Sakurai Problem 1.12: Probability of Getting +h/2

[Bill Foster]

Homework Statement

A spin ½ system is known to be in an eigenstate of $$\textbf{S}\cdot\hat{\textbf{n}}$$ with eigenvalue $\frac{\hbar}{2}$ , where $$\hat{\textbf{n}}$$ is a unit vector lying in the xy-plane that makes an angle γ with the positive z-axis.

a. Suppose $$S_x$$ is measured. What is the probability of getting $+\frac{\hbar}{2}$?

The Attempt at a Solution

$$|\hat{\textbf{n}},+\rangle = \cos\left(\frac{\beta}{2}\right)|+\rangle + \sin\left(\frac{\beta}{2}\right)|-\rangle$$

$$\langle S_x,+|\hat{\textbf{n}},+\rangle = \left(\frac{\langle +|+\langle -|}{\sqrt{2}}\right)\left(\cos\left(\frac{\gamma}{2}\right)|+\rangle + \sin\left(\frac{\gamma}{2}\right)|-\rangle\right)=\frac{1}{\sqrt{2}}\left(\cos\left(\frac{\gamma}{2}\right)+\sin\left(\frac{\gamma}{2}\right)\right)$$

The probability of getting $\frac{\hbar}{2}$ is

$$P=|{\langle S_x,+|\hat{\textbf{n}},+\rangle|^2=\frac{1+\sin\gamma}{2}$$

I don't really understand this solution.

Why are they using $$\langle S_x,+|\hat{\textbf{n}},+\rangle$$ instead of $$\langle S_x|\hat{\textbf{n}}\rangle$$?

And how does $\frac{\hbar}{2}$ come into play here? What if we were looking for the probability of getting something other than $\frac{\hbar}{2}$, like $\frac{\hbar}{4}$ for example? How would that change it?

 

[gabbagabbahey]

What are the eigenvalues and eigenstates of $\textbf{S}\cdot\mathbf{\hat{n}}$ ?

 

[Bill Foster]

What are the eigenvalues and eigenstates of $\textbf{S}\cdot\mathbf{\hat{n}}$ ?

According to the problem statement:

A spin ½ system is known to be in an eigenstate of $$\textbf{S}\cdot\hat{\textbf{n}}$$ with eigenvalue $\frac{\hbar}{2}$ .

So I guess $$\textbf{S}\cdot\hat{\textbf{n}}$$ is the eigenstate, and $\frac{\hbar}{2}$ is its eigenvalue.

 

[gabbagabbahey]

No, $\textbf{S}\cdot\mathbf{\hat{n}}$ is an operator, not an eigenstate. What are its eigenvalues and eigenstates?

 

[Bill Foster]

I don't know.

 

[gabbagabbahey]

You do understand that $\textbf{S}\cdot\mathbf{\hat{n}}$ is the component of the total spin operator along $\mathbf{\hat{n}}$ right?

What are the eigenvalues and eigenvector of any component of the spin operator?

 

[Bill Foster]

You do understand that $\textbf{S}\cdot\mathbf{\hat{n}}$ is the component of the total spin operator along $\mathbf{\hat{n}}$ right?

Yes. But isn't it a state?

What are the eigenvalues and eigenvector of any component of the spin operator?

$$S_x=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)$$
$$S_y=\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)$$
$$S_z=\frac{\hbar}{2}\left(|+\rangle - |-\rangle\right)$$

 

[Bill Foster]

Is this where you're going?

$$\textbf{S}\cdot\hat{\textbf{n}}=\left(S_x\hat{\textbf{x}}+S_y\hat{\textbf{y}}+S_z\hat{\textbf{z}}\right)\cdot\left(\sin{\theta}cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}\right)=S_x\cos{\phi}+S_y\sin{\phi}$$

Because $$\theta=\frac{\pi}{2}$$

So that means

$$\textbf{S}\cdot\hat{\textbf{n}}=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}+\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)\sin{\phi}$$

 

[gabbagabbahey]

Yes. But isn't it a state?

Is $S_x$ a state?

$$S_z=\frac{\hbar}{2}\left(|+\rangle - |-\rangle\right)$$

In the $S_z$ eigenbasis, yes.

More generally, in the $\textbf{S}\cdot\mathbf{\hat{n}}$ eigenbasis, the eigenvalues and corresponding eigenstates of $\textbf{S}\cdot\mathbf{\hat{n}}$ are $\pm\frac{\hbar}{2}$ and $|\mathbf{\hat{n}},\pm\rangle$... Have you really not seen this before?

