It is a Continuous Operator that represents the position of a particle.
In the position basis :
( p ^ ψ ) ( x ) = − i ℏ d d x ψ ( x ) (\hat{p}\psi)(x)=-i\hbar \frac{d}{dx}\psi(x) ( p ^ ψ ) ( x ) = − i ℏ d x d ψ ( x ) Note that
p ^ = ∫ p ∣ p ⟩ ⟨ p ∣ d p \hat{p}=\int p\ket{p}\bra{p}dp p ^ = ∫ p ∣ p ⟩ ⟨ p ∣ d p Which is just like Position Operator but acts on a different basis.
Proof
Let's say we have two bases for the same Hilbert space : position { ∣ x ⟩ } \{\ket{x}\} { ∣ x ⟩ } { ∣ p ⟩ } \{\ket{p}\} { ∣ p ⟩ } ⟨ x ∣ p ⟩ \braket{x|p} ⟨ x ∣ p ⟩
See de Broglie first
Constant C C C
⟨ p ∣ p ′ ⟩ = ⟨ p ∣ I ∣ p ′ ⟩ \braket{p|p'}=\bra{p}I\ket{p'} ⟨ p ∣ p ′ ⟩ = ⟨ p ∣ I ∣ p ′ ⟩ Expand out I I I continuous spectral decomposition. See Continuous Identity .
= ⟨ p ∣ ( ∫ ∣ x ⟩ ⟨ x ∣ d x ) ∣ p ′ ⟩ =\bra{p}\left(\int\ket{x}\bra{x}dx\right)\ket{p'} = ⟨ p ∣ ( ∫ ∣ x ⟩ ⟨ x ∣ d x ) ∣ p ′ ⟩ To normalize it we do
⟨ p ∣ p ′ ⟩ = ∣ C ∣ 2 ∫ − ∞ ∞ e i ( p ′ − p x ) / ℏ d x \braket{p|p'}=|C|^2\int_{-\infty}^\infty e^{i(p'-px)/\hbar}dx ⟨ p ∣ p ′ ⟩ = ∣ C ∣ 2 ∫ − ∞ ∞ e i ( p ′ − p x ) /ℏ d x We Set it to itself so that we can find C C C Normal .
Note the Fourier Transform representation of the Dirac Delta .
δ ( z ) = 1 2 π ∫ − ∞ ∞ ( 1 ) e i y z d y \delta(z)=\frac{1}{2\pi}\int_{-\infty}^\infty (1)e^{iyz}dy δ ( z ) = 2 π 1 ∫ − ∞ ∞ ( 1 ) e i y z d y Substitute it in to get
δ ( p − p ′ ) = ∣ C ∣ 2 ⋅ ℏ ⋅ 2 π δ ( p ′ − p ) \delta (p-p')= |C|^2 \cdot \hbar \cdot 2\pi \, \delta(p' - p) δ ( p − p ′ ) = ∣ C ∣ 2 ⋅ ℏ ⋅ 2 π δ ( p ′ − p ) This forces
∣ C ∣ 2 = 1 / ( 2 π ℏ ) ⇔ C = 1 / 2 π ℏ \boxed{|C|^2=1/(2\pi\hbar)\quad\Leftrightarrow \quad C=1/\sqrt{2\pi\hbar}} ∣ C ∣ 2 = 1/ ( 2 π ℏ ) ⇔ C = 1/ 2 π ℏ Compute ⟨ p ∣ ψ ⟩ \braket{p|\psi} ⟨ p ∣ ψ ⟩
⟨ p ∣ ψ ⟩ = ∫ ⟨ p ∣ x ⟩ ⟨ x ∣ ψ ⟩ d x = 1 2 π ℏ ∫ e − i p x / ℏ ψ ( x ) d x \braket{p|\psi} = \int \braket{p|x}\braket{x|\psi} \, dx = \frac{1}{\sqrt{2\pi\hbar}} \int e^{-ipx/\hbar} \, \psi(x) \, dx ⟨ p ∣ ψ ⟩ = ∫ ⟨ p ∣ x ⟩ ⟨ x ∣ ψ ⟩ d x = 2 π ℏ 1 ∫ e − i p x /ℏ ψ ( x ) d x we compare it to the CTFT we wrote earlier
ψ ~ ( k ) = 1 2 π ∫ − ∞ ∞ e − i k x ψ ( x ) d x \tilde\psi(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} \, \psi(x) \, dx ψ ~ ( k ) = 2 π 1 ∫ − ∞ ∞ e − ik x ψ ( x ) d x so
