Operator algebra: Hermicity and Eigenstates

[physicsxanime]

Homework Statement

See screenshot next

Relevant Equations

hermitian operator $A^\dagger = A$
antihermitian operator $A^\dagger = -A$

A. I can show that A is either hermitian or antihermitian by
$$(B^\dagger B=1-A^2)^\dagger$$
$$B^\dagger B=1-A^\dagger A^\dagger$$
comparing, we know that
$$A^\dagger = \pm A$$
I don't know how I can make use of the communtation relation to get hermiticity of B. But I know that A and B must have same hermiticity because hermitian * antihermitian = 0 and the communtation relation will not make sense.

B. The fact that the question hints on only the other eigenstate is weird, does the above imply A and B are 2x2?
Anyway, consider the state $B|a=0>$
$$AB|a=0> = BA|a=0> + B|a=0> = 0 + B|a=0>$$
Therefore, $B|a=0>$ is a eigenstate of eigenvalue 1.

C. This need result in A.
note that we can work in the basis of A
$|a=0> = \begin{pmatrix} 1\\ 0 \end{pmatrix}$

$B|a=0> = \begin{pmatrix} 0\\ 1 \end{pmatrix}$
$$

We see $B \begin{pmatrix} 0\\ 1 \end{pmatrix} = B^2|a=0> = \pm B^\dagger B|a=0> = \pm(1-A^2)|a=0> = \pm |a=0>$
Very simple matrix to diagonalize

 

[Hill]

Homework Statement: See screenshot next
Relevant Equations: hermitian operator $A^\dagger = A$
antihermitian operator $A^\dagger = -A$

A. I can show that A is either hermitian or antihermitian by
$(B^†B=1−A^2)^†$
$B^†B=1−A^†A^†$
comparing, we know that
$A^†=±A$

I don't see how from $$B^\dagger B=1-A^2$$ you get $$A^\dagger = \pm A$$
Consider the following counterexample:

$A =\begin{pmatrix} 0 &1\\ 0 &0\end{pmatrix}$

$B =\begin{pmatrix} 1 &0\\ 0 &1\end{pmatrix}$

 

[renormalize]

Consider the following counterexample:

$A =\begin{pmatrix} 0 &1\\ 0 &0\end{pmatrix}$

$B =\begin{pmatrix} 1 &0\\ 0 &1\end{pmatrix}$

How is that a counterexample? Those two matrices don't even satisfy the first condition of the problem listed under $2.$: $[A,B]=AB-BA=0\neq B$.

 

[Hill]

How is that a counterexample? Those two matrices don't even satisfy the first condition of the problem listed under $2.$: $[A,B]=AB-BA=0\neq B$.

It is a counterexample to the OP derivation in part A, where the OP does not use the condition $[A,B]=B$.

 

[physicsxanime]

It is a counterexample to the OP derivation in part A, where the OP does not use the condition $[A,B]=B$.

Yes, I should be more careful. The proof is comparing second condition from question with the hermitian conjugate of it, we have:
$$AA = A^\dagger A^\dagger$$
This can be satisfied only with either $AA=A^\dagger A^\dagger=0$ or $A^\dagger = \pm A$
Then I compare and by imposing imposing $A^2 = 0$
$$[A,[A,B]] = B$$
$$-2ABA = B$$
with
$$A[A,B] = AB$$
$$-ABA = AB$$
Since $B^\dagger B = 1$ I can show a contradiction from these.

Therefore, $A^\dagger = \pm A$ is established.


Also, I am thinking I can prove $A$ is symmetric.
Consider
$$[A,B] = B$$
$$AB-BA = B$$
$$AB= B(1+A)$$
$$B^\dagger AB= B^\dagger B(1+A)$$
$$B^\dagger AB= (1+A^2)(1+A)$$
$$(B^\dagger AB= (1+A^2)(1+A))^\dagger$$
$$\pm B^\dagger AB= (1+A^2)(1\pm A)$$

Comparing 5th and last line, $A$ has to be symmetric.

 

[strangerep]

[...] comparing second condition from question with the hermitian conjugate of it, we have:
$$AA = A^\dagger A^\dagger$$
This can be satisfied only with either $AA=A^\dagger A^\dagger=0$ or $A^\dagger = \pm A$

These are not the only possibilities. To see why, decompose A into hermitian and skew-hermitian parts, i.e.,
$$\mbox{Let}~~ A ~=~ X + Y ~,~~~ \mbox{where}~~ X^\dagger = X ~,~~~ \mbox{and}~~ Y^\dagger = -Y ~.$$(Do you know how to prove that A can always be decomposed like this?)

With this decomposition, the equation $\,AA = A^\dagger A^\dagger\,$ gives
$$(X + Y)(X + Y) ~=~(X + Y)^\dagger (X + Y)^\dagger $$$$X^2 + XY + YX + Y^2 ~=~ (X^\dagger)^2 + X^\dagger Y^\dagger + Y^\dagger X^\dagger + (Y^\dagger)^2 ~=~ X^2 - XY - YX + Y^2$$$$\Rightarrow~~ (XY + YX) ~=~ - (XY + YX) ~,~~~ \Rightarrow~ XY + YX = 0 ~.$$I.e., X and Y anticommute.