 

[Bill Foster]

Is $S_x$ a state?

No. But the problem statement says it's in an eigenstate of $$\textbf{S}\cdot\hat{\textbf{n}}$$. What does that mean?

More generally, in the $\textbf{S}\cdot\mathbf{\hat{n}}$ eigenbasis, the eigenvalues and corresponding eigenstates of $\textbf{S}\cdot\mathbf{\hat{n}}$ are $\pm\frac{\hbar}{2}$ and $|\mathbf{\hat{n}},\pm\rangle$... Have you really not seen this before?

Not that I can recall.

 

[gabbagabbahey]

No. But the problem statement says it's in an eigenstate of $$\textbf{S}\cdot\hat{\textbf{n}}$$. What does that mean?

It means that ther system is in some state $|\psi\rangle$ such that $\textbf{S}\cdot\hat{\textbf{n}}|\psi\rangle=\frac{\hbar}{2}|\psi\rangle$...this is the definition of eigenstate.

Not that I can recall.

Do you understand it though?

 

[Bill Foster]

It means that ther system is in some state $|\psi\rangle$ such that $\textbf{S}\cdot\hat{\textbf{n}}|\psi\rangle=\frac{\hbar}{2}|\psi\rangle$...this is the definition of eigenstate.

Do you understand it though?

Is this what you're saying:

$$\textbf{S}\cdot\hat{\textbf{n}}=\textbf{S}|\hat{\textbf{n}}\pm\rangle=\frac{\hbar}{2}|\hat{\textbf{n}}\pm\rangle$$

?

 

[gabbagabbahey]

Is this what you're saying:

$$\textbf{S}\cdot\hat{\textbf{n}}=\textbf{S}|\hat{\textbf{n}}\pm\rangle=\frac{\hbar}{2}|\hat{\textbf{n}}\pm\rangle$$

?

No. Again, $\textbf{S}\cdot\hat{\textbf{n}}$ is an operator, not an eigenstate. $\textbf{S}\cdot\hat{\textbf{n}}$ is an operator just like $S_z\equiv\textbf{S}\cdot\hat{\textbf{z}}$ is an operator.

 

[Bill Foster]

What, then, is it operating on?

 

[gabbagabbahey]

What, then, is it operating on?

In this problem, it isn't acting on any state at all. But, like every other operator it has eigenvalues and eigenstates. The system in this problem is known to be in one of those eigenstates; specifically, the eigenstate with corresponding eigenvalue of $+\frac{\hbar}{2}$.

$$S_z=\frac{\hbar}{2}\left(|+\rangle - |-\rangle\right)$$

I should have caught this mistake last night; but in the $S_z$ eigenbasis,

$$S_z=\frac{\hbar}{2}\left(|+\rangle\langle+| - |-\rangle\langle-|\right)\neq\frac{\hbar}{2}\left(|+\rangle - |-\rangle\right)$$

The first expression is an operator (as it should be), while the second is just a linear combination/superposition of two eigenstates.

Similarly, in the $\textbf{S}\cdot\mathbf{\hat{n}}$ eigenbasis, you have

$$\textbf{S}\cdot\mathbf{\hat{n}}=\frac{\hbar}{2}|\mathbf{\hat{n}},+\rangle\langle\mathbf{\hat{n}},+|-\frac{\hbar}{2}|\mathbf{\hat{n}},-\rangle\langle\mathbf{\hat{n}},-|$$

The extra label $\mathbf{\hat{n}}$ inside the Bras and Kets is just there to make it clear that they are the eigenstates of $\textbf{S}\cdot\mathbf{\hat{n}}$, in its eigenbasis.

 

[Bill Foster]

I find out what is $$\textbf{S}\cdot\hat{\textbf{n}}$$

I get the following:

$$\textbf{S}\cdot\hat{\textbf{n}}=\left(S_x\hat{\textbf{x}}+S_y\hat{\textbf{y}}+S_z\hat{\textbf{z}}\right)\cdot\left(\sin{\theta}cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}\right)=S_x\cos{\phi}+S_y\sin{\phi}$$

$$S_x=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)$$
$$S_y=\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)$$

So

$$\textbf{S}\cdot\hat{\textbf{n}}=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}+\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)\sin{\phi}$$
$$=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}-i\left(|+\rangle\langle -|-|-\rangle\langle +|\right)\sin{\phi}$$


Since the system is in an eigenstate of $$\textbf{S}\cdot\hat{\textbf{n}}$$ with eigenvalue of $\frac{\hbar}{2}$ it has to satisfy this equation:

$$\textbf{S}\cdot\hat{\textbf{n}}|\textbf{S}\cdot\hat{\textbf{n}};+\rangle=\frac{\hbar}{2}|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$$

I understand everything up to here. I want to know why it has to satisfy that last equation. I know the definition of the eigenvalue. I want to know why $$|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$$ is the eigenvector.

