I am trying to find the determinant of the matrix


I expanded the above Matrix using the identity


on expanding I am getting $$\begin{bmatrix} \cos{\frac{\phi}{2}}+in_z\sin{\frac{\phi}{2}} & (n_y+in_x)\sin{\frac{\phi}{2}}\\ (-n_y+in_x)\sin{\frac{\phi}{2}} & \cos{\frac{\phi}{2}}-in_z\sin{\frac{\phi}{2}} \end{bmatrix}$$

Taking determinant of this matrix, I am getting $1+2\sin^2{\frac{\phi}{2}}$ But the actual answer is $\cos{\frac{\phi}{2}}+i\sin{\frac{\phi}{2}}$, Is there any other way for calculating the determinant of this matrix?


closed as off-topic by ACuriousMind Mar 21 '17 at 11:58

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By Jacobi's formula, for any complex square matrix $A$, one has that,

$$\det(e^A) = e^{\mathrm{tr}(A)}.$$

In your case, we have a matrix given by the dot product of the Pauli $\sigma^i$ and a vector $n_i$, for which we obtain that,

$$A := i \frac{\phi}{2}\sigma \cdot \hat{n} = i\frac{\phi}{2}\begin{pmatrix} n_z & n_x-in_y\\ n_x+in_y & -n_z \end{pmatrix}$$

which is a complex, square matrix. Then $\mathrm{tr}(A) = 0$ and we have,

$$\mathrm{det} \exp\left(i \frac{\phi}{2}\sigma \cdot \hat{n}\right) = 1.$$

It turns out matrices of this form are elements of the group $SU(2)$. If we did the calculation your way using your identity, we would have taken,

$$\det \begin{pmatrix} \cos \frac{\phi}{2} + in_z \sin \frac{\phi}{2} & (in_x+n_y)\sin\frac{\phi}{2}\\ (in_x-n_y)\sin\frac{\phi}{2}& \cos\frac{\phi}{2} - in_z \sin \frac{\phi}{2} \end{pmatrix} = \cos^2 \frac{\phi}{2} + (n_x^2+n_y^2+n_z^2)\sin^2\frac{\phi}{2} = 1$$

giving the same result since you assumed we were using a unit vector $\hat n$. Notice using the Jacobi formula we did not need to use the fact that $n$ was a unit vector but using your identity, we had to make this assumption, and so your identity is only valid for unit vectors.

  • $\begingroup$ Yes your argument is correct, the operator is Unitary, so the determinant must be equal to 1, But using explicit calculation I should get the same answer (i.e, determinant=1) right? But what is the error in my calculation? $\endgroup$ – Muthu manimaran Mar 21 '17 at 6:38
  • $\begingroup$ @Muthumanimaran Check my matrix using your identity, and the determinant calculation. You must have just messed up some algebra somewhere. $\endgroup$ – JamalS Mar 21 '17 at 6:41
  • $\begingroup$ Yes I did a miscalculation, I got the answer. $\endgroup$ – Muthu manimaran Mar 21 '17 at 7:12

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