# The formal solution of the time-dependent Schrödinger equation

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?

• (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! Mar 14, 2014 at 23:33

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}\! 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.\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 (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

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

then the factorization (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 (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

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

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

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

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 [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

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

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 in the remainder of this answer.) Let

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

where

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

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 (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 (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] \\ ~&=~ 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

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

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 (U) to check directly that the operator (S) satisfy the TDSE (C).

References:

1. Sidney Coleman, QFT lecture notes, arXiv:1110.5013; p. 77.
• Your (interaction) Hamiltonian $\hat{H}(t)=-f(t)a-\overline{f(t)}a^\dagger$ commutes at different times $\left[\hat{H}(t),\hat{H}(t^\prime)\right]=0$, doesn't this imply that we can drop the time-ordering in your example (Emilio's answer)? Jun 10, 2021 at 16:19
• Hi @bodokaiser. Thanks for the feedback. Answers: 1. The Hamiltonian eq. (L) is only an example. 2. The Hamiltonian eq. (L) actually doesn't necessarily commute at different times. (The function $f$ is not necessarily real.) Jun 10, 2021 at 17:26
• yes, your right, I mixed up the complex conjugation. Actually, even when the classical source term $f(t)$ in the interaction Hamiltonian is real, the commutator doesn't vanish because we have the complex $f(t)e^{-i\omega t}$ in the interaction-picture interaction Hamiltonian. Jun 11, 2021 at 4:52

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 $$t' for which $$\left[\hat H(t),\hat H(t')\right]$$ is not zero, then the whole house of cards comes down.

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).

• This is a terrible answer. Obviously the OP's (2) and (3) coincide for a time-independent hamiltonian; the OP's question isn't whether they're different when $H$ is time-dependent (which the OP is perfectly aware of) but why they're different. Jul 21, 2016 at 12:57