Exponential projection operator in Dirac formalism

[Dixanadu]

Homework Statement

Hey guys.

So here's the situation:
Consider the Hilbert space $H_{\frac{1}{2}}$, which is spanned by the orthonormal kets $|j,m_{j}>$ with $j=\frac{1}{2}, m_{j}=(\frac{1}{2},-\frac{1}{2})$. Let $|+> = |\frac{1}{2}, \frac{1}{2}>$ and $|->=|\frac{1}{2},-\frac{1}{2}>$. Define the following two projection operators:

$\hat{P}_{+}=|+><+|$ and $\hat{P}_{-}=|-><-|$.

Now consider the operator $\hat{O}=e^{i\alpha \hat{P}_{+}+i\beta \hat{P}_{-}}$.

Compute the following:

$<+|\hat{O}|+>$

$<-|\hat{O}|->$

$<-|\hat{O}|+>$

Homework Equations

orthonormality stuff: <+|+> = <-|-> = 1, <-|+> = <+|-> = 0.

Series representation of e^x:

$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k!}$

The Attempt at a Solution

So here's what I've done. Of course you gota represent the operator O in a nicer way, and this is what I need to know if I've done right:

$e^{i\alpha \hat{P}_{+}}=I+(i\alpha) \hat{P}_{+} + \frac{(i\alpha)^{2}}{2}\hat{P}_{+}+...=I+\hat{P}_{+}(i\alpha + \frac{(i\alpha)^2}{2}+...)=I+\hat{P}_{+}(e^{i\alpha}-1)$

Similarly:

$e^{i\beta \hat{P}_{-}}=I+\hat{P}_{-}(e^{i\beta}-1)$

where $I$ is the identity matrix. Now, O is a product of these two:

$\hat{O}=[I+\hat{P}_{+}(e^{i\alpha}-1)][I+\hat{P}_{-}(e^{i\beta}-1)]=I+\hat{P}_{+}(e^{i\alpha}-1)+\hat{P}_{-}(e^{i\beta}-1)$

because the cross-terms in the product vanish as $\hat{P}_{+}\hat{P}_{-}=0$

So that's my expression for O. Using that, I find that

$<+|\hat{O}|+>=e^{i\alpha}$. I haven't done the rest yet, but is this one right guys...?

please tell me if I've made any math errors!

 

[Goddar]

$e^{i\alpha \hat{P}_{+}}=I+(i\alpha) \hat{P}_{+} + \frac{(i\alpha)^{2}}{2}\hat{P}_{+}+...=I+\hat{P}_{+}(i\alpha + \frac{(i\alpha)^2}{2}+...)=I+\hat{P}_{+}(e^{i\alpha}-1)$

Hi.
That's right because ($\hat{P}$$_{\pm}$)$^{n}$ = $\hat{P}$$_{\pm}$, otherwise you would have to determine all exponents of the operators.
You're good so far!