Clarification on a statement about spin 1/2 z component related to x and y

[Sparky_]

Hello,

I have a question about a statement made on a YouTube physics lecture

I was (am) working through chapter 4 section 4 (4.4) - “Spin” of Griffiths. (only because I own this book ) I found the YouTube lectures by searching for phrases like “quantum Griffiths online lectures”. One of the search results was Brant Carlson’s YouTube channel. He is a professor at Carthage college. His quantum lectures are on his channel. He uses Griffiths. I am following the parallel between angular momentum and spin and ladder operators. The mathematics is not an issue, matrix algebra or just algebra.

Within his lecture on spin he states at 21:10 (direct quote) “If I combine 2 states with definite z angular momentum in this very specific superposition, I end up with this state of definite x angular momentum”
My question or confusion is this is his bringing in a statement about “z”. Within the work leading up to this statement, z was never mentioned nor used, meaning it was all x and y within the work:

$$\ S_x |\psi\rangle = \lambda |\psi\rangle \\

\frac {\hbar } {2} \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \begin{pmatrix}
x \\
y \end{pmatrix} = \lambda \begin{pmatrix}
x \\
y \end{pmatrix} \\ ... \\
\lambda = \pm \frac{\hbar}{2} \\ ... \\
\begin{pmatrix}
\frac{\mp\hbar}{2} & \frac{\hbar}{2} \\
\frac{\hbar}{2} & \frac{\mp\hbar}{2}
\end{pmatrix} \begin{pmatrix}
x \\
y \end{pmatrix} = 0 \\
x = \pm y\\
normalized \\
\begin{pmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \end{pmatrix}\\

\chi_+ =
\begin{pmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} \end{pmatrix} \\

\chi_- =
\begin{pmatrix}
\frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} \end{pmatrix} \\$$

His material before his statement was the above, all dealing with x and y eigenvalues and so forth. I do not understand his statement, "“If I combine 2 states with definite z angular momentum in this very specific superposition, I end up with this state of definite x angular momentum” "

Thanks
Sparky_

 

[PeroK]

The two states here $\chi_+$ and $\chi_-$ are eigenstates of the $S_x$ operator. Expressed in the conventional "z-basis".

Neither has definite z angular momentum, as they are a superposition of eigenstates of the $S_z$ operator.

 

[Sparky_]

The two states here $\chi_+$ and $\chi_-$ are eigenstates of the $S_x$ operator. Expressed in the conventional "z-basis".

Neither has definite z angular momentum, as they are a superposition of eigenstates of the $S_z$ operator.

I don't understand "expressed in the conventional z-basis"

if I understand the basis is $$up =
\begin{pmatrix}

1 \\
0 \end{pmatrix} \\
down =
\begin{pmatrix}

0 \\
1 \end{pmatrix}
$$

I'm not sure how to ask the question - I am stuck on "z" what makes this the z-basis? Or is this to be treated as a given?

 

[PeroK]

I don't understand "expressed in the conventional z-basis"

if I understand the basis is $$up =
\begin{pmatrix}

1 \\
0 \end{pmatrix} \\
down =
\begin{pmatrix}

0 \\
1 \end{pmatrix}
$$

I'm not sure how to ask the question - I am stuck on "z" what makes this the z-basis? Or is this to be treated as a given?

"Up" means the state that is an eigenstate of the operator $S_z$ with eigenvalue $+\frac{\hbar}{2}$. And "down" means the same with eigenvalue $-\frac{\hbar}{2}$.

"up" and "down" form a basis for all spin states, and this is generally the conventional basis that is used by default. I sometimes write "z-up" and "z-down" if needed by the context.

In particular, you can express the eigenstates of $S_x$ (and $S_y$) in this basis and this is what is shown in your OP. You could call these "x-up" and "x-down".

 

[PeroK]

PS That Griffiths uses, by default, the standard "z-basis" should have been clear at the outset. This is where it was decided to use $S_z$ along with $S^2$ as, by default, the compatible observables for spin with which to develop the theory.

The theory works just the same with any other direction, but you have to choose one and $z$ is the default all the textbooks I've seen.

