Heisenberg Uncertainty Relation

ΔAΔB≥12∣E[[A^,B^]]∣\Delta A\Delta B\geq \frac{1}{2}\left|\mathbb{E}\left[[\hat{A},\hat{B}]\right]\right| ΔA2ΔB2≥∣⟨ψ∣12j[A^,B^]∣ψ⟩∣2=(12jE[[A^,B^]])2\Delta A^2 \Delta B^2 \geq \left|\left\langle \psi \right\rvert\frac{1}{2j}[\hat{A},\hat{B}]\left\lvert \psi \right\rangle\right|^2 = \left(\frac{1}{2j}\mathbb{E}\left[[\hat{A},\hat{B}]\right]\right)^2

Proof Consider any scalar λ∈C\lambda\in \mathbb{C} and any Vector ∣u⟩,∣v⟩∈Cn\left\lvert u \right\rangle,\left\lvert v \right\rangle\in \mathbb{C}^n and we have two states ∣f⟩\left\lvert f \right\rangle and ∣g⟩\left\lvert g \right\rangle.

We subtract it like so ∣f⟩−λ∣g⟩\left\lvert f \right\rangle-\lambda\left\lvert g \right\rangle is also a vector

Note that for QM systems, ⟨u∣u⟩=1\left\langle u|u \right\rangle=1. This is because of Born Rule where the number it represents is the total probability, which needs to sum to 1.

⟨f−λg∣f−λg⟩≥0\left\langle f-\lambda g|f-\lambda g \right\rangle\geq 0 ⇒⟨f∣f⟩−λ⟨f∣g⟩−λ∗⟨g∣f⟩+(∣λ∣)2⟨g∣g⟩≥0\Rightarrow\left\langle f|f \right\rangle-\lambda \left\langle f|g \right\rangle - \lambda^*\left\langle g|f \right\rangle+(|\lambda|)^2\left\langle g|g \right\rangle\geq 0

We want to choose the λ\lambda that minimizes the expression.

λ=⟨g∣f⟩⟨g∣g⟩⇒λ∗=⟨f∣g⟩⟨g∣g⟩,∣λ∣2=⟨f∣g⟩⟨g∣f⟩⟨g∣g⟩2\lambda=\frac{\left\langle g|f \right\rangle}{\left\langle g|g \right\rangle}\quad \Rightarrow\quad \lambda^*=\frac{\left\langle f|g \right\rangle}{\left\langle g|g \right\rangle},\quad |\lambda|^2=\frac{\left\langle f|g \right\rangle\left\langle g|f \right\rangle}{\left\langle g|g \right\rangle^2}

Which makes ∣f⟩−λ∣g⟩\left\lvert f \right\rangle-\lambda\left\lvert g \right\rangle orthogonal to ∣g⟩\left\lvert g \right\rangle.

⇒⟨f∣f⟩−⟨g∣f⟩⟨g∣g⟩⟨f∣g⟩−⟨f∣g⟩⟨g∣g⟩⟨g∣f⟩+⟨f∣g⟩⟨g∣f⟩⟨g∣g⟩2⟨g∣g⟩≥0\Rightarrow\left\langle f|f \right\rangle-\frac{\left\langle g|f \right\rangle}{\left\langle g|g \right\rangle} \left\langle f|g \right\rangle -\frac{\left\langle f|g \right\rangle}{\left\langle g|g \right\rangle}\left\langle g|f \right\rangle +\frac{\left\langle f|g \right\rangle\left\langle g|f \right\rangle}{\left\langle g|g \right\rangle^2}\left\langle g|g \right\rangle\geq 0

Note that ⟨g∣f⟩⟨f∣g⟩=∣⟨f∣g⟩∣2\left\langle g|f \right\rangle\left\langle f|g \right\rangle=|\left\langle f|g \right\rangle|^2

⇒⟨f∣f⟩−∣⟨f∣g⟩∣2⟨g∣g⟩−∣⟨f∣g⟩∣2⟨g∣g⟩+∣⟨f∣g⟩∣2⟨g∣g⟩≥0\Rightarrow\left\langle f|f \right\rangle-\frac{|\left\langle f|g \right\rangle|^2}{\left\langle g|g \right\rangle}-\frac{|\left\langle f|g \right\rangle|^2}{\left\langle g|g \right\rangle}+\frac{|\left\langle f|g \right\rangle|^2}{\left\langle g|g \right\rangle}\geq 0 ⇒⟨f∣f⟩−∣⟨f∣g⟩∣2⟨g∣g⟩≥0\Rightarrow\left\langle f|f \right\rangle-\frac{|\left\langle f|g \right\rangle|^2}{\left\langle g|g \right\rangle}\geq 0

