Quantum exam practice, operators and eigenstates

• pogs
In summary, the conversation discusses the concept of eigenstates and eigenvalue equations in quantum mechanics, specifically in relation to the operator Lz (angular momentum in the z-direction). The question at hand is whether certain states can be considered eigenstates of Lz and how to determine this without using an operator. The conversation also touches on LaTex code and the correct delimiters for displaying equations. The summary concludes with a clarification on when a linear combination of eigenstates will result in a new eigenstate with the same eigenvalue.
pogs
Homework Statement
We now focus on the Hydrogen atom wavefunctions ##\psi_{nlms}##, where n is the principal quantum number, l and m are the usual quantum numbers associated with orbital angular momentum
(##L^2## and ##L_z## respectively) and s (represented by an up or down-arrow) is the quantum
number associated with ##S_z## . Define J = L + S as the total angular momentum.

1.3. Take the three states :

$$\psi_a = \psi_{321\downarrow}$$
$$\psi_b = \frac{1}{\sqrt2}(\psi_{321↓} + \psi_{321↑})$$
$$\psi_c = \sqrt{\frac{1}{3}}\psi_{211↓} + \sqrt{\frac{2}{3}}\psi_{210↑}$$

For each of them, and for each of the operators $$L_z , J_z$$ and $$S_x$$ (note: x-component!), say whether the state is an eigenstate or not, giving the eigenvalue if it is and calculating the expectation value otherwise. Express your results in the form of a table.
Relevant Equations
$$L_z*f_m^l = \hbar mf_m^l$$
I'm really not sure what the question expects me to do here but here is what I do know. If the state is an eigenstate it should satisfy the eigenvalue equation for example;

$$\hat{H} f_m^l = \lambda f_m^l$$

but is the question asking me to use each operator on each state? How do I know if its an eigenstate without using an operator? IS it just me or is the question not so clear? Also I can't get the LaTex code to parse, what are the correct delimiters?

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pogs said:
Relevant Equations:: $L_z*f = hmf$

I'm really not sure what the question expects me to do here but here is what I do know. If the state is an eigenstate it should satisfy the eigenvalue equation for example;

H * f = lambda * f

but is the question asking me to use each operator on each state? How do I know if its an eigenstate without using an operator?
Look at the relevant equation you cited. If you can label a state as ψ_nlms, what does that tell you about the behaviour of ψ_nlms with respect to, for example, ##L_z##?
pogs said:
Also I can't get the LaTex code to parse, what are the correct delimiters?
Two dollar signs to start and end displayed equations, two hashtags (pound signs) to start and end inlined equations. See https://www.physicsforums.com/help/latexhelp/

Thanks Dr Claude.

I tried editing my question in the homework statement but I hope you can see the LaTex still isn't displaying, but it works elsewhere...

So if I apply Lz to the state I should get back the state times ##\hbar m##
correct?

$$L_z \psi_{nmls} = \hbar m \psi_{nmls}$$

In this case ##m=1## so

$$L_z \psi_{321\downarrow} = \hbar 1 \psi_{321\downarrow}$$

Is that enough to show that its an eigenstate with eigenvalue ##\hbar##?

$$L_z \frac{1}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow}) = \frac{\hbar (m_1+m_2)}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow}) = \sqrt2 \hbar(\psi_{321\downarrow}+ \psi_{321\uparrow})$$

Is this not an eigenstate because of the ##\sqrt 2##? Or it is?

Same for the last one but I get a coefficient of ##\frac{1}{\sqrt3}##. What am I to understand with these coefficients? Do they affect the answer at all?

pogs said:
Is this not an eigenstate because of the 2? Or it is?
Does it satisfy this for some ##\lambda##:
pogs said:
If the state is an eigenstate it should satisfy the eigenvalue equation for example;

H^fml=λfml
?

pogs said:
$$L_z \psi_{321\downarrow} = \hbar 1 \psi_{321\downarrow}$$

Is that enough to show that its an eigenstate with eigenvalue ##\hbar##?
Yes.
pogs said:
$$L_z \frac{1}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow}) = \frac{\hbar (m_1+m_2)}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow}) = \sqrt2 \hbar(\psi_{321\downarrow}+ \psi_{321\uparrow})$$
Your math is incorrect. You can't have ##m_1## and ##m_2## that appear at the front like that.

vanhees71 and Orodruin
DrClaude said:
Your math is incorrect. You can't have ##m_1## and ##m_2## that appear at the front like that.
Good catch! I definitely read that too fast. It is of course a valid expression though, just not the correct one.

But looking at Griffiths book on Quantum mechanics edition 3 p.176 eq.4.174

$$m = m_1 + m_2$$

So ##m_1 = m_2 = 1## ##\implies m =2## no?

