Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
2  
(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 at 23:33

2 Answers 2

up vote 8 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

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

where

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

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

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

where

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

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).

share|improve this answer
    
So you're saying (2) is not a solution? If so, why? –  Emilio Pisanty Mar 14 at 22:41
1  
Because the exponential factor in eq. (2) in general lacks the group-property (C), cf. the BCH formula. –  Qmechanic Mar 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 at 22:50
    
@Emilio Pisanty: I updated the answer with your suggestion. –  Qmechanic Mar 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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.