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Let's have Schrodinger equation (or some equation in Schrodinger form) $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{H} \Psi . $$ One likes to write that it has formal solution $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime}) dt^{\prime}\right]\Psi (0). $$ But usually the solution of $(1)$ is given by the method of successive approximations in a form $$ \tag 3 \Psi (t) ~=~ \hat {T}\exp\left[-i \int \limits_{0}^{t} \hat{H}(t^{\prime}) dt^{\prime}\right]\Psi (0), $$ where $\hat {T}$ is time-ordering operator.

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

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(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
up vote 11 down vote accepted

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

$$\tag{A} \Psi(t_f) ~=~ U(t_f,t_i) \Psi(t_i),$$

where the time-ordered exponentiated Hamiltonian

$$\tag{B} U(t_f,t_i)~=~T\exp\left[-\frac{i}{\hbar}\int_{t_i}^{t_f}\! dt~H(t)\right]$$

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

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

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 time-ordering $T$ in formula (B) is instrumental for the 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).

$$ \frac{\partial }{\partial t_f}U(t_f,t_i) ~\longleftarrow~ \frac{U(t_f+\delta t,t_i) - U(t_f,t_i)}{\delta t} $$ $$\tag{G} ~\stackrel{(F)}{=}~ \frac{U(t_f+\delta t,t_f) - {\bf 1} }{\delta t}U(t_f,t_i) ~\longrightarrow~-\frac{i}{\hbar}H(t_f)U(t_f,t_i)$$

for $\delta t \to 0^{+}$.

Remark: Often the 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 time-ordered exponential (B).

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

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

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 $V(t_f,t_i)$ 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

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

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

$$\tag{K} \frac{dV(t_f,t_i)}{dt_f} ~=~-\frac{i}{\hbar}H(t_f) -\frac{1}{2\hbar^2} \left\{ H(t_f), \int_{t_i}^{t_f}\! dt~H(t) \right\}_{+} +\ldots, $$

where $\{ \cdot, \cdot\}_{+}$ denotes the anti-commutator. The problem is that we would like to have the operator $H(t_f)$ 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 time-ordering in the exponential (H), the $t_f$-dependence is scattered all over the place, so when we differentiate wrt $t_f$, 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

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


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

Note that

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

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

$$ V(t_f,t_i)~=~e^{A(t_f,t_i)}$$ $$\tag{R}~=~\exp\left[-\frac{i}{\hbar}F(t_f,t_i) a^{\dagger}\right]\exp\left[\frac{-1}{2\hbar}|F(t_f,t_i)|^2\right]\exp\left[-\frac{i}{\hbar}\overline{F(t_f,t_i)} a\right].$$

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

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


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

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

$$ \frac{\partial }{\partial t_f}D(t_f,t_i)~=~\frac{{\bf 1}}{2\hbar}\left(\overline{F(t_f,t_i)}f(t_f)-\overline{f(t_f)}F(t_f,t_i)\right)$$ $$\tag{U} ~=~\frac{1}{2}\left[ A(t_f,t_i), \frac{i}{\hbar}H(t_f)\right]~=~\frac{1}{2}\left[\frac{\partial }{\partial t_f}A(t_f,t_i), A(t_f,t_i)\right].$$

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

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So you're saying (2) is not a solution? If so, why? – Emilio Pisanty Mar 14 '14 at 22:41
Because the exponential factor in eq. (2) in general lacks the group-property (C), cf. the BCH formula. – Qmechanic Mar 14 '14 at 22:45
On the other hand, direct differentiation seems to give what it should. This argument could be coupled with a uniqueness-of-the-solution argument to say that it does have the group property. $$\quad $$ Evidently that argument is missing something, and it is probably wrong. I can't see where, though, and I think it is the core of the OP's question. – Emilio Pisanty Mar 14 '14 at 22:50
@Emilio Pisanty: I updated the answer with your suggestion. – Qmechanic Mar 14 '14 at 23:21

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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