Is ${\cal T}\exp\left[\frac{i}{\hbar}\int_0^tH(t')dt'\right]{\cal T}\exp\left[-\frac{i}{\hbar}\int_0^tH(t')dt'\right]$ equal to 1?

I do not think so.

I know 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}

It involves many terms and I do not see them cancel out when calculating the product ${\cal T}\exp\left[\frac{i}{\hbar}\int_0^tH(t')dt'\right]{\cal T}\exp\left[-\frac{i}{\hbar}\int_0^tH(t')dt'\right]$.

  • $\begingroup$ Hi @diff, take a look at my answer below. I dunno what's up with all those people voting to close constructive questions. This question is clearly possible to answer without giving away the derivation. $\endgroup$ – Prof. Legolasov Jan 10 '17 at 6:59

No, it is not equal to $I$.

Supposing that $H(t)$ is hermitian, the evolution operator is unitary:

$$ U = {\cal T}\exp\left[\frac{i}{\hbar}\int_0^tH(t')dt'\right], $$ $$ U^{-1} = U^{\dagger}. $$

When calculating $U^{\dagger}$, you have to take into account that the ordering of terms in the series reverses:

$$ \left( i H(t') H(t'') \right)^{\dagger} = -i H(t'') H(t'). $$

In other words, Hermitian conjugation doesn't commute with time-ordering.

The true inverse of $U$ is given by an operator

$$ U^{-1} = {\cal T^{-1}} \exp \left[ - \frac{i}{\hbar} \int _0 ^t H(t') dt' \right], $$

where ${\cal T^{-1}}$ is reversed time-ordering (orders strings of $H$ just like ${\cal T}$ does, but reverses the result).

Actually, a more general result holds: in case of oriented paths and holonomies of these paths, the inverse holonomy is given by the holonomy of the reversed path. The minus sign in the exponential corresponds to the change of the direction given by $dt'$, and the ${\cal T^{-1}}$ arises naturally through path-reversal.

  • $\begingroup$ Ok i see. I have never seen the symbol of reversed time-ordering before. $\endgroup$ – diff Jan 10 '17 at 8:22
  • 1
    $\begingroup$ @diff it isn't common notation to denote it as ${\cal T^{-1}}$. In fact, I just invented it :P But it doesn't matter how we call the thing, right? $\endgroup$ – Prof. Legolasov Jan 10 '17 at 8:25

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