# Time derivative in Schrödinger equation

In quantum mechanics, a system is descibed by an element $$|\psi\rangle\in\mathcal{H}$$, where $$\mathcal{H}$$ is a Hilbert space.

Then on $$\mathcal{H}$$ (or on a dense subspace of $$\mathcal{H}$$), we can define the Hamiltonian operator $$\mathbf{H}:\mathcal{D}(\mathbf{H})\rightarrow \mathcal{H}$$, where $$\mathcal{D}(\mathbf{H})$$ is a dense subspace of $$\mathcal{H}$$ (called the domain of definition of $$\mathbf{H}$$).

Now, at time $$t_0$$, we consider a system described by $$|\psi(t_0)\rangle\in \mathcal{H}$$, and this system evolves with the Schrödinger equation : $$\forall t\ge t_0,\quad \mathbf{H}|\psi(t)\rangle=i\hbar \frac{\text{d}|\psi(t)\rangle}{\text{d}t},$$ where $$|\psi(t)\rangle$$ is the state of the system at time $$t$$.

But what does $$\frac{\text{d}|\psi(t)\rangle}{\text{d}t}$$ mean ? Because if we take the definition of the derivative, then we would have : $$\frac{\text{d}|\psi(t)\rangle}{\text{d}t}=\lim\limits_{h\rightarrow 0}\frac{|\psi(t+h)\rangle-|\psi(t)\rangle}{h},$$ but how do you define this limit of functions on $$\mathcal{H}$$ ? More precisely, does the sequence $$(|\psi(t)\rangle)_{t\ge t_0}$$ evolves in the same Hilbert space $$\mathcal{H}$$ and what is the domain of definition of $$t\mapsto |\psi(t)\rangle$$ to take the 'derivative' ?

• What is $\mathcal{D}(\mathbf{H})$? Please define notation so that the question can be understood by others. Mar 22, 2020 at 14:41

If you want to be formal, the function $$\psi : \mathbb{R}\to \mathcal{H}, t\mapsto \lvert\psi(t)\rangle$$ needs to be understood as a function between Banach spaces (every Hilbert space is in particular a Banach space). The correct notion of derivative is then the Fréchet derivative.
There are several points: how do you define the derivative? No problem: if $$\psi_1$$ and $$\psi_2$$ are two functions in $$\cal H$$, there is by definition a scalar product $$(\psi_1,\psi_2)$$. If the 2 are functions on $$\mathbb R$$, the scalar product is often defined as $$(\psi_1,\psi_2)=\int_{-\infty}^\infty dx\,\psi_1^*(x)\psi_2(x)$$ But a scalar product is the product of norms with the cosine of the angle, so the norm of $$\psi_1$$, its distance to the origin, is $$||\psi_1||^2=(\psi_1,\psi_1)$$ With this norm you define a distance between two functions $$\psi_1$$ and $$\psi_2$$ as follows $$||\psi_1-\psi_2||=\left[ (\psi_1-\psi_2,\psi_1-\psi_2) \right]^{1/2}$$ A series of vectors $$\psi_n$$ tends to a limit $$\psi_\infty$$ if the distance $$||\psi_n-\psi_\infty||\to0$$ as $$n\to\infty$$.
Now the derivative $$d/dt\left|\psi(t)\right\rangle$$ is what the quotients $$1/h\left[\left|\psi(t+h)\right\rangle-\left|\psi(t)\right\rangle\right]$$ tend to, in the sense defined above. Now if you are asking how to compute this: just take the old-fashioned partial derivative with respect to $$t$$: in every reasonable case, this will be the answer. If it is not, it would mean there is no answer. An example of a case where things go wrong might be $$\psi(x,t)=\frac1{1+x^2}\exp\left[ie^x t \right]$$ The function is in $$\cal H$$, namely $$L^2$$, for all $$t$$, but the partial derivative with respect to $$t$$ definitely is not. This means, I think, that the derivative of $$\psi(x,t)$$ with respect to $$t$$ does not exist.