Or to put it another way, why isn't $$|\textbf{S}\cdot\hat{\textbf{n}};-\rangle$$ the eigenvector? Or why isn't $$|\textbf{S}\cdot\hat{\textbf{n}}\rangle$$ the eigenvector?

 

[Bill Foster]

Also, if we skip to the end of the problem, put another way, the probability of getting $\frac{\hbar}{2}$ when $$S_x$$ is measured is given by

$$|\langle S_x;+|\textbf{S}\cdot\hat{\textbf{n}};+\rangle|^2$$

Why is it that instead of this:

$$|\langle S_x|\textbf{S}\cdot\hat{\textbf{n}}\rangle|^2$$
?

 

[gabbagabbahey]

I find out what is $$\textbf{S}\cdot\hat{\textbf{n}}$$

I get the following:

$$\textbf{S}\cdot\hat{\textbf{n}}=\left(S_x\hat{\textbf{x}}+S_y\hat{\textbf{y}}+S_z\hat{\textbf{z}}\right)\cdot\left(\sin{\theta}cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}\right)=S_x\cos{\phi}+S_y\sin{\phi}$$

$$S_x=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)$$
$$S_y=\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)$$

So

$$\textbf{S}\cdot\hat{\textbf{n}}=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}+\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)\sin{\phi}$$
$$=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}-i\left(|+\rangle\langle -|-|-\rangle\langle +|\right)\sin{\phi}$$

My copy of Sakurai has $\mathbf{\hat{n}}$ lying in the $xz$-plane, making an angle $\gamma$ with the positive $z$-axis

$$\implies\mathbf{\hat{n}}=\sin\gamma\mathbf{\hat{x}}+\cos\gamma\mathbf{\hat{z}}$$

Since the system is in an eigenstate of $$\textbf{S}\cdot\hat{\textbf{n}}$$ with eigenvalue of $\frac{\hbar}{2}$ it has to satisfy this equation:

$$\textbf{S}\cdot\hat{\textbf{n}}|\textbf{S}\cdot\hat{\textbf{n}};+\rangle=\frac{\hbar}{2}|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$$

I understand everything up to here. I want to know why it has to satisfy that last equation. I know the definition of the eigenvalue. I want to know why $$|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$$ is the eigenvector.

Or to put it another way, why isn't $$|\textbf{S}\cdot\hat{\textbf{n}};-\rangle$$ the eigenvector? Or why isn't $$|\textbf{S}\cdot\hat{\textbf{n}}\rangle$$ the eigenvector?

$|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$ is defined as the eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$, with corresponding eigenvalue of $\frac{\hbar}{2}$ (In its eigenbasis!), and $|\textbf{S}\cdot\hat{\textbf{n}};-\rangle$ is defined as the eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$, with corresponding eigenvalue of $-\frac{\hbar}{2}$

So, if the system is known to be in an eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$ with corresponding eigenvalue $\frac{\hbar}{2}$, then it must be in the state $|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$. (If it were instead known to be in an eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$ with corresponding eigenvalue $-\frac{\hbar}{2}$, then it would be in the state $|\textbf{S}\cdot\hat{\textbf{n}};-\rangle$)

Also, writing $|\textbf{S}\cdot\hat{\textbf{n}}\rangle$ makes absolutely no sense. As I said earlier, $\textbf{S}\cdot\hat{\textbf{n}}$ is an operator, not a state; so writing it inside a Ket like this makes no sense.

Also, if we skip to the end of the problem, put another way, the probability of getting $\frac{\hbar}{2}$ when $$S_x$$ is measured is given by

$$|\langle S_x;+|\textbf{S}\cdot\hat{\textbf{n}};+\rangle|^2$$

Why is it that instead of this:

$$|\langle S_x|\textbf{S}\cdot\hat{\textbf{n}}\rangle|^2$$
?

