Prove Pauli Matrices: 65-Character Title

[LagrangeEuler]

Homework Statement

Prove
$$\exp (\alpha \hat{\sigma}_z+\beta \hat{\sigma}_x)=\cosh \sqrt{\alpha^2+\beta^2}+\frac{\sinh \sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}}(\alpha \hat{\sigma}_z+\beta \hat{\sigma}_x)$$

Homework Equations

$$e^{\hat{A}}=\hat{1}+\hat{A}+\frac{\hat{A}^2}{2!}+...$$

The Attempt at a Solution

$$\exp (\alpha \hat{\sigma}_z+\beta \hat{\sigma}_x)=\hat{1}+\alpha \hat{\sigma}_z+\beta \hat{\sigma}_x+\frac{1}{2!}(\alpha^2\hat{\sigma}_z^2+\beta^2\hat{\sigma}_x^2+\alpha \beta \hat{\sigma}_x\hat{\sigma}_z+\alpha \beta \hat{\sigma}_z\hat{\sigma}_x)+...$$
from that
$$\exp (\alpha \hat{\sigma}_z+\beta \hat{\sigma}_x)=\hat{1}+\alpha \hat{\sigma}_z+\beta \hat{\sigma}_x+\frac{1}{2!}(\alpha^2\hat{1}+\beta^2\hat{1}+\alpha \beta \hat{\sigma}_x\hat{\sigma}_z+\alpha \beta \hat{\sigma}_z\hat{\sigma}_x)+...$$
Is this way to go? I'm not sure?

 

[bloby]

Have you seen an expression for $(\vec{n}\cdot \hat{\vec{\sigma}})^k$, with $\vec{n}$ a unit vector?

[edit: no problem]

 

[Oxvillian]

Hi LagrangeEuler!

The series above simplifies drastically when you figure out what to do with $\sigma_x \sigma_z + \sigma_z \sigma_x$.



[edit: sorry Bloby, posted over you by mistake]

 

[SrayD]

Check out the anticommuntation properties {sig_x,sig_z} = 0, so there is a lot of cancelations.

 

[LagrangeEuler]

Have you seen an expression for $(\vec{n}\cdot \hat{\vec{\sigma}})^k$, with $\vec{n}$ a unit vector?

[edit: no problem]

I don't see connection with this problem?

 

[LagrangeEuler]

Hi LagrangeEuler!

The series above simplifies drastically when you figure out what to do with $\sigma_x \sigma_z + \sigma_z \sigma_x$.



[edit: sorry Bloby, posted over you by mistake]

Tnx a lot.

 

[bloby]

I don't see connection with this problem?

With $\sqrt{\alpha^2+\beta^2}=a$, $exp((\beta , 0 , \alpha)\cdot (\hat{\sigma}_x , \hat{\sigma}_y , \hat{\sigma}_z))=exp(\sqrt{\beta^2+\alpha^2}(\hat{n}\cdot \hat{\vec{\sigma}}))=I+a(\hat{n}\cdot\hat{\vec{\sigma}})+\frac{1}{2!}a^2(\quad)^2+...$