Pauli matrix for triplet state? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-19T02:56:33Z https://physics.stackexchange.com/feeds/question/305398 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/305398 3 Pauli matrix for triplet state? Prasad Mani https://physics.stackexchange.com/users/124829 2017-01-16T04:58:41Z 2017-02-05T04:13:15Z <p>Question is, what would be the result of applying the operator $\hat A = [3I + \vec\sigma_1 . \vec\sigma_2]$ on the <strong>|singlet$\rangle$</strong> and |<strong>triplet$\rangle$</strong> states ($\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY), ie, $$\hat A|singlet\rangle=?|singlet\rangle$$</p> <p>and $$\hat A|triplet\rangle=?|triplet\rangle$$</p> <p>I am stuck at the triplet part of the question.</p> <p>For a system of 2 spin half particles, where $\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY, (like adding angular momentum of two electrons) $$\vec\sigma=\vec\sigma_1+\vec\sigma_2$$</p> <p>squaring both sides, $$\vec\sigma^2=(\vec\sigma_1+\vec\sigma_2)^2$$ </p> <p>from which we have$$\vec\sigma_1 . \vec\sigma_2 = (\sigma^2 - \sigma_1^{2} - \sigma_2^{2})/2$$</p> <p>Now, $\sigma_1^{2}=\sigma_{1x}^{2}+\sigma_{1y}^{2}+\sigma_{1z}^{2}=3I$ and similarly, $\sigma_2^{2}=3I$.</p> <p>and that for the singlet state, the value of $\sigma^2=0$, (which i gathered from the total spin being $0$ for the singlet state) which gives $$\vec\sigma_1 . \vec\sigma_2 = (0 - 3I - 3I)/2=-3I$$</p> <p>I dont know what the value of $\sigma^2$ is for the triplet state (i do know that the total spin $S$ is $\sqrt2\hbar$)?</p> <p>I am not able to relate the total spin with the $\vec\sigma$ properly</p> https://physics.stackexchange.com/questions/305398/-/305579#305579 -1 Answer by ZeroTheHero for Pauli matrix for triplet state? ZeroTheHero https://physics.stackexchange.com/users/36194 2017-01-17T02:47:12Z 2017-01-17T02:47:12Z <p>Setting $\hbar=1$ for simplicity, the matrices you need are the $S=1$ matrices. One easily obtains $$S_z=\left(\begin{array}{ccc} 1&amp;0&amp;0\\ 0&amp;0&amp;0\\ 0&amp;0&amp;-1\end{array}\right)\, ,\quad S_+=\sqrt{2}\left(\begin{array}{ccc} 0&amp;1&amp;0\\ 0&amp;0&amp;1\\ 0&amp;0&amp;0\end{array}\right)\, ,\quad S_+=S_-^\dagger,$$ from which one recovers $S_x$ and $S_y$ by inverting $S_\pm=S_x\pm i S_y$. The matrix for $S^2$ will be $2\times I$ where $I$ is the $3\times 3$ unit matrix.</p> https://physics.stackexchange.com/questions/305398/-/309787#309787 3 Answer by Cosmas Zachos for Pauli matrix for triplet state? Cosmas Zachos https://physics.stackexchange.com/users/66086 2017-02-04T23:08:32Z 2017-02-05T04:13:15Z <p>As @rob asked you to, you are meant to simply write down $$\hat{B}\equiv\vec{\sigma}_1\cdot\vec{\sigma}_2 = {\sigma}_1^x {\sigma}_2^x +{\sigma}_1^y {\sigma}_2^y+{\sigma}_1^z {\sigma}_2^z \\= ({\sigma}_1^x+i {\sigma}_1^y)({\sigma}_2^x -i{\sigma}_2^y )/2 +({\sigma}_1^x-i{\sigma}_1^y ) ({\sigma}_2^y +i{\sigma}_2^y)/2+{\sigma}_1^z {\sigma}_2^z\\ \equiv {\sigma}_1^+ {\sigma}_2^- +{\sigma}_1^- {\sigma}_2^+ +{\sigma}_1^z {\sigma}_2^z ~,$$ where $\sigma^+ \uparrow=0$, and $\sigma^+ \downarrow=\uparrow \sqrt{2}$, etc... for both 1 and 2. Recall $$\sigma^+ = \sqrt{2} \begin{pmatrix} 0&amp;1\\ 0&amp;0 \end{pmatrix} .$$</p> <p>Acting on the singlet, $\uparrow \downarrow- \downarrow \uparrow$ , this $\hat B$ has the obvious eigenvalue -3.</p> <p>The triplet is $\uparrow \uparrow$; $(\uparrow\downarrow+\downarrow\uparrow)/\sqrt{2}$; $\downarrow \downarrow$, and so it obviously has eigenvalue 1 under the action of $\hat{B}$. </p> <p>Your $\hat A= 3 1\!\!1 +\hat{B}$ has eigenvalues 0 and 4 respectively, given my normalizations. This is to say, of course, that, for the triplet, $\sigma^2/4=2=(1+1)1$, as expected.</p>