 

[vanhees71]

I don't understand "expressed in the conventional z-basis"

if I understand the basis is $$up =
\begin{pmatrix}

1 \\
0 \end{pmatrix} \\
down =
\begin{pmatrix}

0 \\
1 \end{pmatrix}
$$

I'm not sure how to ask the question - I am stuck on "z" what makes this the z-basis? Or is this to be treated as a given?

You work, by definition in the eigenbasis of $\hat{s}_z$. And that's why in this basis the matrix elements of $\hat{s}_z$ are just $1/2$ and $-1/2$ on the diagonal.

Further you just have verified that the eigenvectors of $\hat{s}_x$ are given by the superpositions
$$|\sigma_x \rangle=\frac{1}{\sqrt{2}} (|\text{\up} \rangle \pm \text{down} \rangle).$$

 

[Sparky_]

Yes -oops I see the
$$ +\frac{\hbar}{2} and - \frac{\hbar}{2} $$

my mistake

I was sort of gathering the statement i was questioning was because he (Griffiths and Carlson) started with $$ S^2 and S_z $$

It seems like there was emphasis on having information about z and ending with the x and y eigenvectors

side note: within Griffith's index I cannot find the word "superposition" - ugh
The MIT online lectures do a good job with superposition.

 

[Sparky_]

Further you just have verified that the eigenvectors of $\hat{s}_x$ are given by the superpositions
$$|\sigma_x \rangle=\frac{1}{\sqrt{2}} (|\text{\up} \rangle \pm \text{down} \rangle).$$

and (am I correct) the $$|\text{\up} \rangle \pm \text{down} \rangle).$$ are the starting point "z" basis (defined or given)?

 

[PeroK]

side note: within Griffith's index I cannot find the word "superposition" - ugh
The MIT online lectures do a good job with superposition.

If you've reached the end of chapter 4 of Griffiths and you're not sure what a superposition is, then maybe you're going to struggle.

 

[Sparky_]

If you've reached the end of chapter 4 of Griffiths and you're not sure what a superposition is, then maybe you're going to struggle.

I agree on the struggle, I can't find where Griffiths explain superposition very well

I do see a little more detail in Griffiths chap 12 - 12.4
I used the MIT material for superposition (took a detour from Griffiths when I was in Chapter 2 and watched several of the MIT lectures) -

 

[vanhees71]

Again a victim of this book! It cannot be that you are in chapter 4 of a QM textbook and don't know what "superposition" means. This should be on page 1 of the book!

 

[Sparky_]

Again a victim of this book! It cannot be that you are in chapter 4 of a QM textbook and don't know what "superposition" means. This should be on page 1 of the book!

and to be clear, I think I know what it means but went different paths to study it. I'm chasing this (study) for fun or a hobby or I'm not a current student. I was introduced to some of this when I was in college and from lay-reading I see the "mysterious" superposition topic brought up, anyway ...

Within this self study, I have found the math to not be the problem, the concepts are sometimes a struggle, thank you all for letting me ask the clarifying questions (not having a teacher or resources to ask)

 

[Sparky_]

ok, been reading and re-reading these few pages, I think I'm getting closer (but not there yet)the state is (the superposition)
$$ \psi = c_1
\begin{pmatrix}
1 \\
0 \end{pmatrix} + c_2
\begin{pmatrix}
0 \\
1 \end{pmatrix}
$$
help me break the statement that is bothering me into pieces: (baby steps)

the first part of the statement:
"If I combine two states with definite z-angular momentum". .
$$\chi_z = \frac{\hbar}{2}
\begin{pmatrix}
\frac{1}{\sqrt{2}} \\
0 \end{pmatrix} +
\frac{\hbar}{2}\begin{pmatrix}
0 \\
\frac{1}{\sqrt{2}} \end{pmatrix}
$$
Is the above the first part of the statement?
(The word "combine is superposition)

I'm and taking the phrase definte z... to mean a measurement has been made and we know ...

The second part of the statement: I end up with this state of definite x angular momentum”
This one I'm still not certain about, the phrase "end up with ... definite x ..."

I would also add to my confusion by saying knowing information about z does nothing for the x or y

I would have said it's not definite, it's 50/50 to be in one state or the other

I admit I am confused.

