Physics Asked on June 15, 2021
Consider the time-dependent 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
$$
tag 2 Psi (t) ~=~ expleft[-i int limits_{0}^{t} hat{ H}(t^{prime}) ~mathrm dt^{prime}right]Psi (0).
$$
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
$$
tag 3 Psi (t) ~=~ hat{mathrm T} expleft[-i int limits_{0}^{t} hat{H}(t^{prime})~mathrm dt^{prime}right]Psi (0), qquad t>0,
$$
where $hat{mathrm T}$ 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?
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
$$begin{align} U(t_2,t_1)~&=~left{begin{array}{rcl} Texpleft[-frac{i}{hbar}int_{t_1}^{t_2}! dt~H(t)right] &text{for}& t_1 ~<~t_2 crcr ATexpleft[-frac{i}{hbar}int_{t_1}^{t_2}! dt~H(t)right] &text{for}& t_2 ~<~t_1 end{array}right.crcr ~&=~left{begin{array}{rcl} lim_{Ntoinfty} expleft[-frac{i}{hbar}H(t_2)frac{t_2-t_1}{N}right] cdotsexpleft[-frac{i}{hbar}H(t_1)frac{t_2-t_1}{N}right] &text{for}& t_1 ~<~t_2 crcr lim_{Ntoinfty} expleft[-frac{i}{hbar}H(t_1)frac{t_2-t_1}{N}right] cdotsexpleft[-frac{i}{hbar}H(t_2)frac{t_2-t_1}{N}right] &text{for}& t_2 ~<~t_1 end{array}right.end{align}tag{B} $$
is formally the unitary evolution operator, which satisfies its own two TDSEs
$$ ihbar frac{partial }{partial t_2}U(t_2,t_1) ~=~H(t_2)U(t_2,t_1),tag{C} $$ $$ihbar frac{partial }{partial t_1}U(t_2,t_1) ~=~-U(t_2,t_1)H(t_1),tag{D} $$
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):
$$begin{array}{ccc} frac{U(t_2+delta t,t_1) - U(t_2,t_1)}{delta t} &stackrel{(F)}{=}& frac{U(t_2+delta t,t_2) - {bf 1} }{delta t}U(t_2,t_1)crcr downarrow & &downarrowcrcr frac{partial }{partial t_2}U(t_2,t_1) && -frac{i}{hbar}H(t_2)U(t_2,t_1).end{array}tag{G}$$
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
$$ V(t_2,t_1)~=~expleft[-frac{i}{hbar}int_{t_1}^{t_2}! dt~H(t)right],tag{H}$$
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
$$ U(t_2,t_1)~=~expleft[-frac{i}{hbar}Delta t~Hright]~=~V(t_2,t_1), qquad Delta t ~:=~t_2-t_1.tag{J}$$
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
$$ frac{partial V(t_2,t_1)}{partial t_2} ~=~-frac{i}{hbar}H(t_2) -frac{1}{2hbar^2} left{ H(t_2), int_{t_1}^{t_2}! dt~H(t) right}_{+} +ldots,tag{K} $$
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 $f:mathbb{R}tomathbb{C}$ 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
$$ C(t,t^{prime})~=~ hbarleft(theta(t-t^{prime})overline{f(t)}f(t^{prime}) +theta(t^{prime}-t)overline{f(t^{prime})}f(t)right) ~{bf 1}tag{N}$$
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
$$ A(t_2,t_1)~=~-frac{i}{hbar}int_{t_1}^{t_2}! dt~H(t) ~=~-frac{i}{hbar}overline{F(t_2,t_1)} a -frac{i}{hbar}F(t_2,t_1) a^{dagger} ,tag{O}$$
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
$$begin{align} V(t_2,t_1)~&=~e^{A(t_2,t_1)} ~&=~expleft[-frac{i}{hbar}F(t_2,t_1) a^{dagger}right]expleft[frac{-1}{2hbar}|F(t_2,t_1)|^2right]expleft[-frac{i}{hbar}overline{F(t_2,t_1)} aright].tag{R} end{align}$$
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
$$begin{align} U(t_2,t_1)~&stackrel{(B)}{=}~Texpleft[-frac{i}{hbar}int_{t_1}^{t_2}! dt~H(t)right] ~&stackrel{(M)}{=}~:expleft[-frac{i}{hbar}int_{t_1}^{t_2}! dt~H(t)right]:~ expleft[frac{-1}{2hbar^2}iint_{[t_1,t_2]^2}! dt~dt^{prime}~C(t,t^{prime})right] ~&=~ e^{A(t_2,t_1)+D(t_2,t_1)}~=~V(t_2,t_1)e^{D(t_2,t_1)}tag{S}, end{align}$$
where
$$ D(t_2,t_1)~=~frac{{bf 1}}{2hbar}iint_{[t_1,t_2]^2}! dt~dt^{prime}~{rm sgn}(t^{prime}-t)overline{f(t)}f(t^{prime})tag{T}$$
is a central element proportional to the identity operator. Note that
$$begin{align} frac{partial }{partial t_2}D(t_2,t_1)~&=~frac{{bf 1}}{2hbar}left(overline{F(t_2,t_1)}f(t_f)-overline{f(t_2)}F(t_2,t_1)right) ~&=~frac{1}{2}left[ A(t_2,t_1), frac{i}{hbar}H(t_2)right]~=~frac{1}{2}left[frac{partial }{partial t_2}A(t_2,t_1), A(t_2,t_1)right].tag{U} end{align}$$
One may use identity (U) to check directly that the operator (S) satisfy the TDSE (C).
References:
Correct answer by Qmechanic on June 15, 2021
The equation
$$partial _{t}psi (t)=-iHpsi (t)$$
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).
Answered by Urgje on June 15, 2021
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 $$ tag 2 Psi (t) ~=~ expleft[-i int_{0}^{t} hat{ H}(t^{prime}) ~mathrm dt^{prime}right]Psi (0) $$ 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 $frac{mathrm d}{mathrm dt} hat A(t)^n stackrel{?}{=} nfrac{mathrm dhat A}{mathrm dt} hat A(t)^{n-1}$ 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: $$ frac{mathrm d}{mathrm dt} hat A(t)^n = frac{mathrm dhat A}{mathrm dt} hat A(t)^{n-1} +hat A(t)frac{mathrm dhat A}{mathrm dt} hat A(t)^{n-2} +hat A(t)^2frac{mathrm dhat A}{mathrm dt} hat A(t)^{n-3} +cdots +hat A(t)^{n-2}frac{mathrm dhat A}{mathrm dt} hat A(t) +hat A(t)^{n-1}frac{mathrm dhat A}{mathrm dt} . $$ This can simplify to just $nfrac{mathrm dhat A}{mathrm dt} hat A(t)^{n-1}$ but only under the condition that $hat A(t)$ commute with its derivative, $$ left[frac{mathrm dhat A}{mathrm dt} , hat A(t)right] stackrel{?}{=} 0, $$ and as a general rule this is not satisfied.
For the particular case in the question, where $hat A(t) = -i int_0^t hat H(tau) mathrm dtau$ and therefore $frac{mathrm dhat A}{mathrm dt} = -i hat H(t)$, we have, by linearity, $$ left[frac{mathrm dhat A}{mathrm dt} , hat A(t)right] = -i int_0^t left[hat H(t),hat H(tau)right] mathrm dtau, $$ 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.
Answered by Emilio Pisanty on June 15, 2021
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