Assuming ⟨g∣g⟩≥0\left\langle g|g \right\rangle\geq 0

⇒⟨f∣f⟩⟨g∣g⟩≥∣⟨f∣g⟩∣2\Rightarrow\left\langle f|f \right\rangle\left\langle g|g \right\rangle\geq |\left\langle f|g \right\rangle|^2

This is Cauchy-Schwarz squared

Consider the Observable that we defined above. This can be written as

∣f⟩=(A^−E[A]I)∣ψ⟩∣g⟩=(B^−E[B]I)∣ψ⟩\left\lvert f \right\rangle=(\hat{A}-\mathbb{E}[A]I)\left\lvert \psi \right\rangle\quad \left\lvert g \right\rangle=(\hat{B}-\mathbb{E}[B]I)\left\lvert \psi \right\rangle

Note that ΔA\Delta A is the Spread of A^\hat{A}.

⟨f∣f⟩=⟨ψ∣(A^−E[A]I)∣ψ⟩\left\langle f|f \right\rangle=\left\langle \psi \right\rvert(\hat{A}-\mathbb{E}[A]I)\left\lvert \psi \right\rangle =⟨ψ∣(A^2−2E[A]A^+E[A]2I)∣ψ⟩= \left\langle \psi \right\rvert\left(\hat{A}^2-2\mathbb{E}[A]\hat{A}+\mathbb{E}[A]^2I\right)\left\lvert \psi \right\rangle =⟨ψ∣A^2∣ψ⟩−2E[A]⟨ψ∣A^∣ψ⟩+E[A]2⟨ψ∣I∣ψ⟩= \left\langle \psi \right\rvert\hat{A}^2\left\lvert \psi \right\rangle-2\mathbb{E}[A]\left\langle \psi \right\rvert\hat{A}\left\lvert \psi \right\rangle+\mathbb{E}[A]^2\left\langle \psi \right\rvert I\left\lvert \psi \right\rangle =E[A^2]−2E[A]2+E[A]2=\mathbb{E}[\hat{A}^2]-2\mathbb{E}[A]^2+\mathbb{E}[A]^2 =E[A^2]−E[A]2=\mathbb{E}[\hat{A}^2]-\mathbb{E}[A]^2 =ΔA2=\Delta A^2

By symmetry

⟨g∣g⟩=ΔB2\left\langle g|g \right\rangle=\Delta B^2

Plug into our equation above

ΔA2ΔB2≥∣⟨f∣g⟩∣2\Delta A^2\Delta B^2\geq |\left\langle f|g \right\rangle|^2 ⇒ΔA2ΔB2≥(Im⟨f∣g⟩)2\Rightarrow\Delta A^2\Delta B^2\geq (Im\left\langle f|g \right\rangle)^2 ⇒ΔA2ΔB2≥(Im(⟨ψ∣(A^−E[A]I)(B^−E[B]I)∣ψ⟩))2\Rightarrow\Delta A^2\Delta B^2\geq (Im(\left\langle \psi \right\rvert(\hat{A}-\mathbb{E}[A]I)(\hat{B}-\mathbb{E}[B]I)\left\lvert \psi \right\rangle))^2 ⇒ΔA2ΔB2≥(Im(E[AB]−2E[A]E[B]−+E[A]E[B]))2\Rightarrow\Delta A^2\Delta B^2\geq (Im(\mathbb{E}[AB]-2\mathbb{E}[A]\mathbb{E}[B]-+ \mathbb{E}[A]\mathbb{E}[B]))^2 ⇒ΔA2ΔB2≥(Im(E[AB]−E[A]E[B]))2\Rightarrow\Delta A^2\Delta B^2\geq (Im(\mathbb{E}[AB]-\mathbb{E}[A]\mathbb{E}[B]))^2

Note that Imz=12i(z−z∗)Im z=\frac{1}{2i}(z-z^*)

⇒ΔA2ΔB2≥(12i(E[AB]−E[BA]))2\Rightarrow\Delta A^2\Delta B^2\geq \left(\frac{1}{2i}(\mathbb{E}[AB]-\mathbb{E}[BA])\right)^2 ⇒ΔA2ΔB2≥(⟨ψ∣12i[A^,B^]∣ψ⟩)2\Rightarrow\Delta A^2\Delta B^2\geq \left(\left\langle \psi \right\rvert\frac{1}{2i}[\hat{A},\hat{B}]\left\lvert \psi \right\rangle\right)^2