 

[PeroK]

ok, been reading and re-reading these few pages, I think I'm getting closer (but not there yet)the state is (the superposition)
$$ \psi = c_1
\begin{pmatrix}
1 \\
0 \end{pmatrix} + c_2
\begin{pmatrix}
0 \\
1 \end{pmatrix}
$$
help me break the statement that is bothering me into pieces: (baby steps)

the first part of the statement:
"If I combine two states with definite z-angular momentum". .
$$\chi_z = \frac{\hbar}{2}
\begin{pmatrix}
\frac{1}{\sqrt{2}} \\
0 \end{pmatrix} +
\frac{\hbar}{2}\begin{pmatrix}
0 \\
\frac{1}{\sqrt{2}} \end{pmatrix}
$$
Is the above the first part of the statement?
(The word "combine is superposition)

I'm and taking the phrase definte z... to mean a measurement has been made and we know ...

The second part of the statement: I end up with this state of definite x angular momentum”
This one I'm still not certain about, the phrase "end up with ... definite x ..."

I would also add to my confusion by saying knowing information about z does nothing for the x or y

I would have said it's not definite, it's 50/50 to be in one state or the other

I admit I am confused.

You said the maths is not a problem. If you forget about QM, can you explain this in terms of linear algebra and a change of (orthonormal) basis?

 

[PeterDonis]

The word "combine is superposition

Yes.

I'm and taking the phrase definte z... to mean a measurement has been made and we know

No, that's not what Griffiths is saying; he's just saying that the two states he's combining (forming a superposition of) are eigenstates of the $S_z$ operator, not that they describe actual particles whose $z$ spin has been measured. It's just math Griffiths is doing here, not an actual physical experiment.

This one I'm still not certain about, the phrase "end up with ... definite x ..."

He just means that the state he ends up with by forming the superposition is an eigenstate of the $S_x$ operator. Check it and see!

 

[vanhees71]

and to be clear, I think I know what it means but went different paths to study it. I'm chasing this (study) for fun or a hobby or I'm not a current student. I was introduced to some of this when I was in college and from lay-reading I see the "mysterious" superposition topic brought up, anyway ...

Within this self study, I have found the math to not be the problem, the concepts are sometimes a struggle, thank you all for letting me ask the clarifying questions (not having a teacher or resources to ask)

I was not critizising you but Griffiths's quantum-mechanics textbook, because it seems to confuse many students, as far as I can judge from the discussions in this forum.

Of course, it's also bad to have strange ideas from popular-"science" textbooks about quantum mechanics. Of course such a text sells better if they claim that there's something "mysterious" about quantum mechanics, but quantum mechanics is the opposite of something mysterious, because it's one of the most fundamental results of the exact quantitative science, physics.

Linear algebra, particularly Hilbert space theory, is far from being mysterious but a very-well understood mathematical discipline, and it's not that complicated after you have get used to it.

Maybe a far better book to start from is Susskinds "Theoretical Minimum - Quantum Mechanics".

 

[Sparky_]

