I have started learning about spin operators and finding their eigenvalues, I have the following question below which will be split into two parts (the first part is needed to solve the second part). I can solve the first part which is shown below:

The single-particle spin operators are defined as follows (omitting the factors of $\hbar$ for clarity):

$$\hat s_z |\uparrow\,\rangle=\frac12|\uparrow\,\rangle\, ,\quad \hat s_z |\downarrow\,\rangle=-\frac12|\downarrow\,\rangle$$ $$\hat s_+ |\uparrow\,\rangle=0\, ,\quad \hat s_+ |\downarrow\,\rangle=|\uparrow\,\rangle$$ $$\hat s_- |\uparrow\,\rangle=|\downarrow\,\rangle\, ,\quad \hat s_- |\downarrow\,\rangle=0$$ Using these operators, and the fact that $\hat s^2$ can be defined as $$\hat s_z^2+\frac12\left(\hat s_-\hat s_++\,\,\hat s_+\hat s_-\right)$$ Show that $\hat s^2$ has the eigenvalue $\frac34$ for both $|\uparrow\,\rangle$ and $|\downarrow\,\rangle$.

Starting with the $|\uparrow\,\rangle$ state I find that $$\hat s_z^2|\uparrow\,\rangle=\hat s_z\hat s_z |\uparrow\,\rangle=\frac12\hat s_z|\uparrow\,\rangle=\frac12\cdot\frac12|\uparrow\,\rangle=\color{blue}{\frac14|\uparrow\,\rangle}$$ $$\hat s_-\hat s_+|\uparrow\,\rangle=\hat s_-(0)=\color{blue}{0}$$ $$\hat s_+\hat s_-|\uparrow\,\rangle=\hat s_+|\downarrow\,\rangle=\color{blue}{|\uparrow\,\rangle}$$

Now substituting the blue parts into $$\begin{align}\hat s^2|\uparrow\,\rangle&=\hat s_z^2|\uparrow\,\rangle+\frac12\left(\hat s_-\hat s_++\,\,\hat s_+\hat s_-\right)|\uparrow\,\rangle\\&=\hat s_z^2|\uparrow\,\rangle+\frac12\hat s_-\hat s_+|\uparrow\,\rangle+\frac12\hat s_+\hat s_-|\uparrow\,\rangle\\&=\frac14|\uparrow\,\rangle+\frac12(0)+\frac12|\uparrow\,\rangle\\&=\frac34|\uparrow\,\rangle\end{align}$$ as required.

For the $|\downarrow\,\rangle$ state I find that $$\hat s_z^2|\downarrow\,\rangle=\hat s_z\hat s_z |\downarrow\,\rangle=-\frac12\hat s_z|\downarrow\,\rangle=-\frac12\cdot-\frac12|\downarrow\,\rangle=\color{red}{\frac14|\downarrow\,\rangle}$$ $$\hat s_-\hat s_+|\downarrow\,\rangle=\hat s_-|\uparrow\,\rangle=\color{red}{|\downarrow\,\rangle}$$ $$\hat s_+\hat s_-|\downarrow\,\rangle=\hat s_+(0)=\color{red}{0}$$

Now substituting the red parts into $$\begin{align}\hat s^2|\downarrow\,\rangle&=\hat s_z^2|\downarrow\,\rangle+\frac12\hat s_-\hat s_+|\downarrow\,\rangle+\frac12\hat s_+\hat s_- |\downarrow\,\rangle\\&=\frac14|\downarrow\,\rangle+\frac12|\downarrow\,\rangle+\frac12(0)\\&=\frac34|\downarrow\,\rangle\end{align}$$ as required.

Here is the second part of the question (which I cannot solve):

Now consider the operators for the joint state of two electrons, e.g. $|\uparrow\uparrow\,\rangle$, where the first arrow indicates the state of spin 1 and the second spin 2. We define the operator for the total spin angular momentum of the system $\hat S=\hat s_1 +\hat s_2$ so we see that $\hat S^2=\hat s_1^2+\hat s_2^2+2\hat s_1\cdot\hat s_2$. We also define the operator for the total projection of the spin on the $z$-axis; $\fbox{$\hat S_z=\hat s_{z_{\large {1}}}+\hat s_{z_{\large {2}}}$}$. By analogy with the expression for $\hat s^2$ we can show that $$\hat s_1\cdot\hat s_2=\hat s_{z_{\large {1}}}\hat s_{z_{\large {2}}}+\frac12\left(\hat s_{1_{\large {-}}}\hat s_{2_{\large {+}}}+\hat s_{1_{\large {+}}}\hat s_{2_{\large {-}}}\right)$$ Using these relations, show that the two states $|\uparrow\uparrow\,\rangle$ and $|\downarrow\downarrow\,\rangle$ both have $S=1$ and find the eigenvalues they have for $\hat S_z$

