2
$\begingroup$

As I've learned the first postulate from Quantum Mechanics can be stated as follows:

The states of a quantum system are described by vectors in a complex Hilbert space $\mathcal{H}$.

The book then emphasises that $\mathcal{H}$ is not necessarily $L^2(\mathbb{R}^n)$ for some $n$, i.e. it is not necessarily the space of wave functions. The author says an element $\left|\psi\right\rangle\in \mathcal{H}$ instead of being a wave function is an abstract object containing all the information about the system in that state.

In that setting, as I understood, wave functions appear as one possible representation of the states of the system in the case where we are dealing with a particle without spin. In that case, for each ket $\left|\psi\right\rangle$ one associates a function $\psi \in L^2(\mathbb{R}^n)$ and one has $\left\langle x\right|$ being defined by

$$\left \langle x |\psi\right\rangle=\psi(x)$$

for all $x\in \mathbb{R}^n$.

Now, the evolution of a wave function is simply given by the Schroedinger equation. In other words, the evolution $t\mapsto \Psi(\cdot ,t)$ with $\Psi(\cdot, t)\in L^2(\mathbb{R}^n)$ is given by

$$-i\hbar \dfrac{\partial \Psi}{\partial t}(x,t)=\hat{H}\Psi(x,t)$$

where $\hat{H} : L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)$ is the Hamiltonian operator for the particle.

But what about the general case? If $\mathcal{H}$ is the Hilbert space for a certain quantum system and if we have one initial state $\left|\psi_0\right\rangle$ given, what is the time evolution in this case? In my opinion it can't be the Schrodinger equation as is, because the kets $\left|\psi\right\rangle\in \mathcal{H}$ are not functions.

Considering this, how does one deals with the time evolution of general quantum systems in this abstract state space formalism?

$\endgroup$
  • $\begingroup$ Short answer is this: $|\psi(t)\rangle$ is a ket valued function of time which associates a ket (state vector) with every value of $t$. The abstract Schrodinger equation is equivalent to $$|\psi(t + dt)\rangle = |\psi(t)\rangle - \frac{i}{\hbar}H|\psi(t)\rangle dt$$ $\endgroup$ – Alfred Centauri Sep 9 '15 at 22:26
3
$\begingroup$

The Schrödinger equation is not limited to any particular kind of Hilbert space. There's no problem with abstract kets.

Given a space of states $\mathcal{H}$, a Schrödinger (or time-dependent) state is given by a (smooth) map $$ \lvert\psi(\dot{})\rangle : \mathbb{R} \to \mathcal{H}, t \mapsto \lvert \psi(t) \rangle$$ so the Schrödinger state $\lvert\psi(\dot{})\rangle$ is in $C^\infty(\mathbb{R},\mathcal{H})$ and time evolution is given by it fulfilling the Schrödinger equation $$ \mathrm{i}\hbar\partial_t\lvert\psi(t)\rangle = H\lvert\psi(t)\rangle$$ where it has to be understood that the $\partial_t$ acts on the function $\mathbb{R}\to\mathcal{H}$ (i.e. $\partial_t\lvert\psi(t)\rangle$ is the derivative of $\lvert\psi(\dot{})\rangle$ evaluated at $t$) and the $H$ acts on its value in $\mathcal{H}$ at the time $t$.

It is important to realize that the time derivative acts on functions into the space of states and is not an operator on the space of states itself, see also this answer of mine.

$\endgroup$
2
$\begingroup$

Who says kets are not functions? A ket can be a function of time: $|\psi(t)\rangle$. Since they are elements of a vector space (with a complete inner product), you can define their derivative:

$$\frac{d|\psi\rangle}{dt} =\lim_{h\rightarrow 0} \frac{|\psi(t+h)\rangle-|\psi(t)\rangle}{h}$$

And given a Hamiltonian $H : \mathcal{H}\to\mathcal{H}$, which is a linear operator on the Hilbert space (and may even be a function of time itself), Schrödinger's equation reads:

$$i\hbar \frac{d|\psi(t)\rangle}{dt} = H |\psi(t)\rangle$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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