Of course, $X=0$ or $Y=0$ are possibilities, but not the most general ones.

 

[DrClaude]

B. The fact that the question hints on only the other eigenstate is weird, does the above imply A and B are 2x2?

The question mentions the other eigenstates of A, so no reason to assume only 2.

 

[strangerep]

@physicsxanime : where did this exercise come from? Please post a link or reference.

 

[physicsxanime]

@physicsxanime : where did this exercise come from? Please post a link or reference.

It is from entrance exam at UBC.
https://phas.ubc.ca/graduate-program-comprehensive-exam-guidelines-phd-students
In particular it is from 2010.
https://phas.ubc.ca/sites/default/files/shared/comp2010.pdf

 

[renormalize]

It turns out that the commutation relation stated above in 2 is that of the generators aff(1) of the 2-parameter, non-abelian 1D affine group Aff(1) (see, e.g., https://math.stackexchange.com/questions/223496/example-of-two-dimensional-non-abelian-lie-algebra). In a 2x2 matrix realization of aff(1), a general element can be written as: $$
\left(\begin{array}{cc}
a & b\\
0 & 0
\end{array}\right)
$$In particular, defining:$$
A\equiv\left(\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right),\:B\equiv\left(\begin{array}{cc}
0 & e^{i\phi}\\
0 & 0
\end{array}\right)
$$it's easy to verify that $\left[A,B\right]=B$ and:$$
B^{\dagger}B=1-A^{2}=\left(\begin{array}{cc}
0 & 0\\
0 & 1
\end{array}\right)
$$as required. So $A$ is hermitian and symmetric, and $B$ is neither hermitian nor antihermitian (as anticipated by @strangerep in a private conversation).

 

[physicsxanime]

It turns out that the commutation relation stated above in 2 is that of the generators aff(1) of the 2-parameter, non-abelian 1D affine group Aff(1) (see, e.g., https://math.stackexchange.com/questions/223496/example-of-two-dimensional-non-abelian-lie-algebra). In a 2x2 matrix realization of aff(1), a general element can be written as: $$
\left(\begin{array}{cc}
a & b\\
0 & 0
\end{array}\right)
$$In particular, defining:$$
A\equiv\left(\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right),\:B\equiv\left(\begin{array}{cc}
0 & e^{i\phi}\\
0 & 0
\end{array}\right)
$$it's easy to verify that $\left[A,B\right]=B$ and:$$
B^{\dagger}B=1-A^{2}=\left(\begin{array}{cc}
0 & 0\\
0 & 1
\end{array}\right)
$$as required. So $A$ is hermitian and symmetric, and $B$ is neither hermitian nor antihermitian (as anticipated by @strangerep in a private conversation).

Thanks so much! If you have spare time, I would really benefit from knowing your motivation on how you come up with a solution and that how one can anticipate the result before the calculation.
For now, it seemed to me that you either know the existence of such group or you don't. At the end of the day what are some ways I can use to figure out the anwser?

 

[strangerep]

Here is my "proof" (from the private conversation mentioned by @renormalize), that if $A$ is hermitian, then $B$ cannot have definite hermiticity...

Suppose $A^\dagger = A$. Then, $$B ~=~ [A,B]~, ~~~\Rightarrow~~ B^\dagger ~=~
[B^\dagger, A^\dagger] ~=~ [B^\dagger, A] ~=~ -[A,B^\dagger] ~.$$So IF B has definite hermiticity, i.e., if $B = \pm B^\dagger$, then we have $\pm B = \mp [A,B]$, hence $B = -[A,B]$, and so $B=0$ (comparing with the original commutation rule).

Instead, if $A^\dagger = -A$, then, $$B ~=~ [A,B]~, ~~~\Rightarrow~~ B^\dagger
~=~ [B^\dagger, A^\dagger] ~=~ [B^\dagger, -A] ~=~ [A,B^\dagger] ~,$$which seems to have no useful implications. I still don't see how one can prove that A must be hermitian.

Separately,... about "you either know the existence of such group or you don't"... yes,... theoretical physicists are supposed to be very familiar with lots of group theory and representations.

 

[renormalize]

For now, it seemed to me that you either know the existence of such group or you don't. At the end of the day what are some ways I can use to figure out the anwser?

Actually, for this problem you don't need any Lie-algebra knowledge once you make the "mental leap" to representing the operators $A,B$ as finite-dimensional square matrices. It's then natural to start playing with the smallest such matrices, which are 2x2. And it doesn't take much experimenting before you realize that a pair of two 2x2 matrices, containing only a single non-zero entry in each, can be chosen to satisfy both the commutation relation $[A,B]=B$ and the constraint $B^{\dagger}B=1-A^{2}$.