3
$\begingroup$

The Hilbert Space is the space where wavefunction live. But how would I describe it in words? Would it be something like:

The infinite dimensional vector space consisting of all functions of position $\psi(\vec x)$ given the conditions that $\psi(\vec x)$ is a smooth, continuous function.

(I am not saying this is right, it is merely an example). Further more how does $\Psi(\vec x, t)$ fit into this? Would it be appropriate to say that at any $\Psi(\vec x, t)$ itself is not in the Hilbert space, but at any given time $t_0$ then the function $\psi_0 (\vec x)\equiv \Psi(\vec X,t_0)$ is a member of the Hilbert Space? And all operators $\hat Q$ have to act on members in the Hilbert Space and therefore cannot have time-derivatives but can have time dependencies.

$\endgroup$
  • $\begingroup$ A very readable explanation here: physicspages.com/2012/08/31/vector-spaces-and-hilbert-space $\endgroup$ – Gert Nov 20 '15 at 19:25
  • 1
    $\begingroup$ You could say that it's the phase space of a system in Quantum Mechanics. Each point in a Hilbert space, given sufficient conditions, represents the state of a quantum mechanical system. $\endgroup$ – Prish Chakraborty Nov 20 '15 at 19:29
9
$\begingroup$

Your idea of what $\psi(\vec x,t)$ is supposed to be is essentially correct. Given a space of states $\mathcal{H}$, the "Schrödinger state" is a map $$ \psi : \mathbb{R}\to\mathcal{H}, t\mapsto\lvert\psi(t)\rangle$$ where $\lvert \psi(t)\rangle\in\mathcal{H}$ for every instant $t$. If $\mathcal{H}$ is a space of functions in a variable $\vec x$, then $\lvert \psi(t)\rangle$ is often written $\psi(\vec x,t)$.

The space of wavefunctions in usual quantum mechanics is crucially not the space of smooth functions $C^\infty(\mathbb{R}^3,\mathbb{C})$, but the space of equivalence classes of square integrable functions $L^2(\mathbb{R^3},\mathbb{C})$(link to Wikipedia article on $L^p$-spaces). This space contains the smooth compactly supported functions $C_c^\infty(\mathbb{R}^3,\mathbb{C})$ (non-compactly supported functions are not necessarily square-integrable) and also all smooth square-integrable functions, but is larger.

The reason for this is twofold: For one, there is no reason to demand a generic wavefunction be continuous - the whole physical content of the wavefunction is encapsulated in the probability density $$ \rho(\vec x) = \lvert\psi(x)\rvert^2$$ and this needs to have $\int\rho(\vec x)\mathrm{d}^3x = 1$ to be probability density, hence $\psi$ must be square integrable, but nothing else.

Another reason is that the smooth compactly supported functions do not form a Hilbert space under the scalar product $$ (f,g) = \int \overline{f(x)}g(x)\mathrm{d}^3 x$$ since they are not complete - there are sequences which are Cauchy with respect to the $L^2$-norm induced by this inner product, but which do not converge. The space of square-integrable functions is precisely the completion of the smooth compactly supported functions under this norm, and hence a Hilbert space. In other words, every wavefunction may be arbitrarily accurately be approximated by a smooth compactly supported function, but is itself not guaranteed to be a smooth function.

Among other difficulties commonly overlooked, this means that, strictly speaking, one cannot evaluate wavefunctions at points, since the $L^2$-space elements are only defined up to a zero measure set, and points are of zero measure. This is again meaningful when considering $\rho(\vec x)$: The value of a probability density on a zero measure set is physically meaningless, since only the integration of it over a set of non-zero measure gives a physically relevant probability.

