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Consider the Schrödinger equation (or some equation in Schrödinger form) written down as $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{H} \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime}) dt^{\prime}\right]\Psi (0). $$ However, this form for the solution of $(1)$ is actually built by the method of successive approximations which actually returns a solution of the form $$ \tag 3 \Psi (t) ~=~ \hat{T} \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime}) dt^{\prime}\right]\Psi (0), \qquad t>0, $$ where $\hat{T}$ is the time-ordering operator.

It seems that $(3)$ doesn't coincide with $(2)$, but formally $(2)$ seems to be perfectly fine: it satisfies $(1)$ and the initial conditions. So where is the mistake?

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5  
(2) does not satisfy (1) as you can see if you carefully compute the derivative without assuming formal (and wrong for operators) arguments. It instead happens if $H(t)$ commutes with $H(t')$ for $t\neq t'$, but it is false in general! – Valter Moretti Mar 14 '14 at 23:33
1  
↑ In particular, it is wrong to assume that $\displaystyle \frac{\mathrm d}{\mathrm dt}\exp(A(t))=\frac{\mathrm dA}{\mathrm dt}\exp(A(t))$. Instead, differentiating the series expansion results in $$\frac{\mathrm d}{\mathrm dt}\sum_{n=0}^\infty \frac{A(t)^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}\sum_{k=1}^nA(t)^{k-1}\frac{\mathrm dA}{\mathrm dt}(t)A(t)^{n-k},$$ and (unless $A(t)$ commutes with its derivative) in general there's not much one can do to simplify that expression. – Emilio Pisanty Mar 11 at 22:53
up vote 14 down vote accepted

I) The solution to the time-dependent Schrödinger equation (TDSE) is

$$\tag{A} \Psi(t_2) ~=~ U(t_2,t_1) \Psi(t_1),$$

where the (anti)time-ordered exponentiated Hamiltonian

$$ U(t_2,t_1)~=~\left\{\begin{array}{rcl}T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right]&\text{for}& t_1 ~<~t_2 \cr\cr AT\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right]&\text{for}& t_2 ~<~t_1 \end{array}\right.$$ $$~=~\lim_{N\to\infty} \exp\left[-\frac{i}{\hbar}H(t_2)\frac{t_2-t_1}{N}\right] \cdots\exp\left[-\frac{i}{\hbar}H(t_1)\frac{t_2-t_1}{N}\right] \tag{B} $$

is formally the unitary evolution operator, which satisfies its own two TDSEs

$$\tag{C} i\hbar \frac{\partial }{\partial t_2}U(t_2,t_1) ~=~H(t_2)U(t_2,t_1), $$ $$\tag{D}i\hbar \frac{\partial }{\partial t_1}U(t_2,t_1) ~=~-U(t_2,t_1)H(t_1), $$

along with the boundary condition

$$\tag{E} U(t,t)~=~{\bf 1}.$$

II) The evolution operator $U(t_2,t_1)$ has the group-property

$$\tag{F} U(t_3,t_1)~=~U(t_3,t_2)U(t_2,t_1). $$

The (anti)time-ordering in formula (B) is instrumental for the (anti)time-ordered expontial (B) to factorize according to the group-property (F).

III) The group property (F) plays an important role in the proof that formula (B) is a solution to the TDSE (C):

$$\begin{array}{ccc} \frac{U(t_2+\delta t,t_1) - U(t_2,t_1)}{\delta t} &\stackrel{(F)}{=}& \frac{U(t_2+\delta t,t_2) - {\bf 1} }{\delta t}U(t_2,t_1)\cr\cr \downarrow & &\downarrow\cr\cr \frac{\partial }{\partial t_2}U(t_2,t_1) && -\frac{i}{\hbar}H(t_2)U(t_2,t_1).\end{array}\tag{G}$$

Remark: Often the (anti)time-ordered exponential formula (B) does not make mathematical sense directly. In such cases, the TDSEs (C) and (D) along with boundary condition (E) should be viewed as the indirect/descriptive defining properties of the (anti)time-ordered exponential (B).

IV) If we define the unitary operator without the (anti)time-ordering in formula (B) as

$$\tag{H} V(t_2,t_1)~=~\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right],$$

then the factorization (F) will in general not take place,

$$\tag{I} V(t_3,t_1)~\neq~V(t_3,t_2)V(t_2,t_1). $$

There will in general appear extra contributions, cf. the BCH formula. Moreover, the unitary operator $V(t_2,t_1)$ will in general not satisfy the TDSEs (C) and (D). See also the example in section VII.