Yes. No, that's not what Griffiths is saying; he's just saying that the two states he's combining (forming a superposition of) are eigenstates of the $S_z$ operator, not that they describe actual particles whose $z$ spin has been measured. It's just math Griffiths is doing here, not an actual physical experiment. He just means that the state he ends up with by forming the superposition is an eigenstate of the $S_x$ operator. Check it and see! PeterDonis - I'm trying to work toward (but am not there yet ) your ending statement ("the state he ends up with by forming the superposition is an eigenState of Sx") I see that this is a key point. - $$\ S_x |\psi\rangle = \lambda |\psi\rangle \\ \begin{pmatrix} 0 & \frac{\hbar}{2} \\ \frac{\hbar}{2} & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \lambda \begin{pmatrix} a \\ b \end{pmatrix} \\ \begin{bmatrix} -\lambda & -\frac{\hbar}{2} \\ \frac{\hbar}{2} &-\lambda \end{bmatrix} = 0\\ \lambda = \pm\frac{\hbar}{2} $$ same eigenvalues as Sz the eigenstates of Sx: $$ \begin{pmatrix} a \\ b \end{pmatrix}\\ \begin{vmatrix} \mp\frac{\hbar}{2} & -\frac{\hbar}{2} \\ \frac{\hbar}{2} &\mp\frac{\hbar}{2} \end{vmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ a = \pm b $$ I know from reading that there is (are) $$\frac{1}{\sqrt{2}}$$ terms involved (after normalization) I need some guidance to show the formed superposition is an eigenstate of Sx Next, just for grins I did the same exercise for Sz below is my work and this generated a question - the eigenvectors seem to be wrong$$\ S_z |\psi\rangle = \lambda |\psi\rangle \\ \begin{pmatrix} \frac{\hbar}{2} & 0\\ 0 & -\frac{\hbar}{2} & 0 \end{pmatrix} \begin{pmatrix} e \\ f \end{pmatrix} = \lambda \begin{pmatrix} e \\ f \end{pmatrix} \\ \begin{vmatrix} \frac{\hbar}{2} - \lambda & 0\\ 0 & -\frac{\hbar}{2} -\lambda \end{vmatrix} = 0\\ (\frac{\hbar}{2}-\lambda)(-\frac{\hbar}{2}-\lambda) = 0 \\ \lambda = \pm\frac{\hbar}{2} $$ the eigenstates of Sz: $$ \begin{pmatrix} e \\ f \end{pmatrix}\\ \begin{bmatrix} \frac{\hbar}{2}\mp\frac{\hbar}{2} & 0 \\ 0 & -\frac{\hbar}{2} \mp\frac{\hbar}{2} \end{bmatrix} \begin{pmatrix} e \\ f \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ e= 0 \\ f = 0 $$ ?? - nonsense? shouldn't I get one state is $$ \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ and the second state $$ \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$

 

[PeroK]

I must admit I have little idea what you are trying to do. Instead:
$$S_x \chi_+ =
\frac{\hbar}{2}\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} \frac{1}{\sqrt 2}
\begin{pmatrix}
1\\
1
\end{pmatrix} = \frac{\hbar}{2}\frac{1}{\sqrt 2}
\begin{pmatrix}
1\\
1
\end{pmatrix}
= \frac{\hbar}{2}\chi_+
\\
$$

 

[Sparky_]

I must admit I have little idea what you are trying to do. Instead:
$$S_x \chi_+ =
\frac{\hbar}{2}\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix} \frac{1}{\sqrt 2}
\begin{pmatrix}
1\\
1
\end{pmatrix} = \frac{\hbar}{2}\frac{1}{\sqrt 2}
\begin{pmatrix}
1\\
1
\end{pmatrix}
= \frac{\hbar}{2}\chi_+
\\
$$

I am not with you, what are you telling me here? I can duplicate that and I see the result within the text,

I am trying to chase what PeterDonis said : "the state he ends up with by forming the superposition is an eigenState of Sx")

I solved for the eigenstates of Sx but don't know where to go from there

 

[PeroK]

I am not with you, what are you telling me here? I can duplicate that and I see the result within the text,

I am trying to chase what PeterDonis said : "the state he ends up with by forming the superposition is an eigenState of Sx")

I solved for the eigenstates of Sx but don't know where to go from there

I can't really say I understand what you do and don't understand here.

We start with eigenstates of $S_z$, we find the eigenstates of $S_x$. The next step, if anything, is to find the eigenstates of $S_y$.

That's what this is all about.

 

[PeterDonis]

?? - nonsense?

Your whole post is so jumbled that I can't tell what you are trying to do, let alone whether it is right or wrong.

Let's start from the basics: in the representation Griffiths is using, we have the spin operator $\hat{S}_z$ represented as

$$
\hat{S}_z = \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & - 1 \end{bmatrix}
$$

and its two eigenvectors are

$$
| z_+ \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$

and

$$
| z_- \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
$$

with eigenvalues $\hbar / 2$ and $- \hbar / 2$. You should verify these eigenvalues yourself by multiplying the vectors $| z_+ \rangle$ and $| z_- \rangle$ by the matrix $\hat{S}_z$ explicitly.

Griffiths then asks you to consider the state

$$
| x_+ \rangle = \frac{1}{\sqrt{2}} \left( | z_+ \rangle + | z_- \rangle \right)
$$

which becomes the column vector

$$
| x_+ \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
$$

He says that this vector is an eigenvector of $\hat{S}_x$. To verify this, you need to find the matrix that represents $\hat{S}_x$ and multiply it by the column vector above, and see if the result is something times the same vector $| x_+ \rangle$. (Hint: this computation has already been done by another poster in this thread.)