Well, the initial state of the system is $|\psi_i\rangle=|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$. If $S_x$ is measured, and the result is $\frac{\hbar}{2}$, what will the final state$|\psi_f\rangle$ of the system be? What is the probability of this outcome? (If you can't immediately answer these question, you need to re-read section 1.4 of Sakurai!)

 

[jdwood983]

$$\implies\mathbf{\hat{n}}=\sin\gamma\mathbf{\hat{x}}+\cos\gamma\mathbf{\hat{z}}$$

While this is generally true, Problem 12 follows from Problem 9, so the angles above should be $$\gamma/2$$

 

[gabbagabbahey]

While this is generally true, Problem 12 follows from Problem 9, so the angles above should be $$\gamma/2$$

Ermm... no, they should still just be $\gamma$...you need only draw a picture to convince yourself of this.

The $\gamma/2$ present in the equation for the eigenstates $|\textbf{S}\cdot\hat{\textbf{n}};\pm\rangle$ results from a straightforward solving of the eigenvalue equation using $\hat{\textbf{n}}=\sin\gamma\hat{\textbf{x}}+\cos\gamma\hat{\textbf{z}}$.

 

[jdwood983]

Ermm... no, they should still just be $\gamma$...you need only draw a picture to convince yourself of this.

The $\gamma/2$ present in the equation for the eigenstates $|\textbf{S}\cdot\hat{\textbf{n}};\pm\rangle$ results from a straightforward solving of the eigenvalue equation using $\hat{\textbf{n}}=\sin\gamma\hat{\textbf{x}}+\cos\gamma\hat{\textbf{z}}$.

Good point. I was thinking

$$|\hat{\mathbf{n}};+\rangle=\cos\frac{\gamma}{2}|+\rangle+\sin\frac{\gamma}{2}|-\rangle$$

 

[Bill Foster]

My copy of Sakurai has $\mathbf{\hat{n}}$ lying in the $xz$-plane, making an angle $\gamma$ with the positive $z$-axis

$$\implies\mathbf{\hat{n}}=\sin\gamma\mathbf{\hat{x}}+\cos\gamma\mathbf{\hat{z}}$$

Wait a second...

Let $$\gamma=\theta$$ (the angle between the unit vector $$\hat{\textbf{n}}$$ and the z-axis), and let $$\phi$$ be the azimuthal angle.

Then in the general sense, the unit vector $$\hat{\textbf{n}}=\sin{\theta}\cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}$$. If $$\hat{\textbf{n}}$$ lies in the xy plane, then it makes an angle with $$\hat{\textbf{z}}$$ of $$\theta=\frac{\pi}{2}$$. So then

$$\hat{\textbf{n}}=\sin{\theta}\cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}=\sin{\frac{\pi}{2}}\cos{\phi}\hat{\textbf{x}}+\sin{\frac{\pi}{2}}\sin{\phi}\hat{\textbf{y}}+\cos{\frac{\pi}{2}}\hat{\textbf{z}}$$
$$=\cos{\phi}\hat{\textbf{x}}+\sin{\phi}\hat{\textbf{y}}$$

$|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$ is defined as the eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$, with corresponding eigenvalue of $\frac{\hbar}{2}$ (In its eigenbasis!), and $|\textbf{S}\cdot\hat{\textbf{n}};-\rangle$ is defined as the eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$, with corresponding eigenvalue of $-\frac{\hbar}{2}$

So, if the system is known to be in an eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$ with corresponding eigenvalue $\frac{\hbar}{2}$, then it must be in the state $|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$. (If it were instead known to be in an eigenstate of $\textbf{S}\cdot\hat{\textbf{n}}$ with corresponding eigenvalue $-\frac{\hbar}{2}$, then it would be in the state $|\textbf{S}\cdot\hat{\textbf{n}};-\rangle$)

Also, writing $|\textbf{S}\cdot\hat{\textbf{n}}\rangle$ makes absolutely no sense. As I said earlier, $\textbf{S}\cdot\hat{\textbf{n}}$ is an operator, not a state; so writing it inside a Ket like this makes no sense.



Well, the initial state of the system is $|\psi_i\rangle=|\textbf{S}\cdot\hat{\textbf{n}};+\rangle$. If $S_x$ is measured, and the result is $\frac{\hbar}{2}$, what will the final state$|\psi_f\rangle$ of the system be? What is the probability of this outcome? (If you can't immediately answer these question, you need to re-read section 1.4 of Sakurai!)