V) In the special (but common) case where the Hamiltonian $H$ does not depend explicitly on time, the time-ordering may be dropped. Then formulas (B) and (H) reduce to the same expression

$$\tag{J} U(t_2,t_1)~=~\exp\left[-\frac{i}{\hbar}\Delta t~H\right]~=~V(t_2,t_1), \qquad \Delta t ~:=~t_2-t_1.$$

VI) Emilio Pisanty advocates in a comment that it is interesting to differentiate eq. (H) w.r.t. $t_2$ directly. If we Taylor expand the exponential (H) to second order, we get

$$\tag{K} \frac{\partial V(t_2,t_1)}{\partial t_2} ~=~-\frac{i}{\hbar}H(t_2) -\frac{1}{2\hbar^2} \left\{ H(t_2), \int_{t_1}^{t_2}\! dt~H(t) \right\}_{+} +\ldots, $$

where $\{ \cdot, \cdot\}_{+}$ denotes the anti-commutator. The problem is that we would like to have the operator $H(t_2)$ ordered to the left [in order to compare with the TDSE (C)]. But resolving the anti-commutator may in general produce un-wanted terms. Intuitively without the (anti)time-ordering in the exponential (H), the $t_2$-dependence is scattered all over the place, so when we differentiate w.r.t. $t_2$, we need afterwards to rearrange all the various contributions to the left, and that process generate non-zero terms that spoil the possibility to satisfy the TDSE (C). See also the example in section VII.

VII) Example. Let the Hamiltonian be just an external time-dependent source term

$$\tag{L} H(t) ~=~ \overline{f(t)}a+f(t)a^{\dagger}, \qquad [a,a^{\dagger}]~=~\hbar{\bf 1},$$

where $f:\mathbb{R}\to\mathbb{C}$ is a function. Then according to Wick's Theorem

$$\tag{M} T[H(t)H(t^{\prime})] ~=~ : H(t) H(t^{\prime}): ~+ ~C(t,t^{\prime}), $$

where the so-called contraction

$$\tag{N} C(t,t^{\prime})~=~ \hbar\left(\theta(t-t^{\prime})\overline{f(t)}f(t^{\prime}) +\theta(t^{\prime}-t)\overline{f(t^{\prime})}f(t)\right) ~{\bf 1}$$

is a central element proportional to the identity operator. For more on Wick-type theorems, see also e.g. this, this, and this Phys.SE posts. (Let us for notational convenience assume that $t_1<t_2$ in the remainder of this answer.) Let

$$\tag{O} A(t_2,t_1)~=~-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t) ~=~-\frac{i}{\hbar}\overline{F(t_2,t_1)} a -\frac{i}{\hbar}F(t_2,t_1) a^{\dagger} ,$$

where

$$\tag{P} F(t_2,t_1)~=~\int_{t_1}^{t_2}\! dt ~f(t). $$

Note that

$$\tag{Q} \frac{\partial }{\partial t_2}A(t_2,t_1)~=~-\frac{i}{\hbar}H(t_2), \qquad \frac{\partial }{\partial t_1}A(t_2,t_1)~=~\frac{i}{\hbar}H(t_1).$$

Then the unitary operator (H) without (anti)time-order reads

\begin{align} V(t_2,t_1)~&=~e^{A(t_2,t_1)} \\ \tag{R}~&=~\exp\left[-\frac{i}{\hbar}F(t_2,t_1) a^{\dagger}\right]\exp\left[\frac{-1}{2\hbar}|F(t_2,t_1)|^2\right]\exp\left[-\frac{i}{\hbar}\overline{F(t_2,t_1)} a\right]. \end{align}

Here the last expression in (R) displays the normal-ordered for of $V(t_2,t_1)$. It is a straightforward exercise to show that formula (R) does not satisfy TDSEs (C) and (D). Instead the correct unitary evolution operator is

\begin{align} U(t_2,t_1)~&\stackrel{(B)}{=}~T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right] \\~&\stackrel{(M)}{=}~:\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right]:~ \exp\left[\frac{-1}{2\hbar^2}\iint_{[t_1,t_2]^2}\! dt~dt^{\prime}~C(t,t^{\prime})\right] \\ \tag{S}~&=~ e^{A(t_2,t_1)+D(t_2,t_1)}~=~V(t_2,t_1)e^{D(t_2,t_1)}, \end{align}

where

$$\tag{T} D(t_2,t_1)~=~\frac{{\bf 1}}{2\hbar}\iint_{[t_1,t_2]^2}\! dt~dt^{\prime}~{\rm sgn}(t^{\prime}-t)\overline{f(t)}f(t^{\prime})$$

is a central element proportional to the identity operator. Note that

\begin{align} \frac{\partial }{\partial t_2}D(t_2,t_1)~&=~\frac{{\bf 1}}{2\hbar}\left(\overline{F(t_2,t_1)}f(t_f)-\overline{f(t_2)}F(t_2,t_1)\right) \\\tag{U} ~&=~\frac{1}{2}\left[ A(t_2,t_1), \frac{i}{\hbar}H(t_2)\right]~=~\frac{1}{2}\left[\frac{\partial }{\partial t_2}A(t_2,t_1), A(t_2,t_1)\right]. \end{align}

One may use identity (U) to check directly that the operator (S) satisfy the TDSE (C).

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The equation

$$\partial _{t}\psi (t)=-iH\psi (t)$$

acting in a Hilbert space with $H$ self-adjoint has the general solution

$$\psi (t)=\exp [-iH(t-t_{0})]\psi (t_{0}),$$

by Stone's theorem. In case $H=H(t)$ depends on $t$ matters change and time ordering becomes relevant. If $H$ does not depend on time your Eq. (3) reduces to (2).

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