I will start by attempting to find the eigenvalues for $\hat S_z$ for the state $|\uparrow\uparrow\,\rangle$ using the boxed formula in the quote above:

$$\begin{align}\hat S_z|\uparrow\uparrow\,\rangle&=\left(\hat s_{z_{\large {1}}}+\hat s_{z_{\large {2}}}\right)|\uparrow\uparrow\,\rangle\\&=\hat s_{z_{\large {1}}}|\uparrow\uparrow\,\rangle+\hat s_{z_{\large {2}}}|\uparrow\uparrow\,\rangle\end{align}$$

Now I am completely stuck as I am not given a relation between the operator $\hat s_{z_{\large {1}}}$ and its eigenvalue (like I was in the first part of the question). In other words $\hat s_{z_{\large {1}}}|\uparrow\uparrow\,\rangle=\mathrm{?}$

The answer simply states that:

Using the defintions given, $$\hat S_z|\uparrow\uparrow\,\rangle=\left(\frac12 +\frac12\right)|\uparrow\uparrow\,\rangle=|\uparrow\uparrow\,\rangle\tag{A}$$ giving an eigenvalue of $M_S=1$ for $|\uparrow\uparrow\,\rangle$.

This answer is of no help whatsoever as I have simply no idea how $(\mathrm{A})$ was obtained. As I am new to this you have probably noticed that I tend to write out all the steps in the calculation (in my answer to the first part of the question).

Could someone please show/explain the intermediate steps in reaching $(\mathrm{A})$ in a similar fashion (if possible) to what I did in the first part?


1 Answer 1


I think the part you're missing is that $\hat{s}_{z_1}$ only acts on the first spin and has the same properties as if the second spin was not present, i.e. $$ \hat{s}_{z_1} |\uparrow s_2\rangle = \frac{1}{2} |\uparrow s_2\rangle$$ and $$ \hat{s}_{z_1} |\downarrow s_2\rangle = -\frac{1}{2} |\downarrow s_2\rangle$$ with $s_2$ being either $\uparrow$ or $\downarrow$. Analogous properties hold for $\hat{s}_{z_2}$. With this I hope you can solve the problem.

To make it even more clear, here is what is really happening behind the notation given. A state like $|s_1 s_2\rangle$ actually means the tensor product of $s_1$ with $s_2$. $$ |s_1 s_2\rangle := |s_1\rangle \otimes |s_2\rangle$$ The operators perviously only acting on one of the spaces, like $\hat{s}_{z_1}$ and $\hat{s}_{z_2}$ are transferred to this tensor space by tensoring them to a unit operator, like so: $$ \hat{s}_{z_1} \otimes \mathbb{I} \qquad \text{and} \qquad \mathbb{I} \otimes \hat{s}_{z_2}$$ The left operator acts on the first state of the tensor product, the second operator on the right state. Then you get things like $$ (\hat{s}_{z_1} \otimes \mathbb{I}) (|s_1\rangle \otimes |s_2\rangle)\\ = (\hat{s}_{z_1}|s_1\rangle) \otimes (\mathbb{I}|s_2\rangle)$$ which is a tensor product of things we already know (from the first part of the question). We define the operators in tensor space like this, to give them precisely this property of only acting on one of the states, which is reasoned in what measurements are. If we measure just the first spin, we let the second one unperturbed. What is given in your example is simply a shorthand notation for the tensor notation used here.

  • $\begingroup$ Thanks for your answer, you mention that "$\hat{s}_{z_1}$ only acts on the first spin and has the same properties as if the second spin was not present". How do you know this? $\endgroup$
    – BLAZE
    Commented Jun 12, 2017 at 11:58
  • $\begingroup$ I extended it a little, hopefully it becomes clearer now. $\endgroup$
    – noah
    Commented Jun 12, 2017 at 12:11
  • 2
    $\begingroup$ If you can just accept the first part of my answer as given, i.e. that the operators only act on their corresponding spins (since the real background is a little too advanced), you are basically there with the try you have given in your question. You only need to apply $\hat{s}_{z_n} |\uparrow \uparrow \rangle$ in the mentioned way. $\endgroup$
    – noah
    Commented Jun 12, 2017 at 12:28

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