Question

Let H be a hermitean(self-adjoint) operator on the 1-Qubit state space, show that e^(-iH) is unitary

if a matrix is Unitary then
U x U* = Identity
If a H matrix is hermitean then
H = H*

Answer 1

okay so im tempted to just say
\[(e^{iH})^*=e^{-iH}\]

dont know if that is right, but lets suppose that it is, then can i really just do
\[e^{iH} ~x~e^{-iH} = e^{i(H-H)} = 1

Answer 2

\[e^{iH} ~*~e^{-iH} = e^{i(H-H)} = 1

Answer 3

where 1 is the identity mat

Answer 4

I think you can say \[(e^{iH})^\dagger = e^{(iH)^\dagger} = e^{-i H}\] but I don't know if they expect you to go more in depth in this

Answer 5

kay cool lemme finish eating before the next qn bby

Answer 6

actually even this might not be explicit enough, but it's better than what I wrote earlier.
\[(e^{iH})^\dagger =\left( \sum_{n=0}^\infty \frac{(iH)^n}{n!}\right)^\dagger= \sum_{n=0}^\infty \frac{(-iH)^n}{n!}=e^{-iH}\]

Answer 7

ya

Answer 8

the explicit way i can think of needs an assuming that there is a P^T D P = H decomposition available

Answer 9

assumption*

Answer 10

Nah, we're almost there let me just write it out and explain each step better

Answer 11

ok cool

Answer 12

also 1 more thing i guess it didnt matter for this question, but since H is a herm opp for a 1 qubbit space, does that mean H is a 2by 2 matrix

Answer 13

gotta show this to be true
(A^n + B^n)*= A^n+B^n if A and B are hermetian

Answer 14

for this i can show it if A and B, are able to be written in P^T D P form but hmm

Answer 15

(symmetric)^n = still symmetric?

Answer 16

Rewrite the e^x in taylor series:
\[(e^{iH})^\dagger =\left( \sum_{n=0}^\infty \frac{(iH)^n}{n!}\right)^\dagger\]
Transposing and complex conjugate both distribute over addition, so we can apply it to each term in the sum:
\[\left( \sum_{n=0}^\infty \frac{(iH)^n}{n!}\right)^\dagger=\sum_{n=0}^\infty \left( \frac{(iH)^n}{n!}\right)^\dagger\]

Do some algebra, I just rewrote H=IH and then multiplied the constants times the identity matrix on the left to separate them out.

\[\sum_{n=0}^\infty \left( \frac{(iH)^n}{n!}\right)^\dagger = \sum_{n=0}^\infty \left[ \left( \frac{i^n}{n!}I \right) \left( H^n\right) \right]^\dagger\]

Now when we transpose we have to write all the matrices in reverse order but since we have powers of the same matrix, each of them transposes simply:

\[\sum_{n=0}^\infty \left[ \left( \frac{i^n}{n!}I \right) \left( H^n\right) \right]^\dagger = \sum_{n=0}^\infty\left( H^n\right)^\dagger \left( \frac{i^n}{n!}I \right) ^\dagger \]

Now conjugate transpose them, but each of the \(H^\dagger = H\) and the only interesting thing about the conjugate transpose of a scalar multiplied by the identity is just \(i^*=-i\).

\[\sum_{n=0}^\infty\left( H^n\right)^\dagger \left( \frac{i^n}{n!}I \right) ^\dagger = \sum_{n=0}^\infty H^n \frac{(-i)^n}{n!}I \]

Scalars commute trivially with all matrices, and the identity matrix absorbs in so we just end up with:
\[ \sum_{n=0}^\infty \frac{(-iH)^n}{n!}=e^{-iH}\]

Done, no need to diagonalize anything, it's simple as that.

Answer 17

oh babyy i just realizedd that

Answer 18

(AB)^T = B^T * A^T

Answer 19

xD

Answer 20

Don't worry about whether or not \((H^n)^\dagger = H^n\) Instead let's look at this:

\[(ABC)^\dagger= C^\dagger B^\dagger A^\dagger\]
now if we independently know \[C^\dagger = C\] and \[B^\dagger = B\] and \[A^\dagger = A\] then we can replace them all:

\[(ABC)^\dagger = C^\dagger B^\dagger A^\dagger = CBA\]

now if C=B=A=H now you have it

\[(H^3)^\dagger = H^3\]

Answer 21

yepp

Answer 22

ok cool lol I actually didn't know this until I typed all this out myself so it was a good exercise for me thanks lol

Answer 23

lol now i kinda realize why the T became dagger haha

Answer 24

they just added like 1 little line to T