0
$\begingroup$

The time-dependent Schrodinger equation is given as $$i \hbar \frac{\mathrm d}{\mathrm dt}| \psi(t) \rangle = \hat{H} | \psi(t) \rangle. $$ To find how the states evolve in time we want to find the linear operator $\hat{U}(t,t_0)$ such that $$| \psi(t) \rangle = \hat{U}(t,t_0)| \psi(t_0) \rangle.$$ Substituting into Schrodinger equation yields $$\frac{\partial \hat{U}(t,t_0)}{ \partial t} = - \frac{i}{\hbar}\hat{H}\hat{U}(t,t_0)$$ this leads to $$\hat{U}(t,t_0) = e^{\frac{-i(t-t_0)\hat{H}}{\hbar}}$$ hence $$| \psi(t) \rangle = e^{-\frac{i(t-t_0)\hat{H}}{\hbar}}|\psi(t_0) \rangle$$ where $e^{a \hat{A}} = \sum \frac{a^n}{n !}\hat{A}^n = \hat{I} + a \hat{A} + \frac{a^2}{2 !} \hat{A}^2 + \frac{a^3}{3 !}\hat{A}^3 +...$

Question:
Given the operator series expansion of $e^{a \hat{A}}$ above, how does it follow that if we consider a time independent Hamiltonian $\hat{H_0}$ where solutions- the eigenvalues $E_n$ and eigenstates $| \psi_n \rangle$ (stationary states)- are equivalently stated as $$e^{\frac{-it \hat{H_0}}{\hbar}}| \psi_n \rangle = e^{-\frac{i E_n t}{\hbar}} | \psi_n \rangle?$$

Thanks.

$\endgroup$
  • 2
    $\begingroup$ Presumably missing a $\partial_t$ in the first term... $\endgroup$ – ZeroTheHero Mar 4 '17 at 13:49
  • $\begingroup$ i'm not sure I follow. You seem to have already done the work: if you take the series and assume the state is an eigenstate of $\hat A$ with eigenvalue $\lambda$, then $\hat A\to \lambda$ everywhere in series, which is then just the series for the exponential function. $\endgroup$ – ZeroTheHero Mar 4 '17 at 13:54
0
$\begingroup$

If $A$ is an arbitrary operator, with eigenbasis $$ A|a\rangle=a|a\rangle\tag{1} $$ then the operator $f(A)$ is defined through $$ f(A)|a\rangle=f(a)|a\rangle\tag{2} $$

If the function $f$ is analytic, $$ f(x)=\sum_n \frac{f^{(n)}(0)}{n!}x^n\tag{3} $$ then you can also define $$ f(A)\equiv \sum_n \frac{f^{(n)}(0)}{n!}A^n\tag{4} $$ and, provided the sum converges, it agrees with $(2)$, as it follows from $A^n|a\rangle=a^n|a\rangle$ (here you have to assume that the Taylor series of $f$ converges for all $x$ in the spectrum of $A$).


In your particular case, $A=H$, $f=\exp$, and $|a\rangle=|\psi_n\rangle$, which means that $$ \mathrm e^{iHt}|\psi_n\rangle=\mathrm e^{-iE_nt}|\psi_n\rangle \tag{5} $$ holds by definition (cf. $(2)$).

$\endgroup$
  • $\begingroup$ Thanks for your answer. Could you elaborate (or reference some result) for the reason why $$f(A) = \sum_{n} \frac{f^{(n)}(0)}{n !} A^n \implies f(A)| a \rangle = f(a)|a \rangle$$ if we assume that $f$ converges for all eigenvalues of $\hat{A}$. Does it follow from a result in functional calculus? $\endgroup$ – user100411 Mar 4 '17 at 14:33
  • 1
    $\begingroup$ @JohnDoe well, if you take $f(A) = \sum_{n} \frac{f^{(n)}(0)}{n !} A^n$ and act on both sides with $|a\rangle$, you get $f(A)|a\rangle = \sum_{n} \frac{f^{(n)}(0)}{n !} A^n|a\rangle$. If you use $ A^n|a\rangle= a^n|a\rangle$, this becomes $f(A)|a\rangle = \sum_{n} \frac{f^{(n)}(0)}{n !} a^n|a\rangle$. Finally, if you factor out $|a\rangle$ on the r.h.s., you get $f(A)|a\rangle = \left[\sum_{n} \frac{f^{(n)}(0)}{n !} a^n\right]|a\rangle=f(a)|a\rangle$, as required. $\endgroup$ – AccidentalFourierTransform Mar 4 '17 at 15:18
  • $\begingroup$ Okay I see, so the only actual functional calculus is really in defining $f(A)$ from the definition of $f(x)$? $\endgroup$ – user100411 Mar 4 '17 at 16:28
  • 1
    $\begingroup$ @JohnDoe yes, correct. $\endgroup$ – AccidentalFourierTransform Mar 4 '17 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy