For every experimental observable there is a corresponding operator in quantum mechanics.
Operators must be linear , because they are derived from the Schrodinger equation, which itself is linear.
Operators must be Hermitian , because only Hermitian operators produce real eigenvalues.
Operators must produce real eigenvalues , because eigenvalues are the only possible values measured in experiments.
Operator commutators show whether two experimental observables can be measured simultaneously . For example, can one simultaneously and precisely determine the position and momentum of an electron?
Commuting operators share eigenfunctions; non-commuting operators have different eigenfunctions.
Operators ¶ In quantum mechanics, operators represent physical observables and are denoted by a hat symbol (^ \hat{} ^
For example, the momentum operator differentiates the function with respect to x x x − i ℏ -i\hbar − i ℏ
p ^ x = − i ℏ d d x \hat{p}_x = -i\hbar\frac{d}{dx} p ^ x = − i ℏ d x d The position operator simply multiplies the function by x x x
In quantum mechanics we use a simple recipe to find operators : take expressions from classical mechanics and replace position and momentum by their respective operator expressions.
Linearity of Operators ¶ A ^ ( ψ 1 + ψ 2 ) = A ^ ψ 1 + A ^ ψ 2 \hat{A}(\psi_1 + \psi_2) = \hat{A}\psi_1 + \hat{A}\psi_2 A ^ ( ψ 1 + ψ 2 ) = A ^ ψ 1 + A ^ ψ 2 A ^ ( c ψ ) = c A ^ ψ \hat{A}(c\psi) = c\hat{A}\psi A ^ ( c ψ ) = c A ^ ψ Here c c c ψ 1 \psi_1 ψ 1 ψ 2 \psi_2 ψ 2 ψ \psi ψ
x ^ \hat{x} x ^ p x ^ \hat{p_x} p x ^ H ^ \hat{H} H ^
Commutations of operators ¶ [ A ^ , B ^ ] f = ( A ^ B ^ − B ^ A ^ ) f {\left[\hat{A},\hat{B}\right]f = \left(\hat{A}\hat{B} - \hat{B}\hat{A}\right)f} [ A ^ , B ^ ] f = ( A ^ B ^ − B ^ A ^ ) f From linear algebra we know that the order of matrix multiplication matters and that A B ≠ B A AB\neq BA A B = B A A A A B B B
Thus we also generally expect A ^ B ^ ≠ B ^ A ^ \hat{A}\hat{B} \neq \hat{B}\hat{A} A ^ B ^ = B ^ A ^
We can quantify the relationship between two operators by computing the commutator .
If the commutator is zero, the order of multiplication of operators or matrices can be changed.
If the commutator is non-zero, the order matters and cannot be changed.
Prove that operators A ^ = x \hat{A} = x A ^ = x B ^ = d / d x \hat{B} = d/dx B ^ = d / d x [ A ^ , B ^ ] ≠ 0 \left[\hat{A}, \hat{B}\right] \ne 0 [ A ^ , B ^ ] = 0
Let f f f A ^ B ^ f \hat{A}\hat{B}f A ^ B ^ f B ^ A ^ f \hat{B}\hat{A}f B ^ A ^ f
A ^ B ^ f = x f ′ ( x ) and B ^ A ^ f = d d x ( x f ( x ) ) = f ( x ) + x f ′ ( x ) \hat{A}\hat{B}f = xf'(x)\textnormal{ and } \hat{B}\hat{A}f = \frac{d}{dx}\left(xf(x)\right) = f(x) + xf'(x) A ^ B ^ f = x f ′ ( x ) and B ^ A ^ f = d x d ( x f ( x ) ) = f ( x ) + x f ′ ( x ) [ A ^ , B ^ ] f = A ^ B ^ f − B ^ A ^ f = − f \left[\hat{A},\hat{B}\right]f = \hat{A}\hat{B}f - \hat{B}\hat{A}f = -f [ A ^ , B ^ ] f = A ^ B ^ f − B ^ A ^ f = − f ⇒ [ A ^ , B ^ ] = − 1 (this is non-zero and the operators do not commute) \Rightarrow \left[\hat{A},\hat{B}\right] = -1\textnormal{ (this is non-zero and the operators do not commute)} ⇒ [ A ^ , B ^ ] = − 1 (this is non-zero and the operators do not commute) Simple rules for commutators ¶ [ A , A ] = [ A , A n ] = [ A n , A ] = 0 {\left[A,A\right] = \left[A,A^n\right] = \left[A^n,A\right] = 0} [ A , A ] = [ A , A n ] = [ A n , A ] = 0 [ A , B ] = − [ B , A ] {\left[A,B\right] = -\left[B,A\right]} [ A , B ] = − [ B , A ] Commutators and experimental measurements ¶ We have seen previously that operators may not always commute (i.e., [ A , B ] ≠ 0 [A, B] \ne 0 [ A , B ] = 0 x ^ \hat{x} x ^ p ^ x \hat{p}_x p ^ x
p ^ x x ^ ψ ( x ) = p ^ x ( x ψ ( x ) ) = ( ℏ i d d x ) ( x ψ ( x ) ) = ℏ x i d ψ ( x ) d x + ℏ i ψ ( x ) {\hat{p}_x\hat{x}\psi(x) = \hat{p}_x\left(x\psi(x)\right) = \left(\frac{\hbar}{i}\frac{d}{dx}\right)\left(x\psi(x)\right) = \frac{\hbar x}{i}\frac{d\psi(x)}{dx} + \frac{\hbar}{i}\psi(x)} p ^ x x ^ ψ ( x ) = p ^ x ( x ψ ( x ) ) = ( i ℏ d x d ) ( x ψ ( x ) ) = i ℏ x d x d ψ ( x ) + i ℏ ψ ( x ) x ^ p ^ x ψ ( x ) = x ( ℏ i d ψ ( x ) d x ) {\hat{x}\hat{p}_x\psi(x) = x\left(\frac{\hbar}{i}\frac{d\psi(x)}{dx}\right)} x ^ p ^ x ψ ( x ) = x ( i ℏ d x d ψ ( x ) ) ⇒ [ p ^ x , x ^ ] ψ ( x ) = ( p ^ x x ^ − x ^ p ^ x ) ψ ( x ) = ℏ i ψ ( x ) {\Rightarrow \left[\hat{p}_x,\hat{x}\right]\psi(x) = \left(\hat{p}_x\hat{x} - \hat{x}\hat{p}_x\right)\psi(x) = \frac{\hbar}{i}\psi(x)} ⇒ [ p ^ x , x ^ ] ψ ( x ) = ( p ^ x x ^ − x ^ p ^ x ) ψ ( x ) = i ℏ ψ ( x ) ⇒ [ p ^ x , x ^ ] = ℏ i {\Rightarrow \left[\hat{p}_x,\hat{x}\right] = \frac{\hbar}{i}} ⇒ [ p ^ x , x ^ ] = i ℏ In contrast, the kinetic energy operator and the momentum operator commute:
[ T ^ , p ^ x ] = [ p ^ x 2 2 m , p ^ x ] = p x 3 2 m − p x 3 2 m = 0 {\left[\hat{T},\hat{p}_x\right] = \left[\frac{\hat{p}_x^2}{2m},\hat{p}_x\right] = \frac{p_x^3}{2m} - \frac{p_x^3}{2m} = 0} [ T ^ , p ^ x ] = [ 2 m p ^ x 2 , p ^ x ] = 2 m p x 3 − 2 m p x 3 = 0 We had the uncertainty principle for the position and momentum operators:
Δ x Δ p x ≥ ℏ 2 \Delta x\Delta p_x \ge \frac{\hbar}{2} Δ x Δ p x ≥ 2 ℏ In general, it turns out that for operators A ^ \hat{A} A ^ B ^ \hat{B} B ^ uncertainty principle applies in the following form:
Δ A Δ B ≥ 1 2 ∣ < [ A ^ , B ^ ] > ∣ {\Delta A\Delta B \ge \frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right|} Δ A Δ B ≥ 2 1 ∣ ∣ ⟨ [ A ^ , B ^ ] ⟩ ∣ ∣ 1 2 ∣ < [ A ^ , B ^ ] > ∣ = 1 2 ∣ < [ x ^ , p ^ x ] > ∣ = 1 2 ∣ < ℏ i > ∣ = 1 2 ∣ < ψ ∣ ℏ i ∣ ψ > ∣ = 1 2 ∣ ℏ i < ψ ∣ ψ > ⏟ = 1 ∣ = ℏ 2 \frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right| = \frac{1}{2}\left|\left<\left[\hat{x},\hat{p}_x\right]\right>\right| = \frac{1}{2}\left|\left<\frac{\hbar}{i}\right>\right|
= \frac{1}{2}\left|\left<\psi\left|\frac{\hbar}{i}\right|\psi\right>\right| = \frac{1}{2}\left|\frac{\hbar}{i}\underbrace{\left<\psi\left|\psi\right.\right>}_{=1}\right| = \frac{\hbar}{2} 2 1 ∣ ∣ ⟨ [ A ^ , B ^ ] ⟩ ∣ ∣ = 2 1 ∣ ⟨ [ x ^ , p ^ x ] ⟩ ∣ = 2 1 ∣ ∣ ⟨ i ℏ ⟩ ∣ ∣ = 2 1 ∣ ∣ ⟨ ψ ∣ ∣ i ℏ ∣ ∣ ψ ⟩ ∣ ∣ = 2 1 ∣ ∣ i ℏ = 1 ⟨ ψ ∣ ψ ⟩ ∣ ∣ = 2 ℏ ⇒ Δ x Δ p x ≥ ℏ 2 \Rightarrow \Delta x\Delta p_x \ge \frac{\hbar}{2} ⇒ Δ x Δ p x ≥ 2 ℏ Commuting operators and simultaneous measurements ¶ [ A ^ , B ^ ] = 0 [\hat{A},\hat{B}]=0 [ A ^ , B ^ ] = 0 A ^ ϕ k = a k ϕ k \hat{A}\phi_k = a_k \phi_k A ^ ϕ k = a k ϕ k B ^ ϕ k = b k ϕ k \hat{B}\phi_k = b_k \phi_k B ^ ϕ k = b k ϕ k
Denote the eigenvalues of A ^ \hat{A} A ^ B ^ \hat{B} B ^ a i a_i a i b i b_i b i ψ i \psi_i ψ i
A ^ ψ i = a i ψ i and B ^ ψ i = b i ψ i \hat{A}\psi_i = a_i\psi_i\textnormal{ and }\hat{B}\psi_i = b_i\psi_i A ^ ψ i = a i ψ i and B ^ ψ i = b i ψ i By using these two equations and expressing the general wavefunction ψ \psi ψ
A ^ B ^ ψ = A ^ ( B ^ ψ ) = A ^ ( B ^ ∑ i = 1 ∞ c i ψ i ) ⏞ complete basis = A ^ ( ∑ i = 1 ∞ c i B ^ ψ i ) ⏞ B ^ linear = A ^ ( ∑ i = 1 ∞ c i b i ψ i ) ⏞ eigenfunction of B ^ \hat{A}\hat{B}\psi = \hat{A}\left(\hat{B}\psi\right) = \hat{A}\overbrace{\left(\hat{B}\sum\limits_{i=1}^{\infty}c_i\psi_i\right)}^{\textnormal{complete basis}}
= \hat{A}\overbrace{\left(\sum\limits_{i=1}^{\infty}c_i\hat{B}\psi_i\right)}^{\hat{B}\textnormal{ linear}} = \hat{A}\overbrace{\left(\sum\limits_{i=1}^{\infty}c_ib_i\psi_i\right)}^{\textnormal{eigenfunction of }\hat{B}} A ^ B ^ ψ = A ^ ( B ^ ψ ) = A ^ ( B ^ i = 1 ∑ ∞ c i ψ i ) complete basis = A ^ ( i = 1 ∑ ∞ c i B ^ ψ i ) B ^ linear = A ^ ( i = 1 ∑ ∞ c i b i ψ i ) eigenfunction of B ^ = ∑ i = 1 ∞ c i b i A ^ ψ i ⏞ A ^ linear = ∑ i = 1 ∞ c i b i a i ψ i ⏞ eigenfunction of A ^ = ∑ i = 1 ∞ c i a i b i ψ i ⏞ a i and b i are constants = ∑ i = 1 ∞ c i a i B ^ ψ i = \overbrace{\sum\limits_{i=1}^{\infty}c_ib_i\hat{A}\psi_i}^{\hat{A}\textnormal{ linear}} = \overbrace{\sum\limits_{i=1}^{\infty}c_ib_ia_i\psi_i}^{\textnormal{eigenfunction of }\hat{A}} = \overbrace{\sum\limits_{i=1}^{\infty}c_ia_ib_i\psi_i}^{a_i\textnormal{ and }b_i\textnormal{ are constants}} = \sum\limits_{i=1}^{\infty}c_ia_i\hat{B}\psi_i = i = 1 ∑ ∞ c i b i A ^ ψ i A ^ linear = i = 1 ∑ ∞ c i b i a i ψ i eigenfunction of A ^ = i = 1 ∑ ∞ c i a i b i ψ i a i and b i are constants = i = 1 ∑ ∞ c i a i B ^ ψ i = B ^ ∑ i = 1 ∞ c i a i ψ i = B ^ ∑ i = 1 ∞ c i A ^ ψ i = B ^ A ^ ∑ i = 1 ∞ c i ψ i = B ^ A ^ ψ = \hat{B}\sum\limits_{i=1}^{\infty}c_ia_i\psi_i = \hat{B}\sum\limits_{i=1}^{\infty}c_i\hat{A}\psi_i = \hat{B}\hat{A}\sum\limits_{i=1}^{\infty}c_i\psi_i = \hat{B}\hat{A}\psi = B ^ i = 1 ∑ ∞ c i a i ψ i = B ^ i = 1 ∑ ∞ c i A ^ ψ i = B ^ A ^ i = 1 ∑ ∞ c i ψ i = B ^ A ^ ψ ⇒ [ A ^ , B ^ ] = 0 \Rightarrow \left[\hat{A},\hat{B}\right] = 0 ⇒ [ A ^ , B ^ ] = 0 Note that the commutation relation must apply to all well-behaved functions and not just for some given subset of functions!
If operators commute, that means we can simultaneously measure the corresponding observables in a single experiment .
For instance, the kinetic energy and momentum operators commute, so we can measure momentum and kinetic energy together. But we cannot do the same for momentum and position.
If we measure observables A A A B B B ϕ k \phi_k ϕ k a k a_k a k b k b_k b k
Expectation expression ¶ The expectation value of an observable A ^ \hat{A} A ^
⟨ A ⟩ = ∫ ψ ∗ A ^ ψ d τ \langle A \rangle = \int \psi^* \hat{A} \psi \, d\tau ⟨ A ⟩ = ∫ ψ ∗ A ^ ψ d τ Special Case : If the wavefunction ψ \psi ψ A ^ \hat{A} A ^ a a a
A ^ ψ = a ψ \hat{A}\psi = a\psi A ^ ψ = a ψ Then the expectation value simplifies to:
⟨ A ⟩ = ∫ ψ ∗ a ψ d τ = a ∫ ψ ∗ ψ d τ = a \langle A \rangle = \int \psi^* a \psi \, d\tau = a \int \psi^*\psi \, d\tau = a ⟨ A ⟩ = ∫ ψ ∗ a ψ d τ = a ∫ ψ ∗ ψ d τ = a Since ∫ ψ ∗ ψ d τ = 1 \int \psi^*\psi \, d\tau = 1 ∫ ψ ∗ ψ d τ = 1 a a a
Dirac Notation ¶ To express quantum states and operators more compactly, we use Dirac (bra–ket) notation .
A state is written as a ket , ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ bra , ⟨ ψ ∣ \langle\psi| ⟨ ψ ∣
The inner product between two states corresponds to the integral over space:
⟨ ϕ ∣ ψ ⟩ = ∫ ϕ ∗ ( r ) , ψ ( r ) , d τ \langle \phi | \psi \rangle = \int \phi^*(r), \psi(r), d\tau ⟨ ϕ ∣ ψ ⟩ = ∫ ϕ ∗ ( r ) , ψ ( r ) , d τ ⟨ A ⟩ = ⟨ ψ ∣ A ^ ∣ ψ ⟩ \langle A \rangle = \langle \psi | \hat{A} | \psi \rangle ⟨ A ⟩ = ⟨ ψ ∣ A ^ ∣ ψ ⟩ This form is elegant and general: it applies to all quantum systems, independent of the particular representation (position, momentum, etc.).
Hermitian Property of Operators ¶ In quantum mechanics, operators often act on complex-valued functions , so we need a notion of “complex conjugate” that applies not just to numbers but to operators.
This leads to the concept of the adjoint operator .
The Adjoint (Conjugate Transpose) ¶ For complex numbers we take the complex conjugate :
( 3 + 2 i ) ∗ = 3 − 2 i (3 + 2i)^* = 3 - 2i ( 3 + 2 i ) ∗ = 3 − 2 i
For matrices or linear operators, the corresponding operation is the adjoint , denoted by the dagger symbol ( † ) (\dagger) ( † )
In matrix element form, taking the adjoint generates different elements:
a j k = ⟨ ψ j ∣ A ^ ∣ ψ k ⟩ ⇒ a k j ∗ = ⟨ ψ k ∣ A ^ † ∣ ψ j ⟩ . a_{jk} = \langle \psi_j | \hat{A} | \psi_k \rangle
\quad \Rightarrow \quad
a^*_{kj} = \langle \psi_k | \hat{A}^\dagger | \psi_j \rangle. a jk = ⟨ ψ j ∣ A ^ ∣ ψ k ⟩ ⇒ a kj ∗ = ⟨ ψ k ∣ A ^ † ∣ ψ j ⟩ . Hermitian (Self-Adjoint) Operators ¶ An operator is Hermitian if it equals its own adjoint:
A ^ = A ^ † . \hat{A} = \hat{A}^\dagger. A ^ = A ^ † . This means the operator behaves the same way when acting on either side of the inner product.
A = A † , a j k = a k j ∗ . A = A^\dagger, \quad a_{jk} = a_{kj}^*. A = A † , a jk = a kj ∗ . ⟨ ϕ ∣ A ^ ψ ⟩ = ⟨ A ^ ϕ ∣ ψ ⟩ , or equivalently, ⟨ j ∣ A ^ ∣ k ⟩ = ⟨ k ∣ A ^ ∣ j ⟩ ∗ . \langle \phi | \hat{A}\psi \rangle = \langle \hat{A}\phi | \psi \rangle,
\qquad \text{or equivalently,} \qquad
\langle j| \hat{A}|k\rangle = \langle k| \hat{A}|j\rangle^*. ⟨ ϕ ∣ A ^ ψ ⟩ = ⟨ A ^ ϕ ∣ ψ ⟩ , or equivalently, ⟨ j ∣ A ^ ∣ k ⟩ = ⟨ k ∣ A ^ ∣ j ⟩ ∗ . In integral form:
∫ ψ j ∗ ( x ) , [ A ^ ψ k ( x ) ] , d x = ∫ ψ k ( x ) , [ A ^ ψ j ( x ) ] ∗ , d x . \int \psi_j^*(x), [\hat{A}\psi_k(x)],dx
= \int \psi_k(x), [\hat{A}\psi_j(x)]^*,dx. ∫ ψ j ∗ ( x ) , [ A ^ ψ k ( x )] , d x = ∫ ψ k ( x ) , [ A ^ ψ j ( x ) ] ∗ , d x . Why Hermitian Operators Matter ¶ Eigenvalues are real : Observables in quantum mechanics (energy, momentum, position, etc.) are represented by Hermitian operators , ensuring all measurement outcomes are real numbers.
A ^ ∣ ψ ⟩ = a ∣ ψ ⟩ ⟹ a ∈ R . \hat{A}|\psi\rangle = a|\psi\rangle \implies a \in \mathbb{R}. A ^ ∣ ψ ⟩ = a ∣ ψ ⟩ ⟹ a ∈ R .
∫ ψ ∗ A ^ ψ d τ = a \int \psi^* \hat{A} \psi \, d\tau = a ∫ ψ ∗ A ^ ψ d τ = a ∫ ψ ( A ^ ψ ) ∗ d τ = a ∗ \int \psi \left(\hat{A} \psi\right)^* \, d\tau = a^* ∫ ψ ( A ^ ψ ) ∗ d τ = a ∗ Eigenfunctions are orthogonal :
⟨ ψ m ∣ ψ n ⟩ = 0 ( m ≠ n ) . \langle \psi_m | \psi_n \rangle = 0 \quad (m \ne n). ⟨ ψ m ∣ ψ n ⟩ = 0 ( m = n ) .
The Hermitian property can also be used to show that eigenfunctions ψ j \psi_j ψ j ψ k \psi_k ψ k a j a_j a j a k a_k a k a j ≠ a k a_j \neq a_k a j = a k
LHS: ∫ ψ j ∗ A ^ ψ k d τ = ∫ ψ j ∗ a k ψ k d τ = a k ∫ ψ j ∗ ψ k d τ \textnormal{LHS: } \int \psi_j^* \hat{A} \psi_k \, d\tau = \int \psi_j^* a_k \psi_k \, d\tau = a_k \int \psi_j^* \psi_k \, d\tau LHS: ∫ ψ j ∗ A ^ ψ k d τ = ∫ ψ j ∗ a k ψ k d τ = a k ∫ ψ j ∗ ψ k d τ RHS: ∫ ψ k ( A ^ ψ j ) ∗ d τ = ∫ ψ k ( a j ψ j ) ∗ d τ = a j ∫ ψ j ∗ ψ k d τ \textnormal{RHS: } \int \psi_k \left(\hat{A} \psi_j \right)^* \, d\tau = \int \psi_k \left(a_j \psi_j \right)^* \, d\tau = a_j \int \psi_j^* \psi_k \, d\tau RHS: ∫ ψ k ( A ^ ψ j ) ∗ d τ = ∫ ψ k ( a j ψ j ) ∗ d τ = a j ∫ ψ j ∗ ψ k d τ ( a k − a j ) ∫ ψ j ∗ ψ k d τ = 0 \left(a_k - a_j \right) \int \psi_j^* \psi_k \, d\tau = 0 ( a k − a j ) ∫ ψ j ∗ ψ k d τ = 0 ∫ ψ j ∗ ψ k d τ = 0 \int \psi_j^* \psi_k \, d\tau = 0 ∫ ψ j ∗ ψ k d τ = 0 This shows that ψ j \psi_j ψ j ψ k \psi_k ψ k
Note : If a j = a k a_j = a_k a j = a k
Which of these matrices is Hermitian?
( 1 2 3 4 ) \begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix} ( 1 3 2 4 ) ( i 0 0 1 ) \begin{pmatrix}
i & 0 \\
0 & 1
\end{pmatrix} ( i 0 0 1 ) ( − 1 − 3 i 3 i 8 ) \begin{pmatrix}
-1 & -3i \\
3i & 8
\end{pmatrix} ( − 1 3 i − 3 i 8 ) ( 1 2 i 2 i 3 ) \begin{pmatrix}
1 & 2i \\
2i & 3
\end{pmatrix} ( 1 2 i 2 i 3 )
For the first matrix we have a 12 = 2 ≠ a 21 ∗ = 3 a_{12}=2\neq a^{*}_{21}=3 a 12 = 2 = a 21 ∗ = 3 non-Hermitian
For the second matrix a 11 ≠ a 11 ∗ = 0 a_{11}\neq a^{*}_{11}=0 a 11 = a 11 ∗ = 0 non-Hermitian
For the third matrix a 12 = − 3 i = a 21 ∗ = ( 3 i ) ∗ = − 3 i a_{12}=-3i =a^{*}_{21} = (3i)^{*}=-3i a 12 = − 3 i = a 21 ∗ = ( 3 i ) ∗ = − 3 i Hermitian
For the fourth matrix a 12 = 2 i ≠ a 21 ∗ = ( 2 i ) ∗ = − 2 i a_{12}=2i \neq a^{*}_{21} = (2i)^{*} = -2i a 12 = 2 i = a 21 ∗ = ( 2 i ) ∗ = − 2 i non-Hermitian
Seeing that differentiation operators are Hermitian requires a little more work.
A trick that helps is integration by parts, where the boundary term is zero because the wavefunction decays to zero at the boundaries (postulate 1, keeping probability finite).
∫ ψ 1 d ψ 2 = − ∫ ψ 2 d ψ 1 + ψ 1 ψ 2 ∣ x m i n x m a x = − ∫ ψ 2 d ψ 1 \int \psi_1 d\psi_2 =- \int \psi_2d\psi_1 + \psi_1\psi_2\Big|_{x_{min}}^{x_{max}} =- \int \psi_2d\psi_1 ∫ ψ 1 d ψ 2 = − ∫ ψ 2 d ψ 1 + ψ 1 ψ 2 ∣ ∣ x min x ma x = − ∫ ψ 2 d ψ 1
∫ − ∞ ∞ ψ j ∗ ( x ) ( − i ℏ d ψ k ( x ) d x ) d x = − i ℏ ∫ − ∞ ∞ ψ j ∗ ( x ) d ψ k ( x ) d x d x = ∫ − ∞ ∞ ψ k ( x ) ( i ℏ d ψ j ∗ ( x ) d x ) d x ⏞ i n t e g r a t i o n b y p a r t s {\int\limits_{-\infty}^{\infty}\psi_j^*(x)\left(-i\hbar\frac{d\psi_k(x)}{dx}\right)dx} = -i\hbar\int\limits_{-\infty}^{\infty}\psi_j^*(x)\frac{d\psi_k(x)}{dx}dx = \\ \overbrace{\int\limits_{-\infty}^{\infty}\psi_k(x)\left(i\hbar\frac{d\psi_j^*(x)}{dx}\right)dx}^{{integration\, by\, parts}} − ∞ ∫ ∞ ψ j ∗ ( x ) ( − i ℏ d x d ψ k ( x ) ) d x = − i ℏ − ∞ ∫ ∞ ψ j ∗ ( x ) d x d ψ k ( x ) d x = − ∞ ∫ ∞ ψ k ( x ) ( i ℏ d x d ψ j ∗ ( x ) ) d x in t e g r a t i o n b y p a r t s = ∫ − ∞ ∞ ψ k ( x ) ( − i ℏ d ψ j ( x ) d x ) ∗ d x ⇒ p ^ x is Hermitian = {\int\limits_{-\infty}^{\infty}\psi_k(x)\left(-i\hbar\frac{d\psi_j(x)}{dx}\right)^*dx} \Rightarrow \hat{p}_x\textnormal{ is Hermitian} = − ∞ ∫ ∞ ψ k ( x ) ( − i ℏ d x d ψ j ( x ) ) ∗ d x ⇒ p ^ x is Hermitian
Geometric Intuition Hermitian operators are the analog of symmetric matrices in real vector spaces. They represent linear transformations that do not rotate vectors into complex directions : they only stretch or compress them along real axes.
Problems ¶ Problem-1: Is the x d / d x xd/dx x d / d x ¶ Check whether the operator A ^ = x d / d x \hat{A} = xd/dx A ^ = x d / d x
∫ a b ψ 1 ∗ ( x ) ( x d d x ψ 2 ( x ) ) d x = ∫ a b ( x d d x ψ 1 ( x ) ) ∗ ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( x \frac{d}{dx} \psi_2(x) \right) \, dx = \int_{a}^{b} \left( x \frac{d}{dx} \psi_1(x) \right)^* \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( x d x d ψ 2 ( x ) ) d x = ∫ a b ( x d x d ψ 1 ( x ) ) ∗ ψ 2 ( x ) d x Note how complex conjugation applies to an expression with an operator inside.
But since our operator contains no imaginary numbers, complex conjugation only applies to the wavefunction.
Step 1: Left-hand side
The left-hand side is:
∫ a b ψ 1 ∗ ( x ) ( x d d x ψ 2 ( x ) ) d x \int_{a}^{b} \psi_1^*(x) \left( x \frac{d}{dx} \psi_2(x) \right) \, dx ∫ a b ψ 1 ∗ ( x ) ( x d x d ψ 2 ( x ) ) d x Step 2: Integration by parts
We apply integration by parts to simplify this expression. Using the product rule for differentiation, we get:
∫ a b ψ 1 ∗ ( x ) ( x d d x ψ 2 ( x ) ) d x = [ x ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b − ∫ a b d d x ( x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( x \frac{d}{dx} \psi_2(x) \right) \, dx = \left[ x \psi_1^*(x) \psi_2(x) \right]_{a}^{b} - \int_{a}^{b} \frac{d}{dx} \left( x \psi_1^*(x) \right) \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( x d x d ψ 2 ( x ) ) d x = [ x ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b − ∫ a b d x d ( x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x The boundary term [ x ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b \left[ x \psi_1^*(x) \psi_2(x) \right]_{a}^{b} [ x ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b
Now, for the remaining integral, we apply the derivative to the product x ψ 1 ∗ ( x ) x \psi_1^*(x) x ψ 1 ∗ ( x )
∫ a b d d x ( x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x = ∫ a b ( ψ 1 ∗ ( x ) + x d d x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x \int_{a}^{b} \frac{d}{dx} \left( x \psi_1^*(x) \right) \psi_2(x) \, dx = \int_{a}^{b} \left( \psi_1^*(x) + x \frac{d}{dx} \psi_1^*(x) \right) \psi_2(x) \, dx ∫ a b d x d ( x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x = ∫ a b ( ψ 1 ∗ ( x ) + x d x d ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x Thus, the left-hand side becomes:
∫ a b ψ 1 ∗ ( x ) ψ 2 ( x ) d x + ∫ a b x d d x ψ 1 ∗ ( x ) ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \psi_2(x) \, dx + \int_{a}^{b} x \frac{d}{dx} \psi_1^*(x) \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ψ 2 ( x ) d x + ∫ a b x d x d ψ 1 ∗ ( x ) ψ 2 ( x ) d x Step 3: Right-hand side
The right-hand side is:
∫ a b ( x d d x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x \int_{a}^{b} \left( x \frac{d}{dx} \psi_1^*(x) \right) \psi_2(x) \, dx ∫ a b ( x d x d ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x Step 4: Comparison
Now, we compare the two expressions. The left-hand side contains the extra term:
∫ a b ψ 1 ∗ ( x ) ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ψ 2 ( x ) d x which is not present in the right-hand side. This means:
∫ a b ψ 1 ∗ ( x ) ( x d d x ψ 2 ( x ) ) d x ≠ ∫ a b ( x d d x ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( x \frac{d}{dx} \psi_2(x) \right) \, dx \neq \int_{a}^{b} \left( x \frac{d}{dx} \psi_1^*(x) \right) \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( x d x d ψ 2 ( x ) ) d x = ∫ a b ( x d x d ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x Problem-2: Is the d 2 / d x 2 d^2/dx^2 d 2 / d x 2 ¶ ∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( d 2 d x 2 ψ 1 ( x ) ) ∗ d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \psi_2(x)\left( \frac{d^2}{dx^2} \psi_1(x) \right)^* \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( d x 2 d 2 ψ 1 ( x ) ) ∗ d x
To show that the operator A ^ = d 2 d x 2 \hat{A} = \frac{d^2}{dx^2} A ^ = d x 2 d 2
∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ( d 2 d x 2 ψ 1 ( x ) ) ∗ ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \left( \frac{d^2}{dx^2} \psi_1(x) \right)^* \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ( d x 2 d 2 ψ 1 ( x ) ) ∗ ψ 2 ( x ) d x Step 1: Left-hand side ¶ The left-hand side is:
∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x Step 2: Integration by parts ¶ We apply integration by parts twice. First, applying integration by parts to the term ψ 1 ∗ ( x ) d 2 d x 2 ψ 2 ( x ) \psi_1^*(x) \frac{d^2}{dx^2} \psi_2(x) ψ 1 ∗ ( x ) d x 2 d 2 ψ 2 ( x )
∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = [ ψ 1 ∗ ( x ) d d x ψ 2 ( x ) ] a b − ∫ a b d d x ψ 1 ∗ ( x ) d d x ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \left[ \psi_1^*(x) \frac{d}{dx} \psi_2(x) \right]_{a}^{b} - \int_{a}^{b} \frac{d}{dx} \psi_1^*(x) \frac{d}{dx} \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = [ ψ 1 ∗ ( x ) d x d ψ 2 ( x ) ] a b − ∫ a b d x d ψ 1 ∗ ( x ) d x d ψ 2 ( x ) d x The boundary term [ ψ 1 ∗ ( x ) d d x ψ 2 ( x ) ] a b \left[ \psi_1^*(x) \frac{d}{dx} \psi_2(x) \right]_{a}^{b} [ ψ 1 ∗ ( x ) d x d ψ 2 ( x ) ] a b
We now apply integration by parts again to the remaining term:
− ∫ a b d d x ψ 1 ∗ ( x ) d d x ψ 2 ( x ) d x = [ d d x ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b − ∫ a b d 2 d x 2 ψ 1 ∗ ( x ) ψ 2 ( x ) d x -\int_{a}^{b} \frac{d}{dx} \psi_1^*(x) \frac{d}{dx} \psi_2(x) \, dx = \left[ \frac{d}{dx} \psi_1^*(x) \psi_2(x) \right]_{a}^{b} - \int_{a}^{b} \frac{d^2}{dx^2} \psi_1^*(x) \psi_2(x) \, dx − ∫ a b d x d ψ 1 ∗ ( x ) d x d ψ 2 ( x ) d x = [ d x d ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b − ∫ a b d x 2 d 2 ψ 1 ∗ ( x ) ψ 2 ( x ) d x Again, the boundary term [ d d x ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b \left[ \frac{d}{dx} \psi_1^*(x) \psi_2(x) \right]_{a}^{b} [ d x d ψ 1 ∗ ( x ) ψ 2 ( x ) ] a b
∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ( d 2 d x 2 ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \left( \frac{d^2}{dx^2} \psi_1^*(x) \right) \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ( d x 2 d 2 ψ 1 ∗ ( x ) ) ψ 2 ( x ) d x Step 3: Conclusion
Since the two sides are equal, we conclude that the operator d 2 d x 2 \frac{d^2}{dx^2} d x 2 d 2
∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ( d 2 d x 2 ψ 1 ( x ) ) ∗ ψ 2 ( x ) d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \left( \frac{d^2}{dx^2} \psi_1(x) \right)^* \psi_2(x) \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ( d x 2 d 2 ψ 1 ( x ) ) ∗ ψ 2 ( x ) d x Problem-3: Is the i d 2 / d x 2 id^2/dx^2 i d 2 / d x 2 ¶ ∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( d 2 d x 2 ψ 1 ( x ) ) ∗ d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \psi_2(x)\left( \frac{d^2}{dx^2} \psi_1(x) \right)^* \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( d x 2 d 2 ψ 1 ( x ) ) ∗ d x
From the last problem we learned that the following condition holds, which makes the second derivative operator d 2 / d x 2 d^2/dx^2 d 2 / d x 2
∫ a b ψ 1 ∗ ( x ) ( d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( d 2 d x 2 ψ 1 ∗ ( x ) ) d x \int_{a}^{b} \psi_1^*(x) \left( \frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \psi_2(x)\left( \frac{d^2}{dx^2} \psi_1^*(x) \right) \, dx ∫ a b ψ 1 ∗ ( x ) ( d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( d x 2 d 2 ψ 1 ∗ ( x ) ) d x ∫ a b ψ 1 ∗ ( x ) ( i d 2 d x 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( i d 2 d x 2 ψ 1 ( x ) ) ∗ d x = − ∫ a b ψ 2 ( x ) ( i d 2 d x 2 ψ 1 ( x ) ∗ ) \int_{a}^{b} \psi_1^*(x) \left( i\frac{d^2}{dx^2} \psi_2(x) \right) \, dx = \int_{a}^{b} \psi_2(x)\left( i\frac{d^2}{dx^2} \psi_1(x) \right)^* \, dx = -\int_{a}^{b} \psi_2(x)\left( i\frac{d^2}{dx^2} \psi_1(x)^* \right) ∫ a b ψ 1 ∗ ( x ) ( i d x 2 d 2 ψ 2 ( x ) ) d x = ∫ a b ψ 2 ( x ) ( i d x 2 d 2 ψ 1 ( x ) ) ∗ d x = − ∫ a b ψ 2 ( x ) ( i d x 2 d 2 ψ 1 ( x ) ∗ ) Problem-4: Identify Hermitian Matrices ¶ A = ( 1 2 2 3 ) A = \begin{pmatrix}
1 & 2 \\
2 & 3
\end{pmatrix} A = ( 1 2 2 3 ) B = ( i 1 − 1 − i ) B = \begin{pmatrix}
i & 1 \\
-1 & -i
\end{pmatrix} B = ( i − 1 1 − i ) C = ( 2 i − i 2 ) C = \begin{pmatrix}
2 & i \\
-i & 2
\end{pmatrix} C = ( 2 − i i 2 )
A Matrix
To check if a matrix is Hermitian, it must satisfy the condition A = A † A = A^\dagger A = A † A † A^\dagger A † A A A
The transpose of A A A
A † = ( 1 2 2 3 )
A^\dagger = \begin{pmatrix}
1 & 2 \\
2 & 3
\end{pmatrix} A † = ( 1 2 2 3 )
Since A = A † A = A^\dagger A = A † A A A Hermitian .
B Matrix
Now, let’s compute the conjugate transpose of B B B
B † = ( − i − 1 1 i ) B^\dagger = \begin{pmatrix}
-i & -1 \\
1 & i
\end{pmatrix} B † = ( − i 1 − 1 i ) Clearly, B ≠ B † B \neq B^\dagger B = B † B B B not Hermitian .
C Matrix
The conjugate transpose of C C C
C † = ( 2 − i i 2 ) C^\dagger = \begin{pmatrix}
2 & -i \\
i & 2
\end{pmatrix} C † = ( 2 i − i 2 ) Since C = C † C = C^\dagger C = C † C C C Hermitian .
Problem-5 Momentum Matrix ¶ Show how the momentum operator looks in matrix form using a finite-dimensional example where you evaluate the wavefunction on 4 points, which will correspond to a 4 × 4 4 \times 4 4 × 4
We can represent the momentum operator p ^ = − i ℏ d d x \hat{p} = -i \hbar \frac{d}{dx} p ^ = − i ℏ d x d
For simplicity, let’s assume we are working in a discrete system, where we approximate the derivative d d x \frac{d}{dx} d x d x n x_n x n
d ψ ( x ) d x ≈ ψ ( x n + 1 ) − ψ ( x n − 1 ) 2 Δ x \frac{d \psi(x)}{dx} \approx \frac{\psi(x_{n+1}) - \psi(x_{n-1})}{2 \Delta x} d x d ψ ( x ) ≈ 2Δ x ψ ( x n + 1 ) − ψ ( x n − 1 ) where Δ x \Delta x Δ x
The corresponding momentum operator matrix in this finite-dimensional space can be written as a skew-symmetric matrix that captures this finite difference behavior.
Here is an example of a 4 × 4 4 \times 4 4 × 4 P P P ℏ = 1 \hbar = 1 ℏ = 1
P = i 2 Δ x ( 0 − 1 0 1 1 0 − 1 0 0 1 0 − 1 − 1 0 1 0 ) P = \frac{i}{2 \Delta x} \begin{pmatrix}
0 & -1 & 0 & 1 \\
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1 \\
-1 & 0 & 1 & 0
\end{pmatrix} P = 2Δ x i ⎝ ⎛ 0 1 0 − 1 − 1 0 1 0 0 − 1 0 1 1 0 − 1 0 ⎠ ⎞ Explanation:
The non-diagonal entries correspond to the finite difference approximation of the derivative.
The factor of i 2 Δ x \frac{i}{2 \Delta x} 2Δ x i
The matrix is anti-Hermitian (i.e., P † = − P P^\dagger = -P P † = − P
This 4 × 4 4 \times 4 4 × 4