Quantum Mechanics - Prof. Eric R. Bittner - University of Houston

The position operator acts on the state |ψ〉 to give the amplitude of the system to be at a
given position:
ˆx|ψ〉 = |x〉〈x|ψ〉 (2.2)
= |x〉ψ(x) (2.3)
We shall call ψ(x) the wavefunction of the system since it is the amplitude of |ψ〉 at point x. Here
we can see that ψ(x) is an eigenstate of the position operator. We also define the momentum
operator ˆp as a derivative operator:
Thus,
ˆp = −i¯h ∂
∂x
(2.4)
ˆpψ(x) = −i¯hψ ′ (x). (2.5)
Note that ψ ′ (x) �= ψ(x), thus an eigenstate of the position operator is not also an eigenstate of
the momentum operator.
We can deduce this also from the fact that ˆx and ˆp do not commute. To see this, first consider
∂
∂x xf(x) = f(x) + xf ′ (x) (2.6)
Thus (using the shorthand ∂x as partial derivative with respect to x.)
[ˆx, ˆp]f(x) = i¯h(x∂xf(x) − ∂x(xf(x))) (2.7)
= −i¯h(xf ′ (x) − f(x) − xf ′ (x)) (2.8)
= i¯hf(x) (2.9)
What are the eigenstates of the ˆp operator? To find them, consider the following eigenvalue
equation:
ˆp|φ(k)〉 = k|φ(k)〉 (2.10)
Inserting a complete set of position states using the idempotent operator
�
I = |x〉〈x|dx (2.11)
and using the “coordinate” representation of the momentum operator, we get
Thus, the solution of this is (subject to normalization)
− i¯h∂xφ(k, x) = kφ(k, x) (2.12)
φ(k, x) = C exp(ik/¯h) = 〈x|φ(k)〉 (2.13)
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