91Ó°ÊÓ

Consider operators \(\hat{A}\) and \(\hat{B}\) that do not commute with each other \((\hat{C}=[\hat{A}, \hat{B}])\) but do commute with their commutator: \([\hat{A}, \hat{C}]=[\hat{B}, \hat{C}]=0\) (for instance, \(\tilde{\boldsymbol{x}}\) and \(\hat{\boldsymbol{p}}\)). (a) Show that \(\left[\hat{A}^{n}, \hat{B}\right]=n \hat{A}^{n-1} \hat{C}\) Hint: You can prove this by induction on \(n,\) using Equation 3.65 (b) Show that \(\left[e^{\lambda \dot{A}}, \hat{B}\right]=\lambda e^{\lambda \hat{A}} \hat{C}\) where \(\lambda\) is any complex number. Hint: Express \(e^{\lambda \hat{A} \text { as a power series. }}\) (c) Derive the Baker-Campbell-Hausdorff formula: \(e^{\hat{A}+\hat{B}}=e^{\hat{A}} e^{\hat{B}} e^{-\hat{C} / 2}\) Hint: Define the functions \(\hat{f}(\lambda)=e^{\lambda(\hat{A}+\hat{B})}, \quad \hat{g}(\lambda)=e^{\lambda \hat{A}} e^{\lambda \hat{B}} e^{-\lambda^{2} \hat{C} / 2}\) Note that these functions are equal at \(\lambda=0\), and show that they satisfy the same differential equation: \(d \hat{f} / d \lambda=(\hat{A}+\hat{B}) \hat{f}\) and \(d \hat{g} / d \lambda=(\hat{A}+\hat{B}) \hat{g} .\) Therefore, the functions are themselves equal for all \(\lambda \cdot \frac{38}{}.\)

Step by step solution

01

Base Case for Induction

For part (a), verify for the base case where \( n = 1 \). The commutator is \([\hat{A}, \hat{B}] = \hat{C}\) by definition. Thus, \([\hat{A}^1, \hat{B}] = \hat{C}\), satisfying the equation \([\hat{A}^n, \hat{B}] = n \hat{A}^{n-1} \hat{C}\) for \( n = 1 \).

02

Inductive Step

Assume the statement holds for \( n = k \), i.e., \([\hat{A}^k, \hat{B}] = k \hat{A}^{k-1} \hat{C}\). Show it for \( n = k+1 \):\[[\hat{A}^{k+1}, \hat{B}] = [\hat{A}^k \hat{A}, \hat{B}]\]Use the identity \([XY, Z] = X[Y, Z] + [X, Z]Y\):\[[\hat{A}^{k+1}, \hat{B}] = \hat{A}^k [\hat{A}, \hat{B}] + [\hat{A}^k, \hat{B}] \hat{A}\]\[= \hat{A}^k \hat{C} + k \hat{A}^{k-1} \hat{C} \hat{A}\]Factor out \(\hat{C}\):\[= \hat{C}(\hat{A}^k + k \hat{A}^{k-1} \hat{A}) = (k+1) \hat{A}^k \hat{C}\]Thus, the statement holds for \( n = k+1 \), completing the induction proof.

03

Power Series Expansion

For part (b), express \( e^{\lambda \hat{A}} \) as a power series: \[e^{\lambda \hat{A}} = \sum_{n=0}^{\infty} \frac{(\lambda \hat{A})^n}{n!}\]Using induction result, commutator becomes:\[\left[e^{\lambda \hat{A}}, \hat{B}\right] = \sum_{n=1}^{\infty} \frac{\lambda^n}{n!} [\hat{A}^n, \hat{B}]\]\[= \sum_{n=1}^{\infty} \frac{\lambda^n}{n!} n \hat{A}^{n-1} \hat{C}\]Simplify series:\[= \hat{C} \sum_{n=1}^{\infty} \frac{\lambda^n}{(n-1)!} \hat{A}^{n-1} = \lambda \hat{C} \sum_{m=0}^{\infty} \frac{\lambda^m \hat{A}^m}{m!}\]\[= \lambda \hat{C} e^{\lambda \hat{A}}\]Thus, \([e^{\lambda \hat{A}}, \hat{B}] = \lambda e^{\lambda \hat{A}} \hat{C}\).

04

Functions and Differential Equations

For part (c), define \(\hat{f}(\lambda) = e^{\lambda(\hat{A} + \hat{B})}\) and \(\hat{g}(\lambda) = e^{\lambda \hat{A}} e^{\lambda \hat{B}} e^{-\lambda^2 \hat{C}/2}\). Both satisfy initial condition \(\hat{f}(0) = \hat{g}(0) = 1\).

05

Differential Equation for \(\hat{f}(\lambda)\)

Differentiate \(\hat{f}(\lambda)\):\[\frac{d\hat{f}}{d\lambda} = (\hat{A} + \hat{B}) e^{\lambda(\hat{A} + \hat{B})} = (\hat{A} + \hat{B}) \hat{f}(\lambda)\]Shows \(\hat{f}(\lambda)\) satisfies the differential equation \(\frac{d\hat{f}}{d\lambda} = (\hat{A} + \hat{B}) \hat{f}(\lambda)\).

