I'm working through a book on Quantum Computing. The section is regarding the Time Evolution Postulate of quantum mechanics, and it has sort of thrown me a curveball. Given time-independent Schrödinger equation: $${d| {\psi(t)} \rangle \over dt} = -\frac{i}{\hbar} H (t) | {\psi (t)} \rangle $$

and supposed particular solution: $$ | {\psi (t_2)} \rangle = e ^ {-\frac{i}{\hbar} H (t_2-t_1)} | {\psi (t_1)} \rangle $$

Show that the above particular solution is indeed a solution to the time-independent Schrödinger equation.

My most fruitful attempt follows:

Noting, that $(t_2 - t_1)$ is not a parameter to H, but rather a duration scalar, and given the Hamiltonian, H, is taken to be constant (re-written K):

Taking the natural logarithm and limit of both sides:

$$ log( | {\psi (t_2)} \rangle ) - log( | {\psi (t_1)} \rangle ) = -\frac{i}{\hbar} K (t_2 - t_1)$$

some scalar division and the limit statement yields:

$$ \lim_{t_1 \rightarrow t_2} {{log( | {\psi (t_2)} \rangle ) - log( | {\psi (t_1)} \rangle )} \over {t_2 - t_1}} = -\frac{i}{\hbar} K $$

based on definition of derivative of vector-valued function is then:

$$ {d \space log( | {\psi (t)} \rangle ) \over {dt}} = -\frac{i}{\hbar} K $$

It would seem that my liberal application of basic algebra has given me a vector that is somehow equal to a matrix.

I'm stuck here, I'm not sure how to proceed, and I'm certain what I've written thus far is even valid, since I obviously haven't drilled down very far into vector calculus for formality. Nothing even remotely similar is present throughout the book (i.e., moving from the specific case we show here back up to the broader generality present in the Schrödinger equation), but nevertheless I want to be thorough because it is a subject I'm very interested in. Obviously the author is trying to help me demonstrate to myself that the quantum information case we are interested in is indeed a subset of the more general claims Quantum Mechanics ultimately makes (i.e., discontinuous time steps, and finite-dimensional state vectors). Help!

  • $\begingroup$ You have garbled $\hbar \to 1/\hbar$. The TISE is an eigenvalue equation. Unless your particular solution were single mode (single energy packet, an extremely specialized type), you wouldn't have a snowball's chance in hell to prove an eigenvalue equation! $\endgroup$ Apr 12 '20 at 21:01

Given $A\in\mathbb C^{n\times n}$ and $x\in\mathbb C^n$, $e^{A}x\in\mathbb C^n$ is well-defined but you tried to apply the multiplication identity $\log(ab)=\log(a)+\log(b)$, $a,b\geq 0$ which does not work here for multiple reasons. The most obvious probably is that the sizes do not match up: $$ \underbrace{\log(e^Ax)}_{\in\mathbb C^n}=\log(e^A)+\log(x)=\underbrace{A}_{\in\mathbb C^{n\times n}}+\underbrace{\log(x)}_{\in\mathbb C^n}\quad $$ Now if the Hamiltonian $H(t)=H$ is constant then the solution of your ordinary differential equation is given by $|\psi(t)\rangle=e^{-\frac{i}{\hbar}Ht}|\psi(0)\rangle$ or, more generally, $|\psi(t)\rangle=e^{-\frac{i}{\hbar} H(t-t_0)}|\psi(t_0)\rangle$ for any initial time $t_0\geq 0$. This is readily verified by differentiating: $$ \begin{align} \frac{d}{dt}|\psi(t)\rangle&=\frac{d}{dt}e^{-\frac{i}{\hbar} H(t-t_0)}|\psi(t_0)\rangle=\Big(\frac{d}{dt}e^{-\frac{i}{\hbar} Ht}\Big)\overbrace{e^{\frac{i}{\hbar} Ht_0}|\psi(t_0)\rangle}^{\text{independent of }t}\\ &=(-\frac{i}{\hbar} H)e^{-\frac{i}{\hbar} H(t-t_0)}|\psi(t_0)\rangle=-\frac{i}{\hbar} H|\psi(t)\rangle \end{align} $$ Indeed because the matrix $H$ in $t\mapsto e^{tH}$ acts like a constant the time-derivative is the same as if $H$ was a scalar.

Side note: if $H=H(t)$ depends on time (in a differentiable manner) then the solution, formally, is given by $$ |\psi(t)\rangle=e^{-\frac{i}{\hbar}\int_{t_0}^t H(t')\,dt'}|\psi(t_0)\rangle\,. $$ Actually for this expression to be correct one needs a time-ordering operator applied to the exponential, but that is way beyond what you asked about here.

  • $\begingroup$ ? What, exactly, do you understand under TISE? Not the celebrated eponymous eigenvalue equation? $\endgroup$ Apr 12 '20 at 20:57
  • $\begingroup$ Given OP's very first equation (and the corresponding solution) is the time-dependent version I strongly assumed that that's what they meant; and them writing "time-independent" twice was a matter of not being familiar with the subject (yet). $\endgroup$ Apr 13 '20 at 6:03
  • $\begingroup$ Your solution certainly answers the title of mine, and I thank you for your time. I haven't just thrown my hands up, and a rather determined me has spent the past week working through "The Feynmann Lectures" @ Caltech. It was a fabulously brief, but helpful introduction to this matter. Having returned to the question though, it appears that they really do only want me to work through the TISE Eigenvalue equation; basically to show that given normal H, that the proposed solution through its time evolution, is still itself normal. Once I've treated it formally I'll post it up here. Thanks! $\endgroup$ Apr 21 '20 at 13:53

Your Answer

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

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