4
$\begingroup$

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' ?

$\endgroup$
1
  • 2
    $\begingroup$ What is $\mathcal{D}(\mathbf{H})$? Please define notation so that the question can be understood by others. $\endgroup$
    – DanielSank
    Commented Mar 22, 2020 at 14:41

4 Answers 4

7
$\begingroup$

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.

Note that this vector-valued function is much easier to differentiate formally than an operator-valued one is in general, see this question and its answers for the rigorous definition of the derivatives of operator-valued functions.

$\endgroup$
3
$\begingroup$

A Hilbert space is, by definition, an inner product space (meaning that it is also a normed space) and it is Cauchy Complete. Completeness means that all the normal definitions of limiting procedures go through as usual (with no more than minor notational changes).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
-2
$\begingroup$

The Schrodinger equation is just a differential equation. Its solution are just differentiable functions. So time derivation is a standard operation.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.