A Integration of Schrodinger equation

[spaghetti3451]

I'm trying to integrate the Schrodinger equation $i\hbar \frac{d}{dt} |\psi(t)\rangle = H |\psi(t)\rangle$ with the initial condition $|\psi(t_{0})\rangle=|\psi_{0}\rangle$

to show that $|\psi(t)\rangle = \exp(\frac{t-t_{0}}{i\hbar}H)|\psi_{0}\rangle$.

I know how to plug in the solution into the Schrodinger equation and differentiate the LHS using the Taylor expansion of the exponential of the Hamiltonian operator in order to show that the two sides of the Schrodinger equation are equal.

However, what I'm looking for is an analog/generalisation of the separation of variables technique (used to integrate ordinary first-order differential equations) that works for operators. Can you help me out with it?

 

[blue_leaf77]

what I'm looking for is an analog/generalisation of the separation of variables technique

I am not sure how solving the above problem can connect to finding a generalization of the separation of variables technique.

 

[vanhees71]

The point of the Dirac formulation is that you don't need to choose a representation (which leads to "wave mechanics" or a mixture of "wave mechanics" and "matrix mechanics", depending on whether your choice of observables to choose the representation contains only observables with a continuous spectrum (e.g., the position or momentum representation of a spin-less particle) or some also having a discrete spectrum (e.g., the energy and orbital angular momentum of a spinless particle in a radial potential)).

The only choice you have to make is that of the picture of time evolution. Now, you've chosen the Schrödinger picture, were the full time dependence is on the states. For a representing state vector the equation reads, as you say in the OP,
$$\mathrm{i} \hbar \mathrm{d}_t |\psi(t) \rangle=\hat{H} |\psi(t),\rangle.$$
The only parameter here is time. So you have an ordinary differential equation for the state vector $|\psi(t) \rangle$. If $\hat{H}$ is independent of time, then it's very easy to solve this equation formally to
$$|\psi(t) \rangle=\exp \left (-\frac{\mathrm{i} t}{\hbar} \hat{H} \right ) |\psi_0 \rangle,$$
where $|\psi_0 \rangle=|\psi(t=0) \rangle$ is the given initial state vector of the system.

This formal solution is only correct when $\hat{H}$ is not time dependent, because otherwise, you'd have to formally integrate over $t$, and for a time-dpendent Hamiltonian $\hat{H}(t_1)$ and $\hat{H}(t_2)$ do not necessarily commute.

 

[MBPTandDFT]

Let's take $\hbar=1$. Then you have:
$i\frac{d\psi}{dt}=H\psi$
The usual separation of variables technique is:
$i\frac{d\psi}{\psi}=Ht$
and integrate:
$i\int_{\psi_0}^{\psi}\frac{d\psi'}{\psi'}=H\int_{t_0}^tt'\longrightarrow\ln\frac{\psi}{\psi_0}=-iH(t-t_0)$
So finally:
$\psi(t)=\psi(t_0)e^{-iH(t-t_0)}$
as expected.

 

[blue_leaf77]

Let's take $\hbar=1$. Then you have:
$i\frac{d\psi}{dt}=H\psi$
The usual separation of variables technique is:
$i\frac{d\psi}{\psi}=Ht$
and integrate:
$i\int_{\psi_0}^{\psi}\frac{d\psi'}{\psi'}=H\int_{t_0}^tt'\longrightarrow\ln\frac{\psi}{\psi_0}=-iH(t-t_0)$
So finally:
$\psi(t)=\psi(t_0)e^{-iH(t-t_0)}$
as expected.

I wouldn't take a direct analogy between vector equation, which applies here, with the usual differential equation applied to functions. In the right side of the equation
$$
i\frac{d\psi}{dt}=H\psi
$$
$H$ is an operator acting on a vector $\psi$. You can imagine this as a square matrix acting on a column matrix. As you may have known, one cannot arbitrarily take a column matrix to the left of a matrix which was originally multiplied with, and then even further bring this column matrix to the denominator in the left hand side, $i\frac{d\psi}{\psi}$.

In order to understand that the solution, especially the time dependency of the state, in the Schroedinger operator equation $i\partial_t\psi = H\psi$ is given by $\psi(t) = \exp(iH(t-t_0))\psi(t_0)$, it's sufficient to understand that this form does satisfy the Schroedinger equation, i.e. when you plug it into Schroedinger equation, the left side matches the right side.

The usual separation of variable arise when you have projected the Schroedinger operator equation into a continuous basis, e.g. position or momentum basis. However, even with this method the solution you get is the normal modes (or eigenstates) of the corresponding Schroedinger equation.

 

[MBPTandDFT]

I agree. I would call it an heuristic derivation, and I would say it holds because H doesn't involve (besides doesn't depend on) time at all, and tha's a differential equation in time, if you close your eyes and don't look at the space-laplacian inside H that would prevent you from dividing by $\psi$.