Show that if H is a hermitian operator, U is unitary

[Vitani11]

Homework Statement

Show that if H is a hermitian operator, then U = e^iH is unitary.

Homework Equations

UU^† = I for a unitary matrix
A^†=A for hermitian operator
I = identity matrix

The Attempt at a Solution

Here is what I have. U = e^iH multiplying both by U^† gives UU^† = e^iHU^† then replacing U^† with U^-1 (a property of unitary matrices) I have UU^† = e^iHU^-1 and so UU^† = e^iHe^-iH = e^i(H-H)=e⁰ = 1 = I. I don't think this is right though...

 

[PeroK]

Here is what I have. U = eiH multiplying both by U† gives UU† = eiHU† then replacing U† with U-1 (a property of unitary matrices) I have UU† = eiHU-1 and so UU† = eiHe-iH = ei(H-H)=e0 = 1 = I. I don't think this is right though...

The bit I've underlined shows you assuming what you are supposed to prove.

You need to show that $UU^{\dagger} = I$ without assuming it!

 

[Vitani11]

Oh crap... okay I think I did it without that assumption. Here: U = e^iH = e^i(H†) = (e^iH)^† = U^† therefore since U = U^† this proves it. What do you think?

 

[PeroK]

Oh crap... okay I think I did it without that assumption. Here: U = e^iH = e^i(H†) = (e^iH)^† = U^† therefore since U = U^† this proves it.

That claims to prove that $U = U^{\dagger}$, which is not what you are trying to prove. What you have is not right. You need to think more carefully about what $(e^{iH})^{\dagger}$ should be.

 

[Vitani11]

Wait, is it even possible to go ahead and write that e^i(H)† is equal to (e^iH)^†? because that is the same as treating the dagger symbol as some exponent. I will continue working on this and be back soon.

 

[Vitani11]

U = e^iH, U^† = (e^-iH)^† = e^-iH so multiplying the first equation by this on both sides yields UU^† = e^iHe^-iH = e^i(H-H)=e⁰ = 1.

 

[PeroK]

U = e^iH, U^† = (e^-iH)^† = e^-iH so multiplying the first equation by this on both sides yields UU^† = e^iHe^-iH = e^i(H-H)=e⁰ = 1.

That looks better. Although, you should show why $H$ must be Hermitian.