The formal solution of the time-dependent Schrödinger equation

Question

Consider the time-dependent Schrödinger equation (or some equation in Schrödinger form) written down as $$ \tag 1 i\hbar \partial_{t} \Psi ~=~ \hat{H} \Psi . $$ Usually, one likes to write that it has a formal solution of the form $$ \tag 2 \Psi (t) ~=~ \exp\left[-\frac{i}{\hbar} \int \limits_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0). $$ 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 $$ \tag 3 \Psi (t) ~=~ \color{red}{\hat{\mathrm T}} \exp\left[-\frac{i}{\hbar} \int \limits_{0}^{t} \hat{H}(t^{\prime})~\mathrm dt^{\prime}\right]\Psi (0), \qquad t>0, $$ where $\color{red}{\hat{\mathrm T}}$ 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?

Answer 1

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

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

where the (anti)time-ordered exponentiated Hamiltonian

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

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

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

along with the boundary condition

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

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

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

The (anti)time-ordering in formula $(\text{B})$ is instrumental for the (anti)time-ordered exponential $(\text{B})$ to factorize according to the group-property $(\text{F})$.

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

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

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

IV) If we define the unitary operator without the (anti)time-ordering in formula $(\text{B})$ as

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

then the factorization $(\text{F})$ will in general not take place,

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

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 $(\text{C})$ and $(\text{D})$. 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 $(\text{B})$ and $(\text{H})$ reduce to the same expression

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

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

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

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 [to compare with the TDSE $(\text{C})$]. However, resolving the anti-commutator may, in general, produce unwanted terms. Intuitively, without the (anti)time-ordering in the exponential $(\text{H})$, 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 generates non-zero terms that spoil the possibility to satisfy the TDSE $(\text{C})$. See also the example in section VII.

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

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

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

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

where the so-called contraction

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

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

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

where

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

Note that

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

Then the unitary operator $(\text{H})$ without (anti)time-order reads

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

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

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

where

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

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

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

One may use identity $(\text{U})$ to check directly that the operator $(\text{S})$ satisfies the TDSE $(\text{C})$.

References:

1. Sidney Coleman, QFT lecture notes, arXiv:1110.5013; p. 77.

Answer 2

The existing answer by Qmechanic is entirely correct and extremely thorough. But it is very long and technical, and there's a danger that the core of the answer can get buried under all of that.

The claim made in the question,

formally $$ \tag 2 \Psi (t) ~=~ \exp\left[-i \int_{0}^{t} \hat{ H}(t^{\prime}) ~\mathrm dt^{\prime}\right]\Psi (0) $$ seems to be perfectly fine: it satisfies $$ \tag 1 i \partial_{0} \Psi ~=~ \hat{ H}~ \Psi $$ and the initial conditions

is incorrect: the wavefunction in $(2)$ does not satisfy the differential equation in $(1)$.

The reason for this is that, as a rule, the exponential of an operator $\hat A(t)$ does not obey the differential equation $$ \frac{\mathrm d}{\mathrm dt}e^{\hat A(t)} \stackrel{?}{=} \frac{\mathrm d \hat{A}}{\mathrm dt} e^{\hat A(t)} $$ that one might naively hope it to satisfy. (It is, after all, the exponential operator, right?)

To see why this doesn't work, consider the series expansion of the exponential: \begin{align} \frac{\mathrm d}{\mathrm dt}e^{\hat A(t)} & = \frac{\mathrm d}{\mathrm dt}\sum_{n=0}^\infty \frac{1}{n!}\hat A(t)^n \\& = \sum_{n=0}^\infty \frac{1}{n!} \frac{\mathrm d}{\mathrm dt} \hat A(t)^n, \end{align} and so far so good. However, if we try to push this further, we don't get a nice derivative of the form $\frac{\mathrm d}{\mathrm dt} \hat A(t)^n \stackrel{?}{=} n\frac{\mathrm d\hat A}{\mathrm dt} \hat A(t)^{n-1}$ like we do for scalar-valued functions.

Instead, when we apply the product rule, we get the individual derivatives of each of the operators in the product, at their place within the product: $$ \frac{\mathrm d}{\mathrm dt} \hat A(t)^n = \frac{\mathrm d\hat A}{\mathrm dt} \hat A(t)^{n-1} +\hat A(t)\frac{\mathrm d\hat A}{\mathrm dt} \hat A(t)^{n-2} +\hat A(t)^2\frac{\mathrm d\hat A}{\mathrm dt} \hat A(t)^{n-3} +\cdots +\hat A(t)^{n-2}\frac{\mathrm d\hat A}{\mathrm dt} \hat A(t) +\hat A(t)^{n-1}\frac{\mathrm d\hat A}{\mathrm dt} . $$ This can simplify to just $n\frac{\mathrm d\hat A}{\mathrm dt} \hat A(t)^{n-1}$ but only under the condition that $\hat A(t)$ commute with its derivative, $$ \left[\frac{\mathrm d\hat A}{\mathrm dt} , \hat A(t)\right] \stackrel{?}{=} 0, $$ and as a general rule this is not satisfied.

For the particular case in the question, where $\hat A(t) = -i \int_0^t \hat H(\tau) \mathrm d\tau$ and therefore $\frac{\mathrm d\hat A}{\mathrm dt} = -i \hat H(t)$, we have, by linearity, $$ \left[\frac{\mathrm d\hat A}{\mathrm dt} , \hat A(t)\right] = -i \int_0^t \left[\hat H(t),\hat H(\tau)\right] \mathrm d\tau, $$ so if there are any times $t'<t$ for which $\left[\hat H(t),\hat H(t')\right]$ is not zero, then the whole house of cards comes down.