 

[Sparky_]

ok,

I think I'm getting it (I was going to say I have it but realized I should not go that far , because you all have seen my struggle)

Below is my explanation to check and see if I have it ( I do agree that the explanation has been within the tread all along) Also for what it's worth the longer rambling above where PeterDonis says is so jumbled is from an online recorded lecture I found where prof is describing how to calculate the eigenstates of Sx but skipped some steps and described them verbally, I would like to delete that rambling to clean up this thread (it's embarrassing)

ok - am I getting it:

my original confusion was on Griffiths, "if I combine two states with z-angular momentum in a superposition, I end up with a state of definite x angular momentum"

1st error: I was taking the phrases "definite z" and "definite x"to mean measured experimentally - wrong

You all have explained that the state formed by the superposition of z is an eigenstate of x

to repeat your help:

consider the superposition:

$$
| \psi \rangle = \frac{1}{\sqrt{2}} \left( | z_+ \rangle + | z_- \rangle \right)
$$

the starting point(s):
$$
| z_+ \rangle =
\begin{pmatrix}
1\\0\end{pmatrix}
$$
and
$$
| z_- \rangle =
\begin{pmatrix}
0\\1 \end{pmatrix}
$$
$$
| \psi \rangle = \frac{1}{\sqrt{2}} (
\begin{pmatrix}
1\\0\end{pmatrix} +
\begin{pmatrix}
0\\1 \end{pmatrix}
)\\
| \psi \rangle = \frac{1}{\sqrt{2}}
\begin{pmatrix}
1\\1\end{pmatrix}
$$

You explained to check (verify) this is an eigenvector or eigenstate of Sx:

$$S_x| x_+ \rangle = \lambda|x_+ \rangle\\
S_x = \frac{\hbar}{2}
\begin{pmatrix}
0&1\\1&0\end{pmatrix}\\
\frac{\hbar}{2}
\begin{pmatrix}
0&1\\1&0\end{pmatrix}
\begin{pmatrix}
1\\1\end{pmatrix} = \lambda
\begin{pmatrix}
1\\1\end{pmatrix} \\
= \frac{\hbar}{2}(0+1+1+0) = 2\lambda\\
\lambda = \frac{\hbar}{2}

$$
So, I had a superposition of z states, that superposition is an eigenvector of Sx?

 

[PeroK]

That's certainly the idea. Note that the eigenstates of $S_z$, namely $|z+ \rangle$ and $|z-\rangle$, form a basis for all spin states. The eigenstates of $S_x$ and $S_y$ must be expressible as some linear combination (superposition) of these basis states.

 

[PeterDonis]

I had a superposition of z states, that superposition is an eigenvector of Sx

Yes, but only that particular superposition. And there is also one other particular superposition of $S_z$ eigenvector that is an $S_x$ eigenvector--can you see what it is?

Any other superposition of $S_z$ eigenvectors, besides the two mentioned above, will not be an $S_x$ eigenvector.

 

[Sparky_]

Yes, but only that particular superposition. And there is also one other particular superposition of $S_z$ eigenvector that is an $S_x$ eigenvector--can you see what it is?

Any other superposition of $S_z$ eigenvectors, besides the two mentioned above, will not be an $S_x$ eigenvector.

well, at the risk of showing that I do not understand ...

either lone state $$
\begin{pmatrix}
1\\0\end{pmatrix} or

\begin{pmatrix}
0\\1 \end{pmatrix}
$$
would work but that would not be a superposition
and the boring matrix$$
\begin{pmatrix}
0\\0\end{pmatrix}
$$ giving 0=0
is not a state

a hint ? regarding another superposition of Sz that also works

 

[Sparky_]

wait .

how about $$
| \psi \rangle = \frac{1}{\sqrt{2}} (
\begin{pmatrix}
1\\0\end{pmatrix} -
\begin{pmatrix}
0\\1 \end{pmatrix}
)\\
| \psi \rangle = \frac{1}{\sqrt{2}}
\begin{pmatrix}
1\\-1\end{pmatrix}
$$

 

[PeterDonis]

how about

Ok, what do you get when you multiply this vector by the matrix $\hat{S}_x$?

