Calculating Commutator of Two Operators

It would probably be helpful to write out explicitly the matrix elements of A and B as well.In summary, the complementary unitary operators U and V for a system of N eigenstates can be expressed as exponentials of complementary self-adjoint operators A and B. The commutator [A,B] can be calculated by subtracting the matrix elements of AB from BA. Further calculations may be necessary to fully solve the problem.
  • #1
jfy4
649
3

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}
AB &= \sum_{n}|n_u\rangle\langle n_u |\frac{n}{N}\sum_{m}|m_v\rangle\langle m_v |\frac{m}{N} \\
&= \sum_{n}\sum_{m}|n_u\rangle\langle n_u|m_v\rangle\langle m_v| \frac{nm}{N^2} \\
&= \sum_{n}\sum_{m}|n_u\rangle \frac{1}{\sqrt{N}}e^{i 2\pi nm/N}\langle m_v |\frac{nm}{N^2}
\end{align}
[/tex]
for BA:
[tex]
\begin{align}
BA &=\sum_{m}\sum_{n}|m_v\rangle\langle m_v | n_u\rangle\langle n_u |\frac{nm}{N^2} \\
&= \sum_{m}\sum_{n}|m_v\rangle \frac{1}{\sqrt{N}}e^{-i2\pi nm/N}\langle n_u |\frac{nm}{N^2}
\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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  • #2
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.
 

1. What is the definition of a commutator of two operators?

The commutator of two operators is given by the expression [A,B] = AB - BA, where A and B are operators in a mathematical space. This operation measures the extent to which the two operators do not commute, or in other words, the extent to which their order matters.

2. What is the significance of the commutator in quantum mechanics?

In quantum mechanics, the commutator plays a crucial role in determining the uncertainty principle. The uncertainty principle states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The commutator of position and momentum operators gives rise to this fundamental principle.

3. How is the commutator related to the eigenvalues of operators?

The commutator of two operators is related to the eigenvalues of those operators. If the commutator of two operators is zero, then they share a set of eigenvalues. If the commutator is non-zero, then the operators do not share a complete set of eigenvalues. This relationship is important in understanding the properties of quantum systems.

4. Can the commutator of two operators have a physical interpretation?

Yes, the commutator of two operators can have a physical interpretation. In quantum mechanics, the commutator of position and momentum operators has a physical interpretation as the uncertainty in position multiplied by the uncertainty in momentum. This interpretation is derived from the uncertainty principle mentioned earlier.

5. How is the commutator used in solving quantum mechanical problems?

The commutator is used in solving quantum mechanical problems by allowing us to determine which operators are compatible with each other. If two operators commute, then they can be measured simultaneously and share a complete set of eigenvalues. This information can be used to simplify and solve complex quantum mechanical problems.

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