⟨ p ∣ ψ ⟩ = ψ ˉ ( p ) = 1 ℏ ψ ~ ( k = p / ℏ ) \braket{p|\psi}=\bar{\psi}(p)=\frac{1}{\sqrt{\hbar}} \tilde\psi(k = p/\hbar) ⟨ p ∣ ψ ⟩ = ψ ˉ ( p ) = ℏ 1 ψ ~ ( k = p /ℏ ) Note that
p ^ ∣ p ⟩ = p ∣ p ⟩ \hat{p}\ket{p}=p\ket{p} p ^ ∣ p ⟩ = p ∣ p ⟩ take Hermitian adjoint of both sides. Because p ^ \hat{p} p ^ Hermitian and real ,
⟨ p ∣ p ^ † = p ∗ ⟨ p ∣ ⟹ ⟨ p ∣ p ^ = p ⟨ p ∣ \bra{p}\hat{p}^\dagger = p^* \bra{p} \Longrightarrow \bra{p}\hat{p} = p\bra{p} ⟨ p ∣ p ^ † = p ∗ ⟨ p ∣ ⟹ ⟨ p ∣ p ^ = p ⟨ p ∣ Because we want to evaluate p ^ ∣ ψ ⟩ \hat{p}\ket{\psi} p ^ ∣ ψ ⟩ ⟨ p ∣ p ^ ∣ ψ ⟩ \bra{p}\hat{p}\ket{\psi} ⟨ p ∣ p ^ ∣ ψ ⟩
⟨ p ∣ p ^ ∣ ψ ⟩ = p ⟨ p ∣ ψ ⟩ = p ψ ˉ ( p ) \braket{p|\hat{p}|\psi} = p \braket{p|\psi} = p \, \bar\psi(p) ⟨ p ∣ p ^ ∣ ψ ⟩ = p ⟨ p ∣ ψ ⟩ = p ψ ˉ ( p ) In the p representation
( p ^ ψ ˉ ) ( p ) = p ψ ˉ ( p ) (\hat{p}\bar\psi)(p) = p \, \bar\psi(p) ( p ^ ψ ˉ ) ( p ) = p ψ ˉ ( p ) In the k representation
( p ^ ψ ~ ) ( k ) = ℏ k ψ ~ ( k ) (\hat{p}\tilde\psi)(k) = \hbar k \, \tilde\psi(k) ( p ^ ψ ~ ) ( k ) = ℏ k ψ ~ ( k ) In the x representation?
Let's find out p ^ ∣ ψ ⟩ \hat{p}\ket{\psi} p ^ ∣ ψ ⟩ ⟨ x ∣ p ^ ∣ ψ ⟩ \bra{x}\hat{p}\ket{\psi} ⟨ x ∣ p ^ ∣ ψ ⟩
⟨ x ∣ p ^ ∣ ψ ⟩ = ∫ − ∞ ∞ p ⟨ x ∣ p ⟩ ⟨ p ∣ ψ ⟩ d p \braket{x|\hat{p}|\psi} = \int_{-\infty}^{\infty} p \braket{x|p}\braket{p|\psi}\,dp ⟨ x ∣ p ^ ∣ ψ ⟩ = ∫ − ∞ ∞ p ⟨ x ∣ p ⟩ ⟨ p ∣ ψ ⟩ d p Since we know p p p x x x ∂ ∂ x \frac{\partial}{\partial x} ∂ x ∂
= ∫ − ∞ ∞ ( ℏ i ∂ ∂ x ⟨ x ∣ p ⟩ ) ⟨ p ∣ ψ ⟩ d p = \int_{-\infty}^{\infty} \left(\frac{\hbar}{i}\frac{\partial}{\partial x}\braket{x|p}\right)\braket{p|\psi}\,dp = ∫ − ∞ ∞ ( i ℏ ∂ x ∂ ⟨ x ∣ p ⟩ ) ⟨ p ∣ ψ ⟩ d p = ℏ i ∂ ∂ x ∫ − ∞ ∞ ⟨ x ∣ p ⟩ ⟨ p ∣ ψ ⟩ d p = \frac{\hbar}{i}\frac{\partial}{\partial x} \int_{-\infty}^{\infty} \braket{x|p}\braket{p|\psi}\,dp = i ℏ ∂ x ∂ ∫ − ∞ ∞ ⟨ x ∣ p ⟩ ⟨ p ∣ ψ ⟩ d p Recall from Continuous Operator
= ℏ i ∂ ∂ x ⟨ x ∣ ψ ⟩ = \frac{\hbar}{i}\frac{\partial}{\partial x}\braket{x|\psi} = i ℏ ∂ x ∂ ⟨ x ∣ ψ ⟩ = ℏ i ∂ ∂ x ψ ( x ) = \frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x) = i ℏ ∂ x ∂ ψ ( x ) Hence
( p ^ ψ ) ( x ) = ℏ i ∂ ∂ x ψ ( x ) (\hat{p} \psi)(x) = \frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x) ( p ^ ψ ) ( x ) = i ℏ ∂ x ∂ ψ ( x ) ( p ^ ψ ) ( x ) = − i ℏ ∂ ∂ x ψ ( x ) \boxed{(\hat{p} \psi)(x) = -i\hbar \frac{\partial}{\partial x}\psi(x)} ( p ^ ψ ) ( x ) = − i ℏ ∂ x ∂ ψ ( x ) Physical Wave Functions