 

[Sparky_]

Ok, what do you get when you multiply this vector by the matrix $\hat{S}_x$?

hmm,

multiplying Sx and the above I get 0

$$
\begin{pmatrix}
0&1\\1&0\end{pmatrix} \begin{pmatrix}
1\\-1\end{pmatrix}\\
= 0-1+1+0
$$

 

[PeterDonis]

multiplying Sx and the above I get

No, you don't. Multiplying a column vector by a matrix gives you another column vector, not a number. The number you should get is the eigenvalue, not the product. You did it correctly in post #22 for the column vector

$$
\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
$$

so you should be able to easily do it for the very similar column vector

$$
\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ - 1 \end{bmatrix}
$$

 

[PeterDonis]

You did it correctly in post #22

Actually, I take that back; you didn't do it entirely correctly in post #22, you just got lucky.

The correct part of what you did in post #22 is this equation:

$$
\frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \lambda \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
$$

where I have put back the factors of $1 / \sqrt{2}$ on each side to make it clear exactly what we are doing, namely, writing the eigenvalue equation for $\hat{S}_x$ and $| x_+ \rangle$. But this gives two equations for $\lambda$, not one: one equation for the upper component of the RHS, the second for the lower component of the RHS. Those equations are

$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} \left( 0 \times 1 + 1 \times 1 \right) = \lambda \times \frac{1}{\sqrt{2}} \times 1
$$

$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} \left( 1 \times 1 + 0 \times 1 \right) = \lambda \times \frac{1}{\sqrt{2}} \times 1
$$

Both of these equations give the same value for $\lambda$, namely $\hbar / 2$.

Now do the same with the column vector

$$
\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}
$$

 

[Sparky_]

Actually, I take that back; you didn't do it entirely correctly in post #22, you just got lucky.

The correct part of what you did in post #22 is this equation:

$$
\frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \lambda \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
$$

where I have put back the factors of $1 / \sqrt{2}$ on each side to make it clear exactly what we are doing, namely, writing the eigenvalue equation for $\hat{S}_x$ and $| x_+ \rangle$. But this gives two equations for $\lambda$, not one: one equation for the upper component of the RHS, the second for the lower component of the RHS. Those equations are

$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} \left( 0 \times 1 + 1 \times 1 \right) = \lambda \times \frac{1}{\sqrt{2}} \times 1
$$

$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} \left( 1 \times 1 + 0 \times 1 \right) = \lambda \times \frac{1}{\sqrt{2}} \times 1
$$

Both of these equations give the same value for $\lambda$, namely $\hbar / 2$.

Now do the same with the column vector

$$
\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}
$$

shoot, (wasn't being careful) , should have looked up the actual matrix operation -

$$
\frac{1}{\sqrt{2}}\begin{pmatrix}
0&\frac{\hbar}{2}\\ \frac{\hbar}{2}&0\end{pmatrix} \begin{pmatrix}
1\\-1\end{pmatrix} = \lambda \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix}
$$
2 equations:
$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} = \frac{\lambda}{\sqrt{2}} \\
-\frac{\hbar}{2} \frac{1}{\sqrt{2}} = -\frac{\lambda}{\sqrt{2}} \\

\lambda = \frac{\hbar}{2}
$$

 

[PeterDonis]

2 equations

You have the signs wrong. The eigenvalue should be $- \hbar / 2$.

 

[Sparky_]

crap, went too fast

(I did on scratch paper earlier the "characteristic equation" deal for finding eigenvalues - the determinant resulting from subtracting the eigenvalues times the identity matrix and setting to zero ...

there I got $$ \pm \frac{\hbar}{2} $$

I should have caught that. I knew there was a $$-\frac{\hbar}{2}$$

corrected:
$$
\frac{1}{\sqrt{2}}\begin{pmatrix} 0&\frac{\hbar}{2}\\ \frac{\hbar}{2}&0\end{pmatrix} \begin{pmatrix} 1\\-1\end{pmatrix} = \lambda \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix} \\

-\frac{\hbar}{2} \frac{1}{\sqrt{2}} = \frac{\lambda}{\sqrt{2}} \\
\frac{\hbar}{2} \frac{1}{\sqrt{2}} = -\frac{\lambda}{\sqrt{2}} \\

\lambda = -\frac{\hbar}{2}
$$