Hilbert spaces in quantum mechanics

We identify quantum states as functions $\psi : \mathbb{R}^n \to \mathbb{C}$, where $n$ is the dimensionality of the configuration space for the given physical problem. We interpret the states as probability amplitudes, giving probability densities for the particle's location by

(1)
\begin{align} PDF = \lvert \psi \rvert ^2 \quad \implies \quad P \left( \text{particle is in E} \subset \mathbb{R}^n \right) = \int_{E} \lvert \psi \rvert ^2 d^n x \end{align}

This naturally gives the normalization condition

(2)
\begin{align} \int_{\mathbb{R}^n} \lvert \psi \rvert ^2 d^n x = 1 \quad \text{(The particle is definitely somewhere.)} \end{align}

So if two functions $\psi : \mathbb{R}^n \to \mathbb{C}$ and $\phi : \mathbb{R}^n \to \mathbb{C}$ are related by a constant multiple, we can think of them as representing the same state, since they must both be normalized to the same function in order to be interpreted as a probability density.

We can think of

(3)
\begin{align} \| \psi \| _2 \equiv \left( \int \lvert \psi \rvert ^2 \right) ^{1/2} \end{align}

as a norm, and functions satisfying the normalization condition as vectors of norm 1. Notice that we can also set up a vector space of functions having finite norm $\| \cdot \| _2$; this vector space of functions is known as $L^2$ (or more specifically, for functions $\mathbb{R}^n \to \mathbb{C}$ as $L^2 \left( \mathbb{R}^n \right)$).

(4)
\begin{align} L^2 \left( \mathbb{R}^n , d^n x \right) = \left\{ \psi : \mathbb{R}^n \to \mathbb{C} : \int \lvert \psi \rvert ^2 < \infty \right\} \end{align}

The norm $\| \cdot \|_2$ comes from the inner product $\langle \phi , \psi \rangle = \int \bar{\phi} \psi$. Exercise: check that this is a sesquilinear form

Also, note that we can use this new terminology to express our definition of expectation values of quantum operators:

From before,

(5)
\begin{align} \hat A &= \text{ quantum operator corresponding to classical observable } A = A \left( x, p \right) \\ \langle \hat A \rangle & \equiv \int_{\mathbb{R}^n} \bar{\psi} \hat A \psi d^n x = \langle \psi , \hat A \psi \rangle \end{align}

If $\psi$ is an eigenstate of $\hat A$, then $\hat A \psi = a \cdot \psi$ and $\langle \hat A \rangle = \langle \psi , a \psi \rangle = a$. Recall that if an operator is self-adjoint ($A = A^*$, where $A^*$ is defined to be such that $\langle \psi , A \phi \rangle = \langle A^* \psi , \phi \rangle$), its eigenvalues are real (work this out for yourself), so since we want expectation values of quantum observables to be real, we impose the requirement that all physical observables have to promote to quantum operators which are self-adjoint with respect to the $L^2$ inner product. Exercise: Work this out for the quantum operators we defined earlier for position and momentum:

(6)
\begin{align} \hat x : \psi \left( x \right) & \to x \psi \left( x \right) \\ \hat p : \psi \left( x \right) & \to - i \hbar \partial_x \psi \left( x \right) \end{align}

Important point:

Notice that the choice of integration measure affects which operators on functions $\hat A : L^2 \left( \mathbb{R} \right) \to L^2 \left( \mathbb{R} \right)$ are self-adjoint. Notice that on $L^2 \left( \mathbb{R}, dx \right)$, the derivative operator $\hat A = -i \hbar \partial_x : L^2 \left( \mathbb{R} \right) \to L^2 \left( \mathbb{R} \right)$ is self-adjoint, while on $L^2 \left( \mathbb{R} , f \left( x \right) dx \right)$ it is not.

For instance, the modified operator $-i \hbar \left( 1 / f \left( x \right) \right) \partial_x f \left( x \right)$ is self-adjoint with respect to the $L^2$ inner product defined by the measure $f \left( x \right) dx$.

Clarification:

(7)
\begin{align} -i \hbar \frac{1}{ f \left( x \right)} \partial_x f \left( x \right) : \psi \left( x \right) \to -i \hbar \frac{1}{ f \left( x \right)} \partial_x \left[ f \left( x \right) \psi \left( x \right) \right] \end{align}
page revision: 0, last edited: 05 Apr 2014 14:48