OK, thanks.

 

[gabbagabbahey]

Wait a second...

Let $$\gamma=\theta$$ (the angle between the unit vector $$\hat{\textbf{n}}$$ and the z-axis), and let $$\phi$$ be the azimuthal angle.

Then in the general sense, the unit vector $$\hat{\textbf{n}}=\sin{\theta}\cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}$$. If $$\hat{\textbf{n}}$$ lies in the xy plane, then it makes an angle with $$\hat{\textbf{z}}$$ of $$\theta=\frac{\pi}{2}$$. So then

$$\hat{\textbf{n}}=\sin{\theta}\cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}=\sin{\frac{\pi}{2}}\cos{\phi}\hat{\textbf{x}}+\sin{\frac{\pi}{2}}\sin{\phi}\hat{\textbf{y}}+\cos{\frac{\pi}{2}}\hat{\textbf{z}}$$
$$=\cos{\phi}\hat{\textbf{x}}+\sin{\phi}\hat{\textbf{y}}$$

Sure, but in this problem $\hat{\textbf{n}}$ lies in the $xz$-plane, not the $xy$-plane.

 

[Bill Foster]

Sure, but in this problem $\hat{\textbf{n}}$ lies in the $xz$-plane, not the $xy$-plane.

Well I'll be darned...you're right. That means $$\phi=0$$ or $$\phi=\pi$$

 

[gabbagabbahey]

I would assume that $\gamma$ is measured from the positive z-axis, counterclockwise towards the positive x-axis...making it slightly different than the definition of the polar angle, and resulting in the straightforward $\mathbf{\hat{n}}=\sin\gamma\mathbf{\hat{x} }+\cos\gamma\mathbf{\hat{z}}$.

 

[Bill Foster]

I've got this problem figured out.

However, in part b I'm confused.

I'm trying to calculate the following:

$$S_x^2=\frac{\hbar}{2}\left(|+\rangle\langle-|+|-\rangle\langle+|\right)\frac{\hbar}{2}\left(|+\rangle\langle-|+|-\rangle\langle+|\right)$$
$$=\frac{\hbar^2}{4}\left(|-\rangle\langle-|+|+\rangle\langle+|\right)$$
$$=\frac{\hbar^2}{4}\left(|+\rangle+|-\rangle\right)$$

But it's supposed to be just $\frac{\hbar^2}{4}$

What am I doing wrong that end up with a factor of $$|+\rangle+|-\rangle$$?

 

[gabbagabbahey]

$$S_x^2=\frac{\hbar}{2}\left(|+\rangle\langle-|+|-\rangle\langle+|\right)\frac{\hbar}{2}\left(|+\rangle\langle-|+|-\rangle\langle+|\right)$$
$$=\frac{\hbar^2}{4}\left(|-\rangle\langle-|+|+\rangle\langle+|\right)$$
$$=\frac{\hbar^2}{4}\left(|+\rangle+|-\rangle\right)$$

How do you go from an operator $S_x^2=\frac{\hbar^2}{4}\left(|-\rangle\langle-|+|+\rangle\langle+|\right)$ to an expression for a state $\frac{\hbar^2}{4}\left(|+\rangle+|-\rangle\right)$?!

 

[Bill Foster]

$$|-\rangle\langle-| = \left(\begin{array}{cc}0\\1\end{array}\right)\left(\begin{array}{cc}0&1\end{array}\right)=\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$$

$$|+\rangle\langle+| = \left(\begin{array}{cc}1\\0\end{array}\right)\left(\begin{array}{cc}1&0\end{array}\right)=\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$

Therefore,

$$|-\rangle\langle-| + |+\rangle\langle+|=\left(\begin{array}{cc}0&0\\0&1\end{array}\right) + \left(\begin{array}{cc}1&0\\0&0\end{array}\right) = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$$

For some reason I was thinking that

$$\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=|+\rangle + |-\rangle$$

But

$$|+\rangle + |-\rangle=\left(\begin{array}{cc}1\\1\end{array}\right)$$

 

[gabbagabbahey]

Therefore,

$$|-\rangle\langle-| + |+\rangle\langle+|=\left(\begin{array}{cc}0&0\\0&1\end{array}\right) + \left(\begin{array}{cc}1&0\\0&0\end{array}\right) = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$$

Right, which is the identity operator in the $S_z$ eigenbasis.