Generalised uncertainty principle

The Generalized Uncertainty Principle is a cornerstone of quantum mechanics that extends beyond just position and momentum. It provides a mathematical limit on the precision with which any pair of incompatible observables can be known simultaneously. This post details the rigorous derivation of this principle using the geometric properties of Hilbert space, specifically the Cauchy-Schwarz inequality.

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Derivation of Generalized Uncertainty Principle via Cauchy-Schwarz Inequality

1. Mathematical Preliminaries

Before deriving the inequality, we must define the statistical tools used in quantum mechanics. Let \( \hat{A} \) and \( \hat{B} \) be two Hermitian operators corresponding to physical observables.

Operators and Expectation Values

The expectation value of an operator \( \hat{A} \) in the state \( | \psi \rangle \) is defined as:

\[ \langle \hat{A} \rangle = \langle \psi | \hat{A} | \psi \rangle \]

We define the deviation operator \( \Delta\hat{A} \), which measures the difference between the operator and its mean value:

\[ \Delta\hat{A} = \hat{A} - \langle \hat{A} \rangle \]

The variance (uncertainty squared) is the expectation value of the square of this deviation:

\[ \sigma_A^2 = \langle (\Delta\hat{A})^2 \rangle = \langle \psi | (\Delta\hat{A})^2 | \psi \rangle \]

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2. The Step-by-Step Derivation

The derivation relies on defining two auxiliary state vectors and applying the Cauchy-Schwarz inequality to them.

Step 1: Define Auxiliary Vectors

Let us define two vectors (kets) in the Hilbert space based on the deviation operators applied to the state \( | \psi \rangle \):

\[ | f \rangle = \Delta\hat{A} | \psi \rangle \] \[ | g \rangle = \Delta\hat{B} | \psi \rangle \]

Step 2: Apply the Cauchy-Schwarz Inequality

The inequality states that the square of the inner product of two vectors is less than or equal to the product of their squared norms:

\[ | \langle f | g \rangle |^2 \le \langle f | f \rangle \langle g | g \rangle \]

Step 3: Analyze the Norms (Right-Hand Side)

Since \( \hat{A} \) and \( \hat{B} \) are Hermitian observables, their deviation operators \( \Delta\hat{A} \) and \( \Delta\hat{B} \) are also Hermitian (\( (\Delta\hat{A})^\dagger = \Delta\hat{A} \)).

Therefore, the inner product \( \langle f | f \rangle \) becomes:

\[ \langle f | f \rangle = \langle \psi | (\Delta\hat{A})^\dagger (\Delta\hat{A}) | \psi \rangle = \langle \psi | (\Delta\hat{A})^2 | \psi \rangle = \sigma_A^2 \]

Similarly for \( | g \rangle \):

\[ \langle g | g \rangle = \sigma_B^2 \]

Thus, the RHS of our inequality is \( \sigma_A^2 \sigma_B^2 \).

Step 4: Analyze the Inner Product (Left-Hand Side)

Now we evaluate \( \langle f | g \rangle \):

\[ \langle f | g \rangle = \langle \psi | (\Delta\hat{A})^\dagger \Delta\hat{B} | \psi \rangle = \langle \psi | \Delta\hat{A} \Delta\hat{B} | \psi \rangle = \langle \Delta\hat{A} \Delta\hat{B} \rangle \]

The product of two operators \( \Delta\hat{A} \Delta\hat{B} \) can be decomposed into a symmetric part (anti-commutator) and an antisymmetric part (commutator):

\[ \Delta\hat{A} \Delta\hat{B} = \frac{1}{2} \{ \Delta\hat{A}, \Delta\hat{B} \} + \frac{1}{2} [ \Delta\hat{A}, \Delta\hat{B} ] \]

Note that the commutator of deviations is the same as the commutator of the operators themselves: \( [ \Delta\hat{A}, \Delta\hat{B} ] = [ \hat{A}, \hat{B} ] \). Thus:

\[ \langle f | g \rangle = \frac{1}{2} \langle \{ \Delta\hat{A}, \Delta\hat{B} \} \rangle + \frac{1}{2} \langle [ \hat{A}, \hat{B} ] \rangle \]

Step 5: Real and Imaginary Components

This is a crucial step. Because \( \hat{A} \) and \( \hat{B} \) are Hermitian:

• The expectation value of the anti-commutator \( \{ \Delta\hat{A}, \Delta\hat{B} \} \) is purely Real.
• The expectation value of the commutator \( [ \hat{A}, \hat{B} ] \) is purely Imaginary.

The squared modulus \( |z|^2 \) of a complex number \( z = x + iy \) is \( x^2 + y^2 \). Therefore:

\[ | \langle f | g \rangle |^2 = \left| \frac{1}{2} \langle \{ \Delta\hat{A}, \Delta\hat{B} \} \rangle \right|^2 + \left| \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle \right|^2 \]

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3. The Robertson-Schrödinger Relation

Substituting the results from Steps 3 and 5 back into the original Cauchy-Schwarz inequality (\( \sigma_A^2 \sigma_B^2 \ge | \langle f | g \rangle |^2 \)):

\[ \sigma_A^2 \sigma_B^2 \ge \left( \frac{1}{2} \langle \{ \Delta\hat{A}, \Delta\hat{B} \} \rangle \right)^2 + \left( \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle \right)^2 \]

This is the Robertson-Schrödinger Uncertainty Relation. It is stronger than the standard Heisenberg form because it includes the covariance term (the anti-commutator part).

Reduction to Heisenberg Principle

If we ignore the correlation term (which is non-negative), we get the weaker, more familiar form:

\[ \sigma_A^2 \sigma_B^2 \ge \left( \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle \right)^2 \]

Taking the square root:

\[ \sigma_A \sigma_B \ge \frac{1}{2} | \langle [ \hat{A}, \hat{B} ] \rangle | \]

For position (\( \hat{x} \)) and momentum (\( \hat{p} \)), the commutator is \( [ \hat{x}, \hat{p} ] = i\hbar \). Plugging this in yields the famous result:

\[ \sigma_x \sigma_p \ge \frac{\hbar}{2} \]