Given the TDSE for the evolution operator $U(t,t_0)$: $$ {\displaystyle i\hbar {\frac {d}{dt}}U(t, t_0)={\hat {H}}U(t,t_0)} $$ and assuming a time-independent Hamiltonian $\hat H$, and the boundary condition $U(t_0, t_0)=\mathbb{I}$, the general solution takes the form: $$ U(t,t_0) = e^{-\frac i \hbar \hat H(t-t_0)}. $$

In all the references I've come across so far, this solution is given without any proof, for it's probably seen as an obvious extension of a linear ordinary differential equation of order 1, which involves operators instead of functions.

I'm sorry, but I can't see all this resemblance. For example, how should the following be interpreted if one blindly applies the separation of variables method? $$ {\frac {d U(t,t_0)}{U(t,t_0)}}=-\frac{i}{\hbar}{\hat {H}}dt. $$ You can't divide for a vector, so just imagine doing it with an operator...

  • 4
    $\begingroup$ you would not divide by $U$ but multiply by $U^{-1}$. $\endgroup$ Commented Dec 2, 2022 at 18:37
  • $\begingroup$ If you plug in the expression for $U$, do you agree that it satisfies the TDSE? Or are you unclear on why it does? If you’re unclear, do you understand how the exponential of an operator is defined? $\endgroup$
    – Ghoster
    Commented Dec 2, 2022 at 18:49
  • $\begingroup$ Applying to both sides $U^{-1}$, either to the left or to the right, doesn't seem helpful. In the first case, you obtain $\displaystyle i\hbar U^{-1}\partial_t U = U^{-1}HU$, which appears more complicated than the original... @Ghoster I perfectly agree that $U$ is a solution, but I can't see a clear way to derive it. $\endgroup$
    – ric.san
    Commented Dec 2, 2022 at 19:39
  • $\begingroup$ you want $i\hbar (\partial_t U) U^{-1}=H$…. The left hand side is perfectly well defined… $\endgroup$ Commented Dec 2, 2022 at 20:01
  • $\begingroup$ Guessing solutions to differential equations — in this case by obvious analogy — and then verifying them is a valid way to obtain them. If you took a course on differential equations, didn’t you solve some of them by starting with an ansatz? Uniqueness theorems make how you get the solution irrelevant. Analogy and intuition are critical skills for mathematics. $\endgroup$
    – Ghoster
    Commented Dec 2, 2022 at 20:03

1 Answer 1


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I think that for a time-independent Hamiltonian $\:\hat{\mr H}\plr{t}\e\hat{\mr H}\:$ we could derive the exponential expression of the time evolution operator \begin{equation} \mr U\plr{t,t_0}\e e^{\m i\,\hat{\mr H}\plr{t\m t_0}/\hbar} \tl{01} \end{equation} using its property \begin{equation} \mr U\plr{t,t_0}\e \mr U\plr{t,t_k}\mr U\plr{t_k,t_0} \tl{02} \end{equation} and the identity \begin{equation} \lim_{n\bl\rightarrow\infty}\plr{1\p\dfrac{z}{n}\,}^n \e e^z\qquad \plr{n\bl\in \mathbb N\,, z\bl\in \mathbb C} \tl{03} \end{equation} So, suppose that we insert in the time interval $\:\plr{t_0,t}\:$ a number of $\:\plr{n\m 1}\:$ time moments $\: \plr{n\bl\in\mathbb N}\:$ dividing it into $\:n\:$ sub-intervals

\begin{equation} t_0\les t_1\les t_2\les\cdots\les t_k\les\cdots\les t_n\bl\equiv t \tl{04} \end{equation} Using the property \eqref{02} we can build the operator $\:\mr U\plr{t,t_0}\:$ from a product of $\:n\:$ operators $\:\mr U\plr{t_k,t_{k\m 1}}\:$ \begin{align} \mr U\plr{t,t_0}&\e\mr U\plr{t_n,t_0} \e\mr U\plr{t_n,t_{n\m 1}}\mr U\plr{t_{n\m 1},t_0}\e\mr U\plr{t_n,t_{n\m 1}}\mr U\plr{t_{n\m 1},t_{n\m 2}}\mr U\plr{t_{n\m 2},t_0} \nonumber\\ &\e\mr U\plr{t_n,t_{n\m 1}}\mr U\plr{t_{n\m 1},t_{n\m 2}}\cdots\mr U\plr{t_k,t_{k\m 1}}\cdots\mr U\plr{t_3,t_2}\mr U\plr{t_2,t_1}\mr U\plr{t_1,t_0} \tl{05} \end{align} that is \begin{equation} \mr U\plr{t,t_0}\e \prod\limits_{k\e 1}^{k\e n}\mr U\plr{t_k,t_{k\m 1}} \tl{06} \end{equation} Now, make the $\:\plr{n\m 1}\:$ time moments equidistant \begin{align} t_k\m t_{k\m 1}&\e \Delta t \vp \tl{07a}\\ \Delta t&\e\dfrac{t\m t_0}{n} \tl{07b} \end{align} Taking $\:n\:$ extremely very large the time interval $\:\Delta t\:$ is infinitesimally very small so we could write \begin{align} t_k\m t_{k\m 1}&\e \mr dt \vp \tl{08a}\\ \mr dt&\e\dfrac{t\m t_0}{n}\qquad \plr{n\gr\gr 1} \tl{08b} \end{align} and equation \eqref{06} yields \begin{equation} \mr U\plr{t,t_0}\e \prod\limits_{k\e 1}^{k\e n}\mr U\plr{t_{k\m 1}\p \mr dt ,t_{k\m 1}} \tl{09} \end{equation} But every factor in the rhs of \eqref{09} is an infinitesimal time evolution operator of the form $\:\mr U\plr{t\p \mr dt ,t}\:$ which satisfies the following equation \begin{equation} \mr U\plr{t\p \mr dt ,t}\bl\approx 1\!\!1\m\dfrac{i}{\hbar}\hat{\mr H}\,\mr dt \tl{10} \end{equation} as proved and explained in the APPENDIX. Note that $\:1\!\!1\:$ is the identity operator.

