k\Delta t)$$ Here, all the $j$ ($t'$ in the continuous version) terms come before the $k$ ($t''$), regardless if $t' > t''$.

On the other hand, the same integral with the time-ordering symbol would be \begin{align} \mathcal{T}\left(\int_0^t dt' H(t') \right)^2 & = \mathcal{T}\left[\int_0^t dt' \int_0^t dt'' H(t') H(t'')\right] \\ & \to \sum_{j=0}^{t/\Delta t} \sum_{k=0}^{t/\Delta t} (\Delta t)^2 \mathcal{T} [H(j\Delta t) H(k\Delta t)] \\ & = \sum_{j=0}^{t/\Delta t} \sum_{k=0}^{j} (\Delta t)^2 H(j\Delta t) H(k\Delta t) \\ & \qquad + \sum_{j=0}^{t/\Delta t} \sum_{k=j}^{t/\Delta t} (\Delta t)^2 H(k\Delta t) H(j\Delta t) \end{align} Now, the two Hamiltonians are correctly time-ordered.

The $\mathcal{T}$ in your case comes from the series expansion of the time-ordered exponential $$\mathcal{T}\exp\left(-\frac{i}{\hbar}\int_0^t dt' H(t)\right).$$

tl;dr: The $\mathcal{T}$ determines the prescription for the Riemann sums when you're integrating non-commuting objects.