Time-dependent Schrodinger equation from variational principle

In the paper, "Density-functional theory for time-dependent systems" Physical Review Letters 52 (12): 997 the authors mentioned that the action $$A= \int_{t_0}^{t_1} \mathrm dt \langle \Phi(t) | i \hbar\;\partial / \partial t - \hat{H}(t) | \Phi(t) \rangle \tag{1}$$

provides the solution of time-dependent Schrödinger equation at its stationary point. Wikipedia called (1) as the Dirac action without further reference.

If I do a variation, indeed the stationary point of action (1) gives $$i \hbar\;\partial / \partial t | \Phi(t) \rangle = \hat{H}(t) | \Phi(t) \rangle$$

However, from path-integral point of view, the least action principle is only a limiting case when $\hbar \rightarrow 0$. In general, there is no least action principle in quantum mechanics.

My question is, how to reconcile these two aspects? What does vary of action (1) mean?

• This is simply another variational principle, not necessarily connected with the classical action. From the Ritz variational functional $E[\psi] = \int d^3x \psi^*(x) H \psi(x)$ you will get the time indepent Schrödigner equation, this is just a natural extension of this. May 23, 2015 at 23:37
• What, exactly, is the question here? You vary $(1)$, you obtain the Schrödinger equation. What is the question about that? Jan 21, 2016 at 21:00
• Action is just a term used for certain kind of functionals. The qm action and classical action are different things, in different spaces. Thus, there is nothing to resolve. (Assuming that I understood the question right) Jan 21, 2016 at 22:58

If you are interested in an path integral with the action: $$\mathcal{S}= \int_{t_0}^{t_1} dt \langle \Phi(t) | i \hbar\partial / \partial t - \hat{H}(t) | \Phi(t) \rangle \tag{1}$$ then $\Phi(k, t)=\langle k| \Phi(t) \rangle$ is now an operator or a mude variable inside the path integral. The bridge between the operator and path integral linguage is:

$$\langle \alpha|\mathcal{T}\left(...\hat{\Phi}(k,t)...\right)|\beta\rangle=\int \mathcal{D}\phi(k,t)\mathcal{D}\bar{\phi}(k,t)\left(...\phi(k,t)...\right)e^{\frac{i}{\hbar}\mathcal{S}}$$

And the Action is now written as: $$\mathcal{S}= \int_{t_0}^{t_1} dt \int dk\, \bar{\phi}(k,t) ( i \hbar\partial / \partial t - \hat{H}_k(t) ) \phi(k, t)$$ with $\hat{H}_k(t)$ being an linear operator acting on functions $\phi(k,t)$.

This is the second quantization. Now we have an complex quantum field theory. Taking the canonical momentum of the fields and using the Dirac rule of quantization: $$\left[\hat{\Phi}(k,t),\hat{\Phi}^{\dagger}(k',t')\right]_\pm=\delta(k'-k)\delta(t'-t)$$ This is the algebra of annihilation and creation operators. Because the theory are linear ($\mathcal{L}$ bilinear in $\Phi$) the number operator $$N=\sum_{k}\Phi(k, t)^{\dagger}\Phi(k, t)$$ commutes with $\mathcal{H}$, the hamiltonian related to the action $\mathcal{S}$. This imply that $N$ is a constant of motion. The schrodinger equation (equation of motion) is obeyed by a field operator: $$i \hbar\frac{\partial}{\partial t} \hat{\Phi}(k,t)= \hat{H}_k(t)\hat{\Phi}(k,t)$$ and if you find eigenfunctions of $u_n(k,t)$ for the $\hat{H}_k(t)$ we have:

$$\hat{\Phi}(k,t)=\sum_{n}u_n(k,t)\hat{c}_n$$

where $\hat{c}_n$ is the annihilation operator of a particle. You see that your prediction would be the same.

• How does this answer the question? Jan 21, 2016 at 21:00
• The question are concerned with the histories that do not obey the classical equation of motion of the action. Here I show that this histories only give us an discription where the number of particles can be dynamicals. But, because we have a symmetry (bilinear terms only), we have a symmetry that is the conservation of number of particles. So, pratically we have the same thing but with different discriptions. Jan 21, 2016 at 21:20

There definitely is a least action principle in quantum mechanics, indeed, the path-integral method is based on it. Feynman's doctoral thesis is titled:" the least action principle in quantum mechanics". Please see, e.g., http://cds.cern.ch/record/101498/files/?ln=en

• Link-only answers generally are frowned upon, as links can - as I'd put it - "die". Can you explain this more? Dec 12, 2014 at 19:47