$\endgroup$
  • $\begingroup$ Just a precision: $L^2$ doesn't contain all $C^\infty$ functions. Only those with finite energy. $\endgroup$ – nabla Nov 20 '15 at 21:28
  • $\begingroup$ @gerd: Ah, yes, I forgot to add the condition of compact support. Thanks! $\endgroup$ – ACuriousMind Nov 20 '15 at 21:40
  • $\begingroup$ The good ol' continuous wavefunction...we could put a tag also for that :-D Just to mess things up a little bit, of course the map $\psi:\mathbb{R}\to \mathcal{H}$ is itself continuous (at least if you do not want to consider non-self-adjoint-generated evolutions). $\endgroup$ – yuggib Nov 21 '15 at 8:58
  • 1
    $\begingroup$ Just to clarify something the space $L^2(\mathbb{R^3},\mathbb{C})$ is the space of square integrable functions that take an argument from $\mathbb{R^3}$ and produce a value in $\mathbb{C}$ or something different? $\endgroup$ – Quantum spaghettification Nov 23 '15 at 13:45
  • $\begingroup$ @Joseph: Yes, exactly. (In general, $L^p(X,Y)$ is the space of $p$-integrable functions on $X$ that take values in $Y$) $\endgroup$ – ACuriousMind Nov 23 '15 at 14:07
1
$\begingroup$

In the context of quantum mechanics, Hilbert spaces usually refer to the infinite-dimensional space of solutions to the time-dependent Schrodinger equation $$ i\frac{d}{dt} \left|\psi (t)\right\rangle = H(t) \left|\psi(t)\right\rangle $$ for the state vector $\left|\psi (t)\right\rangle$. This space is completely determined by the (in general) time-dependent Hamiltonian $H(t)$. An initial condition $\left|\psi (t_0)\right\rangle$ determines a time-parameterized trajectory of the vector $\left|\psi (t)\right\rangle$ through the Hilbert space of $H(t)$. Projecting the Hilbert space vectors into a coordinate representation $$ \psi(\mathbf{x},t)\equiv \langle \mathbf{x} \left|\psi (t)\right\rangle $$ we can refer to the Hilbert space as the space of position-space functions that satisfies the position-space representation of Schrodinger's equation. So, to answer your question about the time variable: yes, the variable $t$ simply parametrizes the trajectory through Hilbert space. One does not usually include the space of functions of time in the Hilbert space.

One illustrative caveat is the case of time-periodic Hamiltonians. One can choose the operator $$ \mathcal{H}(t) = i\frac{d}{dt}-H(t) $$ to define your Hilbert space. In that case, the Hilbert space can naturally be defined as the product space of the space of all functions of position and the space of all functions of time that satisfy Schrodinger's equation. In that case, time no longer parametrizes trajectories in Hilbert space, but plays a similar role to $\mathbf{x}$ in a normal Hamiltonian Hilbert space. See for example the pioneering work by Jon Shirley (I apologize if the paywall is an issue): http://journals.aps.org/pr/abstract/10.1103/PhysRev.138.B979

Going back to the conventional treatment where we define our Hilbert space in terms of $H(t)$:

Very importantly, Hilbert spaces are a type of metric space. This means that their is an inner product function that gives the "distance" between two vectors in hilbert space. This same function, when applied to two identical vectors gives the squared "norm" (or length) of that vector which must be positive definite. This property of Hilbert spaces places strong constraints on the vectors in the space, including that position-space representations of state-vectors must be continuous, as stated in your definition. As for smoothness, solutions need not be smooth. Two contrived examples are the $\delta$-function potential and the infinite square-well potential, both of which have discontinuities in derivatives.

As for your statement about operators $\hat Q$, we can answer that easily within the current framework. Since our Hilbert space treats $t$ as a "parameter" time derivatives are not operators in the normal sense of what we mean by "operator" in the context of quantum mechanics. Rather, acting time derivatives on state vectors tells us something about the trajectory through Hilbert space.

You probably found this already, but in case not, Wikipedia summarizes the formal properties of Hilbert spaces: https://en.wikipedia.org/wiki/Hilbert_space#Definition

$\endgroup$

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.