## Homework Statement

Let U and A be two linear maps related by U=e^iA. Show that U is unitary if A is self-adjoint. Give a counterexample to show that U can be unitary if A is not self-adjoint.

## The Attempt at a Solution

OK, so I had no problem with the first part. It's easy to show that U dagger=U inverse. However, I'm having some trouble coming up with a counter example for the second part. I tried something anti-Hermitian, but that just shows that U is self-adjoint, not that it is unitary.

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Write out what it means that $UU^*=I=U^*U$. What could go wrong?

Wait. What do you mean "what could go wrong"? I need to show that U can be unitary even if A isn't self-adjoint by way of an example.

Wait. What do you mean "what could go wrong"? I need to show that U can be unitary even if A isn't self-adjoint by way of an example.
I didn't phrase that the best way as I could.

So, indeed, you need to find a non-self-adjoint matrix A such that $U=e^{iA}$ is unitary. So, assume that U is unitary, then $UU^*=I=U^*U$. What does that mean for A??

I didn't phrase that the best way as I could.

So, indeed, you need to find a non-self-adjoint matrix A such that $U=e^{iA}$ is unitary. So, assume that U is unitary, then $UU^*=I=U^*U$. What does that mean for A??
$UU^*=(e^{iA})(e^{iA})^+=(e^{iA})^+(e^{iA})=U^*U$

I feel like this is really simple; I wonder why I'm not seeing it.

$UU^*=(e^{iA})(e^{iA})^+=(e^{iA})^+(e^{iA})=U^*U$

I feel like this is really simple; I wonder why I'm not seeing it.
OK, so you want $UU^*=I=UU^*$. You indeed say that this is the same as

$$(e^{iA})(e^{iA})^+=I=(e^{iA})^+(e^{iA})$$

Now, do you know what $(e^{iA})^+$ is?? Can you write it in another way??

OK, so you want $UU^*=I=UU^*$. You indeed say that this is the same as

$$(e^{iA})(e^{iA})^+=I=(e^{iA})^+(e^{iA})$$

Now, do you know what $(e^{iA})^+$ is?? Can you write it in another way??
$(e^{iA})^+=e^{-iA^+}$

OK, so now you have

$$e^{iA}e^{-iA^+}=I=e^{-iA^+}e^{iA}$$

Now, what happens if you use $e^{A+B}=e^A e^B$?? (I know the rule isn't always valid for matrices, but let's assume $AA^+=A^+A$, in that case the rule is valid).

OK, so now you have

$$e^{iA}e^{-iA^+}=I=e^{-iA^+}e^{iA}$$

Now, what happens if you use $e^{A+B}=e^A e^B$?? (I know the rule isn't always valid for matrices, but let's assume $AA^+=A^+A$, in that case the rule is valid).
You get $$e^{i(-A^++A)}=I=e^{i(A-A^+)}$$.

OK, so what did we find now? We found out that if A isn't self-adjoint, then there are two ways that $e^{iA}$ could still be unitary.

Either, $B=A+A^+$ is a matrix such that $e^{iB}=I$.
Or $AA^+\neq A^+A$.

One of those two things must hold for A.

Let's focus on the second one, can you find a matrix such that $AA^+\neq A^+A$.

OK, so what did we find now? We found out that if A isn't self-adjoint, then there are two ways that $e^{iA}$ could still be unitary.

Either, $B=A+A^+$ is a matrix such that $e^{iB}=I$.
Or $AA^+\neq A^+A$.

One of those two things must hold for A.

Let's focus on the second one, can you find a matrix such that $AA^+\neq A^+A$.
I think I can come up with a matrix, but can't we just treat these as operators, too? How about the ladder operator in QM. It is non-Hermitian; i.e. $AA^+\neq A^+A$.

I think I can come up with a matrix, but can't we just treat these as operators, too? How about the ladder operator in QM. It is non-Hermitian; i.e. $AA^+\neq A^+A$.
The condition $AA^+=A^+A$ is known as "normal".
But anyway, what do you mean with the ladder operator??
Matrices are fine examples of operators. Matrices are exactly the operators on a finite dimensional space. So any counterexamples you find as matrices will be a counterexample of operators.

The condition $AA^+=A^+A$ is known as "normal".
But anyway, what do you mean with the ladder operator??
Matrices are fine examples of operators. Matrices are exactly the operators on a finite dimensional space. So any counterexamples you find as matrices will be a counterexample of operators.

I am terrible at Latex, so I'm just linking to Wikipedia. I think this makes sense, right?

I am terrible at Latex, so I'm just linking to Wikipedia. I think this makes sense, right?
Oh, those. I call them shift operators

Anyway, maybe those ladder operators work, but for that you got to calculate $e^{iA}$ and that seems a little tedious.
The easiest calculations are, in my opinion, with matrices.

Oh, those. I call them shift operators

Anyway, maybe those ladder operators work, but for that you got to calculate $e^{iA}$ and that seems a little tedious.
The easiest calculations are, in my opinion, with matrices.
OK, how about let A=({2,3}, {-2,1})? I think that works.

OK, how about let A=({2,3}, {-2,1})? I think that works.
OK, but the calculations are a bit annoying. Can't we find a matrix whose exponential can easily be calculated? For example, try a nilpotent matrix.

OK, but the calculations are a bit annoying. Can't we find a matrix whose exponential can easily be calculated? For example, try a nilpotent matrix.