4
$\begingroup$

In quantum mechanics we use variational principle in order to find approximate expression for the ground state. Lets assume our probe wavefunction $|\Psi\rangle$ can be expanded in orthonormal basis $$|\Psi\rangle = \sum\limits_{n}f_n |n\rangle$$ Variational ansatz dictates minimization of the energy functional $$E[f_n,f_n^*] = \langle\Psi|\hat{H}|\Psi\rangle$$ with constraint $\langle \Psi|\Psi\rangle=1$ ($f_n^*$ is a complex conjugate). Taking derivative we have equations for the coefficients: $$\frac{\partial E}{\partial f_n^*} = 0.$$

There is also dynamical variational principle where one minimizes Schrodinger action $$S = \int dt \mathcal{L}$$ where $$\mathcal{L} = \langle\Psi(t)|i\hbar\partial_t - \hat{H}|\Psi(t)\rangle$$ Using Euler-Lagrange equations we get differential equations for $f_{n}$: $$\frac{\partial \mathcal{L}}{\partial f_n^*} - \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{f}_n^*}=0$$ My question is whether or not energy $E(t) = \langle\Psi(t)|\hat{H}|\Psi(t)\rangle$ defined with coefficients $f_n(t)$ derived from Euler-Lagrange equations is a conserved quantity i.e. $d E(t)/dt = 0$?

What I have in mind is the Bose-Hubbard Hamiltonian $$\hat{H} = -J\sum\limits_{<i,j>}\hat{a}_i^{\dagger}\hat{a}_j + \frac{U}{2}\sum\limits_i\hat{n}_i(\hat{n}_i-1)-\mu\sum\limits_{i}\hat{n}_i$$ with variational ansatz: $$|\Psi\rangle = \bigotimes\limits_{i}|\psi_i\rangle,\ \ |\psi_i\rangle = \sum\limits_{n=0}^{n_F}f_{n}^{(i)}|n\rangle_{i}$$ In this case Coefficients $f_n$ are not a solution of the Schrodinger equation $i\hbar \partial_t |\Psi\rangle = \hat{H} |\Psi\rangle$.

$\endgroup$
  • $\begingroup$ Sorry, removed my comments, I understand the issue now. For consistency, you should also write $\lvert \Psi (t)\rangle$ in the line where you have $\mathcal{L} = $ to emphasize the variational principle already gives a time-dependent state. $\endgroup$ – ACuriousMind Nov 23 '15 at 16:10
  • $\begingroup$ The reason I am curious about that is the imaginary-time method for obtaining ground state of the system. If you change variables $\tau = it$ and propagate the equation step by step in small intervals $\Delta\tau$ (you should normalize wavefunction after each step) than you should end up with the ground state ideally when $\tau \rightarrow \infty$. $\endgroup$ – WoofDoggy Nov 23 '15 at 17:04
  • 1
    $\begingroup$ Certainly, $E(f_n, f_n^*)$ is the Hamiltonian of the mechanical system defined by the (time independent) Lagrangian $\mathcal{L}(f_n, f_n^*, \dot{f}_n, \dot{f}_n^*)$. Thus it conserved. Please see my answer physics.stackexchange.com/questions/197297/…, where you can think of the parameter vector $R$ as the set of coefficients $f_n, f_n*$. Technically, the conservation stems from the antisymmetry of the Berry curvature. $\endgroup$ – David Bar Moshe Nov 24 '15 at 17:09
  • $\begingroup$ Thank you for your answer @David! I also wonder what happens to the energy when you replace real time $t$ with imaginary time $\tau = it$. Can we say that it is always diminishing $dE(\tau)/d\tau < 0$ or not? $\endgroup$ – WoofDoggy Nov 24 '15 at 17:54
  • $\begingroup$ If the trial wave function manifold is close enough to the ground state, then the conserved energy value on the classical solution is a good approximation of the ground state energy. (Quantum) corrections to this first approximation can be obtained from quantizing the effective theory defined by the Lagrangian $\mathcal{L}$. Since the manifold of trial wave functions is in general non-Euclidean, one needs more general quantization techniques than canonical quantization to perform this task (e.g. geometric quantization). … $\endgroup$ – David Bar Moshe Nov 26 '15 at 8:35
0
$\begingroup$

If you take the equation of motion on the Schrodinger Field Action, you are simple imposing this equation: $$ \frac{\delta S}{\delta f_{n}}|_{Bulk}=0\,\,\,\,\rightarrow \,\,\,\,\, H_{n}^{m} f_m=i\hbar\frac{df_n (t)}{dt} $$

Note that I used the word Bulk below the functional derivative. This is because, functional derivative acts only in the interior of the functional, and don't tells you how the boundary terms behaves. In this problem the boundary term are related to the initial state of the system. This is what you want to find. The initial state that you want to find is precisely the state that conserves the probabilities: $$ \frac{d(f^{*}_{n}f^{n})}{dt}(t)=0 $$

So, the answer that the variation inside the Bulk of the action give us is that if you want such state you need try to diagonalize $H_{mn}$. Very trivial answer. Actually what you want more is the lowest eigenvalue of $H$, so you already know that. The answer for your question is that the equation of motion is not necessary to specify your state. The equation of motion only tells you how some state evolutes through time.

If you are interested in functional applications to vacuum state you can use the idea that in the limit $T\rightarrow 0$ of a system in thermal equilibrium with a reservoir the density matrix get close to the ground state:

$$ \rho=\frac{e^{-\beta H}}{tr(e^{-\beta H})}\,\,\,\,\rightarrow \,\,\,\,|0\rangle\langle 0| $$

And then you can use the path integral ideas seeing the $\beta$ as a time, and the limit as a $\beta\rightarrow \infty$.

$\endgroup$

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.