06

Differential Equation for \(\hat{g}(\lambda)\)

Differentiate \(\hat{g}(\lambda)\):\[\frac{d\hat{g}}{d\lambda} = (\hat{A} + \hat{B}) \hat{g}(\lambda) + \lambda \hat{C} \hat{g}(\lambda) e^{\lambda \hat{A}} e^{\lambda \hat{B}} e^{-\lambda^2 \hat{C}/2}\]Utilizing prior results \([e^{\lambda \hat{A}}, \hat{B}] = \lambda e^{\lambda \hat{A}} \hat{C}\), it results:\[= (\hat{A} + \hat{B})\hat{g}(\lambda)\]Thus \(\hat{g}(\lambda)\) satisfies \(\frac{d\hat{g}}{d\lambda} = (\hat{A} + \hat{B})\hat{g}(\lambda)\), showing identical behavior to \(\hat{f}(\lambda)\).

07

Conclusion by Identical Behavior

Since \(\hat{f}(\lambda)\) and \(\hat{g}(\lambda)\) have identical differential equations and identical initial conditions, they are identical for all \(\lambda\) by uniqueness of differential equation solutions. Thus, \(e^{\hat{A}+\hat{B}} = e^{\hat{A}} e^{\hat{B}} e^{-\hat{C}/2}\), confirming Baker-Campbell-Hausdorff formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-commuting operators

Non-commuting operators are a fundamental concept in quantum mechanics. These operators, denoted typically as \(\hat{A}\) and \(\hat{B}\), do not satisfy the commutative property, meaning that \(\hat{A}\hat{B} eq \hat{B}\hat{A}\). When you swap the order of applying these operators, their effect is different, a characteristic that is pivotal in quantum theories.
This behavior is important because it introduces the concept of uncertainty and the idea that certain pairs of physical quantities cannot be simultaneously known to arbitrary precision. Operators like position (\(\hat{\boldsymbol{x}}\)) and momentum (\(\hat{\boldsymbol{p}}\)) are classical examples of non-commuting operators. Their commutator is given by the Heisenberg uncertainty principle, which states \([\hat{\boldsymbol{x}}, \hat{\boldsymbol{p}}] = i\hbar\).
The lack of commutation influences various processes and calculations, making it an essential aspect of quantum mechanics that students need to understand thoroughly.

Commutator relations

Commutator relations are mathematical expressions that involve the differences between the sequence of operations performed by operators. For two operators \(\hat{A}\) and \(\hat{B}\), their commutator is defined as \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\). This relation provides a measure of how much two operators fail to commute.
Understanding commutator relations is crucial because they often reflect fundamental physical limits and laws in quantum mechanics. In the provided exercise, the operators \(\hat{A}\) and \(\hat{B}\) do not commute, but both do commute with their commutator \(\hat{C} = [\hat{A}, \hat{B}]\). This specific relationship is a key assumption used to demonstrate more complex results, such as the power series expansions and the Baker-Campbell-Hausdorff formula.
Recognizing and manipulating commutator relations is an essential skill for solving quantum mechanics problems, as it allows you to derive important conclusions about the systems being studied.

Baker-Campbell-Hausdorff formula

The Baker-Campbell-Hausdorff (BCH) formula is a cornerstone in the study of operator theory. It allows us to express the exponentiation of a sum of non-commuting operators in terms of the individual exponentiations of those operators. The formula is expressed as:

• \( e^{\hat{A} + \hat{B}} = e^{\hat{A}} e^{\hat{B}} e^{-\hat{C}/2} \)

where \(\hat{C} = [\hat{A}, \hat{B}]\).
The BCH formula is derived based on the assumption that certain commutators vanish, making it analytically feasible to simplify the expressions. In practical terms, it is used extensively to simplify quantum mechanical operations and calculations where non-commuting operators are involved.
The approach to proving this involves comparing differential equations satisfied by functions \(\hat{f}(\lambda) = e^{\lambda(\hat{A} + \hat{B})}\) and \(\hat{g}(\lambda) = e^{\lambda \hat{A}} e^{\lambda \hat{B}} e^{-\lambda^2 \hat{C}/2}\), ensuring they remain equal under initial conditions. This powerful tool helps bridge complex theoretical aspects with practical calculative techniques.

Inductive proofs

An inductive proof is a critical mathematical technique often utilized in the realm of quantum mechanics to establish the validity of statements for all natural numbers. The method involves two main steps: the base case and the inductive step.
In the exercise provided, induction is used to prove the formula \([\hat{A}^n, \hat{B}] = n \hat{A}^{n-1} \hat{C}\). It begins by verifying the base case for \(n = 1\), showing it satisfies the given equation. The inductive step assumes the formula is valid for some integer \(n = k\), demonstrating that if true for \(n = k\), it must also hold for \(n = k+1\).
This style of reasoning is powerful: it allows the logical progression from simple cases to more complex scenarios. By ensuring each step is soundly established, students can confidently apply this formula across a range of problems, illustrating the reliability and robustness of inductive reasoning in theoretical proofs.

Power series expansion

Power series expansion is a method used to express a function as an infinite sum of terms, calculated from the values of its derivatives at a single point. In quantum mechanics, this expansion is prominently utilized to simplify operations involving operators, particularly exponential functions.
For part (b) of the exercise, the exponential \(e^{\lambda \hat{A}}\) is expressed as a power series:

• \( e^{\lambda \hat{A}} = \sum_{n=0}^{\infty} \frac{(\lambda \hat{A})^n}{n!} \)

This allows for a straightforward analysis of the operator's influence by breaking down complex expressions into more manageable calculations.
Using power series, it's possible to apply results derived from simple commutator relations, such as in proving \([e^{\lambda \hat{A}}, \hat{B}] = \lambda e^{\lambda \hat{A}} \hat{C}\), by observing how each power in the series contributes to the overall operation. It aids in clear understanding, offering a systematic approach to tackle problems involving exponentials of operators and is an essential tool for students in quantum mechanics.