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Question is, what would be the result of applying the operator $\hat A = [3I + \vec\sigma_1 . \vec\sigma_2]$ on the |singlet$\rangle$ and |triplet$\rangle$ states ($\vec\sigma_1$ acts on the 1st particle and $\vec\sigma_2$ acts on the second particle ONLY), ie, $$\hat A|singlet\rangle=?|singlet\rangle$$

and $$\hat A|triplet\rangle=?|triplet\rangle$$

I am stuck at the triplet part of the question.

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) $$\vec\sigma=\vec\sigma_1+\vec\sigma_2$$

squaring both sides, $$\vec\sigma^2=(\vec\sigma_1+\vec\sigma_2)^2$$

from which we have$$\vec\sigma_1 . \vec\sigma_2 = (\sigma^2 - \sigma_1^{2} - \sigma_2^{2})/2$$

Now, $\sigma_1^{2}=\sigma_{1x}^{2}+\sigma_{1y}^{2}+\sigma_{1z}^{2}=3I$ and similarly, $\sigma_2^{2}=3I$.

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 $$\vec\sigma_1 . \vec\sigma_2 = (0 - 3I - 3I)/2=-3I$$

I dont know what the value of $\sigma^2$ is for the triplet state (i do know that the total spin $S$ is $\sqrt2\hbar$)?

I am not able to relate the total spin with the $\vec\sigma$ properly

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    $\begingroup$ Have a go at writing out the matrix and the triplet state vector. Edit this question to place it in the reopen queue if you have a conceptual question afterwards. $\endgroup$ – rob Jan 16 '17 at 5:16
  • $\begingroup$ @rob , please be a little understanding. You may not realize this but i have spent a couple of hours trying to understand how to go about writing out the total pauli matrix for triplet state. All i have is scattered knowledge of pauli matrices and and singlet and triplet states. It is a shame that a user of your level cannot have a little empathy towards user of mine. Just because you have the power to put questions on hold, it doesnt mean you should put it if it doesnt fit your interpretation of the high standards of this site. Give others a chance to answer my doubts. $\endgroup$ – Prasad Mani Jan 16 '17 at 6:31
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    $\begingroup$ Sorry if I came off as gruff, @Prasad — showing empathy without getting wordy is tricky. Your edited post should be in the reopen queue, and this is the sort of borderline-homework-like question that the folks in Physics Chat might help with. $\endgroup$ – rob Jan 16 '17 at 6:40
  • $\begingroup$ @rob Is it too much to ask to just give a quick answer? The answer given is $8I$....it goes like this...$\sigma=2S$ where $S=\sqrt2$ and hence $\sigma^2=4S^2=8I$. What i dont understand is how $\sigma=2S$ for the triplet. $\endgroup$ – Prasad Mani Jan 16 '17 at 6:47
  • $\begingroup$ sorry but you question is not clear at all. Sure, $\sigma_1$ acts on 1st particle etc, but what does $\sigma^2$ act on? Presumably this is the total spin so $\sigma^2$ is just the unit operator (which you seem to call $I$ ) multiplied by $S(S+1)=1\times 2=2$ for the triplet. $\endgroup$ – ZeroTheHero Jan 17 '17 at 0:15
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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 $$ \sigma^+ = \sqrt{2} \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} . $$

Acting on the singlet, $\uparrow \downarrow- \downarrow \uparrow$ , this $\hat B$ has the obvious eigenvalue -3.

The triplet is $\uparrow \uparrow$; $(\uparrow\downarrow+\downarrow\uparrow)/\sqrt{2}$; $\downarrow \downarrow$, and so it obviously has eigenvalue 1 under the action of $\hat{B}$.

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.

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  • $\begingroup$ Thank you, this clears it up. Only one more thing. Isnt $\sigma^+$ = $\sigma^x+i\sigma^y$? What is the extra factor of $1/\sqrt{2}$. I understand you had to write it that way. But $\sigma^+$ is defined as $\sigma^x+i\sigma^y$ right? $\endgroup$ – Prasad Mani Feb 5 '17 at 4:03
  • $\begingroup$ I edited my answer to display explicitly $σ^+$ as normalized in the preceding expression for the dot term, so then $1/\sqrt{2}$ of what you write. I am using Pauli matrices, not spin matrices, which are half of the Pauli matrices. So the answer is just a brute force evaluation of the matrix tensor product. $\endgroup$ – Cosmas Zachos Feb 5 '17 at 4:17
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Setting $\hbar=1$ for simplicity, the matrices you need are the $S=1$ matrices. One easily obtains $$ S_z=\left(\begin{array}{ccc} 1&0&0\\ 0&0&0\\ 0&0&-1\end{array}\right)\, ,\quad S_+=\sqrt{2}\left(\begin{array}{ccc} 0&1&0\\ 0&0&1\\ 0&0&0\end{array}\right)\, ,\quad S_+=S_-^\dagger, $$ 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.

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  • $\begingroup$ Sorry, thats what i thought. But the answer i have is $8I$. I dont know which is right. See the comments where i told rob how that came about $\endgroup$ – Prasad Mani Jan 17 '17 at 2:51
  • $\begingroup$ I don't how you deduce that $\sigma=2S$. $\sigma$ by itself is not a number, nor is it a single operator. $\sigma^2$ makes sense as the sum of squares of all component but what is $\sigma$? A matrix? $\endgroup$ – ZeroTheHero Jan 17 '17 at 3:04
  • $\begingroup$ You wanted the matrices for the triplet states, they are as above. There is nothing more. $\endgroup$ – ZeroTheHero Jan 17 '17 at 3:06
  • $\begingroup$ I did not deduce that......i mentioned whatever was given in my book....i thought it didnt make sense ...so i asked here. So to sum it up, $\sigma=S$ without the $\hbar$ right? I dont know why they took 2S $\endgroup$ – Prasad Mani Jan 17 '17 at 3:21
  • $\begingroup$ Sorry you still have not clarified what is $\sigma$. Is it an operator, a vector of operators, or a number? $\endgroup$ – ZeroTheHero Jan 17 '17 at 3:25

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