- #1

Matthew_

- 5

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- TL;DR Summary
- I have seen different formulations of the TDVP. My issue is that it seems to me that they are perfectly equivalent whenever the trial state spans the whole Hilbert space, but are not equivalent whenever one uses them to approach problems that are not solvable exactly.

To my understanding, the most general formulation of the TDVP relies on the effective Action

$$\begin{equation}\mathcal{S}=\int_{t_1}^{t_2}dt\:\mathcal{L}', \hspace{15pt} \mathcal{L}'= \dfrac{i\hbar}{2}\dfrac{\braket{\Psi|\dot{\Psi}}-\braket{\dot{\Psi}|\Psi}}{\braket{\Psi|\Psi}}-\dfrac{\braket{\Psi|\hat{H}|\Psi}}{\braket{\Psi|\Psi}}\end{equation}$$

Starting from this definition, which does not rely on the proper normalization of the trial state, one can get the time-dependent SE (or, rather, an equation which is related to the SE via

$$\mathcal{S}=\int_{t_1}^{t_2}dt\left(i\hbar\braket{\Psi|\partial_t|\Psi}-\braket{\Psi|\hat{H}|\Psi}\right),$$

which, by inspection, gives exactly the SE by imposing the stationarity condition for variations ##\bra{\Psi}\rightarrow\bra{\Psi}+\bra{\delta\Psi}##.

Now, when going the other way around, i.e. when trying to "derive" a variational principle that works starting from the Schrödinger equation, my lecturer started from the so called "weak version" of the Schrödinger equation

$$

\braket{\Phi_t|i\hbar\partial_t-\hat{H}|\Phi_t}=0.

$$

He used the trial state ##\ket{\Phi_t}=\exp(i\mathcal{S}/\hbar)\ket{Z}##, where ##\ket{Z}## is a properly normalized state that depends on the set of variational parameters ##\mathbf{z}##. The previous equation can be written as:

$$

\begin{align*}

&\braket{\Phi_t|i\hbar\partial_t-\hat{H}|\Phi_t}=\bra{Z}e^{-i\mathcal{S}/\hbar}e^{i\mathcal{S}/\hbar}\left(-\dot{\mathcal{S}}\ket{Z}+i\hbar\partial_t\ket{Z}-\hat{H}\ket{Z}\right)=0\Rightarrow\\

\Rightarrow\:&\dot{\mathcal{S}}=i\hbar\braket{Z|\partial_t|Z}-\braket{Z|\hat{H}|Z}\Rightarrow \mathcal{S}=\int_{t_1}^{t_2}dt\left(i\hbar\braket{Z|\partial_t|Z}-\braket{Z|\hat{H}|Z}\right).

\end{align*}

$$

What bothers me is that now we apply the variational principle to a trial state that gets multiplied by a function of time (which is in-fact the exponentiated action). Now, to my understanding of the problem, assuming that ##\ket{\Psi}## is an exact solution of the SE, substitution gives ##\mathcal{L}=0## identically, therefore at the end of the process ##\ket{\Phi_t}=\ket{Z}##.

Secondly, assuming I am right on why this process gets you the right expression for the action, I wonder what happens whenever one uses the variational principle as intended, i.e. to approximate problems. In this context, the end result is not an exact solution of the full SE, therefore i do not see a reason for ##\ket{\Phi_t}=\ket{Z}##, but that is absurd as now I have two different variational principles that lead to different end results (which differ one to another of a time-dependent complex phase).

$$\begin{equation}\mathcal{S}=\int_{t_1}^{t_2}dt\:\mathcal{L}', \hspace{15pt} \mathcal{L}'= \dfrac{i\hbar}{2}\dfrac{\braket{\Psi|\dot{\Psi}}-\braket{\dot{\Psi}|\Psi}}{\braket{\Psi|\Psi}}-\dfrac{\braket{\Psi|\hat{H}|\Psi}}{\braket{\Psi|\Psi}}\end{equation}$$

Starting from this definition, which does not rely on the proper normalization of the trial state, one can get the time-dependent SE (or, rather, an equation which is related to the SE via

*a posteriori*normalization of the trial state). Whenever the trial state is by construction properly normalized at all times, the effective action reduces to$$\mathcal{S}=\int_{t_1}^{t_2}dt\left(i\hbar\braket{\Psi|\partial_t|\Psi}-\braket{\Psi|\hat{H}|\Psi}\right),$$

which, by inspection, gives exactly the SE by imposing the stationarity condition for variations ##\bra{\Psi}\rightarrow\bra{\Psi}+\bra{\delta\Psi}##.

Now, when going the other way around, i.e. when trying to "derive" a variational principle that works starting from the Schrödinger equation, my lecturer started from the so called "weak version" of the Schrödinger equation

$$

\braket{\Phi_t|i\hbar\partial_t-\hat{H}|\Phi_t}=0.

$$

He used the trial state ##\ket{\Phi_t}=\exp(i\mathcal{S}/\hbar)\ket{Z}##, where ##\ket{Z}## is a properly normalized state that depends on the set of variational parameters ##\mathbf{z}##. The previous equation can be written as:

$$

\begin{align*}

&\braket{\Phi_t|i\hbar\partial_t-\hat{H}|\Phi_t}=\bra{Z}e^{-i\mathcal{S}/\hbar}e^{i\mathcal{S}/\hbar}\left(-\dot{\mathcal{S}}\ket{Z}+i\hbar\partial_t\ket{Z}-\hat{H}\ket{Z}\right)=0\Rightarrow\\

\Rightarrow\:&\dot{\mathcal{S}}=i\hbar\braket{Z|\partial_t|Z}-\braket{Z|\hat{H}|Z}\Rightarrow \mathcal{S}=\int_{t_1}^{t_2}dt\left(i\hbar\braket{Z|\partial_t|Z}-\braket{Z|\hat{H}|Z}\right).

\end{align*}

$$

What bothers me is that now we apply the variational principle to a trial state that gets multiplied by a function of time (which is in-fact the exponentiated action). Now, to my understanding of the problem, assuming that ##\ket{\Psi}## is an exact solution of the SE, substitution gives ##\mathcal{L}=0## identically, therefore at the end of the process ##\ket{\Phi_t}=\ket{Z}##.

*I think*that this is the ultimate reason why the above derivation "just works" (as a matter of fact, it seems to me that this all breaks down whenever ##\ket{Z}## is not assumed as properly normalized, and you can't get ##(1)## as it covers cases where ##\ket{\Phi_t}\neq\ket{Z}##). Am I correct?Secondly, assuming I am right on why this process gets you the right expression for the action, I wonder what happens whenever one uses the variational principle as intended, i.e. to approximate problems. In this context, the end result is not an exact solution of the full SE, therefore i do not see a reason for ##\ket{\Phi_t}=\ket{Z}##, but that is absurd as now I have two different variational principles that lead to different end results (which differ one to another of a time-dependent complex phase).