Calculating Commutator of Two Operators

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SUMMARY

The discussion focuses on calculating the commutator of two complementary unitary operators, U and V, defined for a system of N eigenstates. The operators are expressed as exponentials of self-adjoint operators A and B, with A and B defined using the eigenstates |n_u⟩ and |m_v⟩. The commutator [A, B] is derived from the expressions for AB and BA, leading to the equation AB - BA = ∑(nm/N²)(1/√N)(e^(i2πnm/N)|n_u⟩⟨m_v| - |m_v⟩⟨n_u|e^(-i2πnm/N)). The discussion highlights the complexity of the calculation and suggests evaluating matrix elements in either basis for further insights.

PREREQUISITES
  • Understanding of quantum mechanics, specifically unitary operators and eigenstates.
  • Familiarity with the concept of commutators in operator algebra.
  • Knowledge of matrix representations of operators and their eigenvalues.
  • Proficiency in mathematical notation and manipulation of complex exponentials.
NEXT STEPS
  • Study the properties of commutators in quantum mechanics to understand their implications.
  • Learn about the spectral theorem and its application to self-adjoint operators.
  • Explore the derivation of matrix elements for operators in different bases.
  • Investigate the role of complementary observables in quantum mechanics.
USEFUL FOR

Quantum mechanics students, physicists working with operator theory, and researchers analyzing unitary transformations in quantum systems will benefit from this discussion.

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Homework Statement


Let [itex]U[/itex] and [itex]V[/itex] be the complementary unitary operators for a system of [itex]N[/itex] eingenstates as discussed in lecture. Recall that they both have eigenvalues [itex]x_n=e^{2\pi in/N}[/itex] where [itex]n[/itex] is an integer satisfying [itex]0\leq n\leq N[/itex]. The operators have forms
[tex] U=\sum_{n}|n_u\rangle\langle n_u |e^{2\pi in/N}\quad\quad V=\sum_{m}|m_v\rangle\langle m_v |e^{2\pi i m/N}[/tex]
These operators can be expressed as exponentials of complementary self-adjoint operators [itex]A[/itex] and [itex]B[/itex]:
[tex] U=e^{i2\pi A}\quad\quad V=e^{i2\pi B}[/tex]
where the operators [itex]A[/itex] and [itex]B[/itex] are
[tex] A=\frac{n}{N}\sum_{n}|n_u\rangle\langle n_u| \quad\quad B=\frac{m}{N}\sum_{m}|m_v\rangle\langle m_v |[/tex]
Calculate the commutator [itex][A,B][/itex].


Homework Equations


[tex] \mathbb{I}=\sum |n\rangle\langle n|[/tex]

for complementary observables
[tex] \frac{1}{\sqrt{N}}e^{i2\pi mn/N}=\sum_{n}\sum_{m}\langle n_u |m_v\rangle[/tex]

The Attempt at a Solution


First I have tried to work out AB and BA separately then combine them. Here is AB
[tex] \begin{align}<br /> AB &= \sum_{n}|n_u\rangle\langle n_u |\frac{n}{N}\sum_{m}|m_v\rangle\langle m_v |\frac{m}{N} \\<br /> &= \sum_{n}\sum_{m}|n_u\rangle\langle n_u|m_v\rangle\langle m_v| \frac{nm}{N^2} \\<br /> &= \sum_{n}\sum_{m}|n_u\rangle \frac{1}{\sqrt{N}}e^{i 2\pi nm/N}\langle m_v |\frac{nm}{N^2}<br /> \end{align}[/tex]
for BA:
[tex] \begin{align}<br /> BA &=\sum_{m}\sum_{n}|m_v\rangle\langle m_v | n_u\rangle\langle n_u |\frac{nm}{N^2} \\<br /> &= \sum_{m}\sum_{n}|m_v\rangle \frac{1}{\sqrt{N}}e^{-i2\pi nm/N}\langle n_u |\frac{nm}{N^2}<br /> \end{align}[/tex]
Then
[tex] AB-BA=\sum_{m}\sum_{n}\frac{nm}{N^2}\frac{1}{\sqrt{N}}\left( e^{i2\pi nm/N}|n_u\rangle\langle m_v |-|m_v\rangle\langle n_u |e^{-i2\pi nm/N}\right)[/tex]
I'm stuck here more or less. I can put either the u basis vectors into the v basis or visa versa, but I don't know if that is right. Where should I go from here?

Thanks,
 
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Hm... well, I haven't done the problem myself so I can't guarantee that this will work, but in general the commutator of operators is itself an operator. So I would suggest calculating the matrix elements of [itex]AB - BA[/itex] in either basis, and see if the result suggests anything.
 

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