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Operators
An operator,
(say), is a mathematical entity which transforms
one function into another: i.e.,
![\begin{displaymath}
O(f(x))\rightarrow g(x).
\end{displaymath}](img525.png) |
(184) |
For instance,
is an operator, since
is a different function
to
, and is fully specified once
is given. Furthermore,
is also an operator, since
is a different function
to
, and is fully specified once
is given.
Now,
![\begin{displaymath}
x \frac{df}{dx} \neq \frac{d}{dx}\left(x f\right).
\end{displaymath}](img529.png) |
(185) |
This can also be written
![\begin{displaymath}
x \frac{d}{dx} \neq \frac{d}{dx} x,
\end{displaymath}](img530.png) |
(186) |
where the operators are assumed to act on everything to
their right, and a final
is understood [where
is a general function]. The above expression illustrates
an important point: i.e., in general, operators do not
commute. Of course, some operators do commute: e.g.,
![\begin{displaymath}
x x^2 = x^2 x.
\end{displaymath}](img531.png) |
(187) |
Finally, an operator,
, is termed linear if
![\begin{displaymath}
O(c f(x)) =c O(f(x)),
\end{displaymath}](img532.png) |
(188) |
where
is a general function, and
a general complex number.
All of the operators employed in quantum mechanics are linear.
Now, from Eqs. (158) and (174),
These expressions suggest a number of things. First, classical dynamical
variables, such as
and
, are represented in quantum mechanics
by linear operators which act on the wavefunction. Second,
displacement is represented by the algebraic operator
,
and momentum by the differential operator
: i.e.,
![\begin{displaymath}
p \equiv -{\rm i} \hbar \frac{\partial}{\partial x}.
\end{displaymath}](img538.png) |
(191) |
Finally, the expectation value of some dynamical variable represented by
the operator
is simply
![\begin{displaymath}
\langle O \rangle = \int_{-\infty}^{\infty}\psi^\ast(x,t) O(x) \psi(x,t) dx.
\end{displaymath}](img540.png) |
(192) |
Clearly, if an operator is to represent a dynamical variable which has
physical significance then its expectation value must be real.
In other words, if the operator
represents a physical variable
then we require that
, or
![\begin{displaymath}
\int_{-\infty}^{\infty} \psi^\ast (O \psi) dx = \int_{-\infty}^{\infty}(O \psi)^\ast \psi dx,
\end{displaymath}](img542.png) |
(193) |
where
is the complex conjugate of
. An operator which
satisfies the above constraint is called an Hermitian operator.
It is easily demonstrated that
and
are both Hermitian.
The Hermitian conjugate,
, of
a general operator,
, is defined as follows:
![\begin{displaymath}
\int_{-\infty}^{\infty} \psi^{\ast} (O \psi) dx=\int_{-\infty}^\infty
(O^\dag \psi)^\ast \psi dx.
\end{displaymath}](img545.png) |
(194) |
The Hermitian conjugate of an Hermitian operator is the same as the operator
itself: i.e.,
. For a non-Hermitian operator,
(say),
it is easily demonstrated that
, and that the operator
is Hermitian.
Finally, if
and
are two operators, then
.
Suppose that we wish to find the operator which corresponds to the
classical dynamical variable
. In classical mechanics, there
is no difference between
and
. However, in quantum
mechanics, we have already seen that
. So,
should be choose
or
? Actually, neither of these combinations
is Hermitian. However,
is Hermitian.
Moreover,
, which neatly resolves
our problem of which order to put
and
.
It is a reasonable guess that the operator corresponding to energy (which is
called the Hamiltonian, and conventionally denoted
) takes the form
![\begin{displaymath}
H \equiv \frac{p^2}{2 m} + V(x).
\end{displaymath}](img557.png) |
(195) |
Note that
is Hermitian. Now, it follows from Eq. (191) that
![\begin{displaymath}
H \equiv -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x).
\end{displaymath}](img558.png) |
(196) |
However, according to Schrödinger's equation, (137), we have
![\begin{displaymath}
-\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x) = {\rm i} \hbar \frac{\partial}{\partial t},
\end{displaymath}](img559.png) |
(197) |
so
![\begin{displaymath}
H \equiv {\rm i} \hbar \frac{\partial}{\partial t}.
\end{displaymath}](img560.png) |
(198) |
Thus, the time-dependent Schrödinger equation can be written
![\begin{displaymath}
{\rm i} \hbar \frac{\partial\psi}{\partial t} = H \psi.
\end{displaymath}](img561.png) |
(199) |
Finally, if
is a classical dynamical variable which is
a function of displacement, momentum, and energy, then a reasonable
guess for the corresponding operator in quantum mechanics is
, where
, and
.
Next: Momentum Representation
Up: Fundamentals of Quantum Mechanics
Previous: Ehrenfest's Theorem
Richard Fitzpatrick
2010-07-20