So, we have \begin{equation} \mr U\plr{t_{k\m 1}\p \mr dt ,t_{k\m 1}}\bl\approx 1\!\!1\m\dfrac{i}{\hbar}\hat{\mr H}\,\mr dt\e 1\!\!1\m\dfrac{i\,\hat{\mr H}\plr{t\m t_0}/\hbar}{n}\qquad \plr{k\e1,2,\cdots,n} \tl{11} \end{equation} and from \eqref{09} we obtain \begin{equation} \mr U\plr{t,t_0}\bl\approx \blr{1\!\!1\m\dfrac{i\,\hat{\mr H}\plr{t\m t_0}/\hbar}{n}}^n \tl{12} \end{equation} Using the identity \eqref{03} we prove the exponential expression \eqref{01} of the time evolution operator \begin{equation} \mr U\plr{t,t_0}\e \lim_{n\bl\rightarrow\infty}\blr{1\!\!1\m\dfrac{i\,\hat{\mr H}\plr{t\m t_0}/\hbar}{n}}^n \e e^{\m i\,\hat{\mr H}\plr{t\m t_0}/\hbar} \tl{13} \end{equation}


APPENDIX : $\:\texttt{The infinitesimal time evolution operator }\mr U\plr{t\p \mr dt ,t}$

The Schrodinger equation \begin{equation} i\,\hbar\dfrac{\mr d\vra{\psi\plr{t}}}{\mr dt}\e \hat{\mr H}\plr{t}\vra{\psi\plr{t}} \tl{A-01} \end{equation} is linear in $\:\vra{\psi\plr{t}}\:$ so we expect this state to be obtained from an initial state $\:\vra{\psi\plr{t_0}}\:$ at time $\:t_0\:$ via a linear operator \begin{equation} \vra{\psi\plr{t}}\e \mr U\plr{t,t_0}\vra{\psi\plr{t_0}} \tl{A-02} \end{equation} the time evolution operator $\:\mr U\plr{t,t_0}$. Since \eqref{A-02} would be valid equally well for the evolution of $\:\vra{\psi\plr{t}}\:$ from time $\:t\:$ to time $\:t\p\mr dt\:$ we have \begin{equation} \boxed{\:\:\vra{\psi\plr{t\p\mr dt}}\e \mr U\plr{t\p\mr dt,t}\vra{\psi\plr{t}}\:\:\Vp{\tfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}} \tl{A-03} \end{equation} The operator $\:\mr U\plr{t\p\mr dt,t}\:$ is the infinitesimal time evolution operator.

From equation \eqref{A-01} we have \begin{equation} \!\!\!\!\!\!\!\!i\hbar\dfrac{\mr d\vra{\psi\plr{t}}}{\mr dt}\e \hat{\mr H}\plr{t}\vra{\psi\plr{t}}\bl\implies\vra{\psi\plr{t\p\mr dt}}\m\vra{\psi\plr{t}}\bl\approx \m \dfrac{i}{\hbar}\,\hat{\mr H}\plr{t}\vra{\psi\plr{t}}\mr dt \tl{A-04} \end{equation} so \begin{equation} \boxed{\:\:\vra{\psi\plr{t\p\mr dt}}\bl\approx \blr{1\!\!1\m \dfrac{i}{\hbar}\,\hat{\mr H}\plr{t}\mr dt}\vra{\psi\plr{t}}\:\:\Vp{\tfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}} \tl{A-05} \end{equation} where $\:1\!\!1\:$ is the identity operator. Comparing equations \eqref{A-03},\eqref{A-05} we obtain \begin{equation} \boxed{\:\:\mr U\plr{t\p\mr dt,t}\bl\approx 1\!\!1\m \dfrac{i}{\hbar}\,\hat{\mr H}\plr{t}\mr dt\:\:\Vp{\tfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}} \tl{A-06} \end{equation} We note that because of \eqref{A-06} equation \eqref{10} is valid for time-dependent Hamiltonian in general.


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