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Unitary and self-adjoint operators

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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.


Homework Equations


Self-adjoint: A*=A


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.
 

Answers and Replies

  • #2
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Write out what it means that [itex]UU^*=I=U^*U[/itex]. What could go wrong?
 
  • #3
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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.
 
  • #4
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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 [itex]U=e^{iA}[/itex] is unitary. So, assume that U is unitary, then [itex]UU^*=I=U^*U[/itex]. What does that mean for A??
 
  • #5
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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 [itex]U=e^{iA}[/itex] is unitary. So, assume that U is unitary, then [itex]UU^*=I=U^*U[/itex]. What does that mean for A??
[itex]UU^*=(e^{iA})(e^{iA})^+=(e^{iA})^+(e^{iA})=U^*U[/itex]

I feel like this is really simple; I wonder why I'm not seeing it.
 
  • #6
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[itex]UU^*=(e^{iA})(e^{iA})^+=(e^{iA})^+(e^{iA})=U^*U[/itex]

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

[tex](e^{iA})(e^{iA})^+=I=(e^{iA})^+(e^{iA})[/tex]

Now, do you know what [itex](e^{iA})^+[/itex] is?? Can you write it in another way??
 
  • #7
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OK, so you want [itex]UU^*=I=UU^*[/itex]. You indeed say that this is the same as

[tex](e^{iA})(e^{iA})^+=I=(e^{iA})^+(e^{iA})[/tex]

Now, do you know what [itex](e^{iA})^+[/itex] is?? Can you write it in another way??
[itex](e^{iA})^+=e^{-iA^+}[/itex]
 
  • #8
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OK, so now you have

[tex]e^{iA}e^{-iA^+}=I=e^{-iA^+}e^{iA}[/tex]

Now, what happens if you use [itex]e^{A+B}=e^A e^B[/itex]?? (I know the rule isn't always valid for matrices, but let's assume [itex]AA^+=A^+A[/itex], in that case the rule is valid).
 
  • #9
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OK, so now you have

[tex]e^{iA}e^{-iA^+}=I=e^{-iA^+}e^{iA}[/tex]

Now, what happens if you use [itex]e^{A+B}=e^A e^B[/itex]?? (I know the rule isn't always valid for matrices, but let's assume [itex]AA^+=A^+A[/itex], in that case the rule is valid).
You get [tex]e^{i(-A^++A)}=I=e^{i(A-A^+)}[/tex].
 
  • #10
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OK, so what did we find now? We found out that if A isn't self-adjoint, then there are two ways that [itex]e^{iA}[/itex] could still be unitary.

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

One of those two things must hold for A.

Let's focus on the second one, can you find a matrix such that [itex]AA^+\neq A^+A[/itex].
 
  • #11
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OK, so what did we find now? We found out that if A isn't self-adjoint, then there are two ways that [itex]e^{iA}[/itex] could still be unitary.

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

One of those two things must hold for A.

Let's focus on the second one, can you find a matrix such that [itex]AA^+\neq A^+A[/itex].
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. [itex]AA^+\neq A^+A[/itex].
 
  • #12
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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. [itex]AA^+\neq A^+A[/itex].
The condition [itex]AA^+=A^+A[/itex] 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.
 
  • #13
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The condition [itex]AA^+=A^+A[/itex] 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.
http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Ladder_operator_method

I am terrible at Latex, so I'm just linking to Wikipedia. I think this makes sense, right?
 
  • #14
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  • #15
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Oh, those. I call them shift operators :smile:

Anyway, maybe those ladder operators work, but for that you got to calculate [itex]e^{iA}[/itex] 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.
 
  • #16
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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.
 
  • #17
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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, how about ({0,1}, {0,0})?
 
  • #18
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OK, how about ({0,1}, {0,0})?
Yes, try that one! What do you get?
 
  • #19
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Yes, try that one! What do you get?
You get A dagger does not commute with A, as we wanted. What else do I need to show?
 
  • #20
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What do I need to show from there?
 

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