I guess you're suggesting I leave the m's for each state as they are like so.

$$L_z \frac{1}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow}) = \frac{\hbar }{\sqrt2}(m_1 \psi_{321\downarrow}+ m_2 \psi_{321\uparrow}) = \frac{\hbar}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow})$$

Which gives us another eigenstate with ##\lambda = \frac{\hbar}{\sqrt2}##. Is that correct?

pogs said:
But looking at Griffiths book on Quantum mechanics edition 3 p.176 eq.4.174

$$m = m_1 + m_2$$

So ##m_1 = m_2 = 1## ##\implies m =2## no?
I don't have that edition of Griffiths, but this looks like the addition of angular momenta, which is not what you have at this stage (this can come in when considering ##J_z##).
pogs said:
$$L_z \frac{1}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow}) = \frac{\hbar }{\sqrt2}(m_1 \psi_{321\downarrow}+ m_2 \psi_{321\uparrow}) = \frac{\hbar}{\sqrt2}(\psi_{321\downarrow}+ \psi_{321\uparrow})$$

Which gives us another eigenstate with ##\lambda = \frac{\hbar}{\sqrt2}##. Is that correct?
No, the ##1 / \sqrt2## is already present in the wave function, so you are left with ##\lambda = \hbar## only.

To summarize, if you have an operator ##\hat{A}## with eigenvalues ##a##,
$$\hat{A} \phi_i = a_i \phi_i$$
and you apply it to a sum of eigenfunctions,
$$\psi = c_1 \phi_1 + c_2 \phi_2$$
(with ##c_1## and ##c_2## complex numbers) you need to apply it to each wave function individually:
\begin{align*} \hat{A} \psi &= \hat{A} \left( c_1 \phi_1 + c_2 \phi_2 \right) \\ &= c_1 \hat{A} \phi_1 + c_2 \hat{A} \phi_2 \\ &= c_1 a_1 \phi_1 + c_2 a_2 \phi_2 \end{align*}
Only in the case where ##a_1 = a_2 = a## do you get that ##\psi## is also an eigenfunction:
\begin{align*} \hat{A} \psi &= c_1 a \phi_1 + c_2 a \phi_2 \\ &= a \left( c_1 \phi_1 + c_2 \phi_2 \right) \\ &= a \psi \end{align*}

OK, that makes sense, thanks very much, and yes it was the addition of angular momenta for a two particle system, my mistake.

For the third state ##\psi_c = \sqrt{\frac{1}{3}}\psi_{211\downarrow} + \sqrt{\frac{2}{3}}\psi_{210\uparrow}##

##m_1 \neq m_2 ## so I would guess its not an eigenstate i.e.

$$L_z \psi_{c} = \hbar (m_1 \sqrt{\frac{1}{3}}\psi_{211\downarrow} + m_2\sqrt{\frac{2}{3}}\psi_{210\uparrow})$$.

Is that correct?

If so I have to calculate the expectation value I make the following calculation, ignoring the cross terms as they're orthogonal and taking ##L_z## outside.

$$(\sqrt{\frac{1}{3}}\psi_{211\downarrow}^* + \sqrt{\frac{2}{3}}\psi_{210\downarrow}^*) \cdot L_z \cdot (\sqrt{\frac{1}{3}}\psi_{211\downarrow} + \sqrt{\frac{2}{3}}\psi_{210\downarrow})$$

Leaves this, where the product of the states with themselves (conjugated) give 1. Although honestly, again , I'm not totally certain about ##m_1 ## and ## m_2##. Is it the case that as ##m_2=0## the second term disappears or is the whole expression just equal to zero?

$$\hbar [m_1 \frac{1}{3}\psi_{211\downarrow}^* \psi_{211\downarrow} + m_2\frac{2}{3}\psi_{210\uparrow}^*\psi_{210\uparrow}] = \langle L_z \rangle = \hbar$$

DrClaude said:
I don't have that edition of Griffiths, but this looks like the addition of angular momenta, which is not what you have at this stage (this can come in when considering Jz).
Most likely. I have the first edition so equation numbering is a bit different, but indeed chapter 4 discusses addition of angular momenta around eq 4.175.

pogs said:
For the third state ψc=13ψ211↓+23ψ210↑

m1≠m2 so I would guess its not an eigenstate i.e.

Lzψc=ℏ(m113ψ211↓+m223ψ210↑).

Is that correct?
Yes. In general, a linear combination of any eigenstates of an operator with the same eigenvalue will result in a new eigenstate with the same eigenvalue. Any linear combination of eigenstates with different eigenvalues will not be an eigenstate.

great thanks Oroduin, can you comment on the last part of the question too please?

pogs said:
$$\hbar [m_1 \frac{1}{3}\psi_{211\downarrow}^* \psi_{211\downarrow} + m_2\frac{2}{3}\psi_{210\uparrow}^*\psi_{210\uparrow}] = \langle L_z \rangle = \hbar$$
When you integrate, ##\psi_{211\downarrow}^* \psi_{211\downarrow}## and ##\psi_{210\uparrow}^*\psi_{210\uparrow}## equal 1, so you are left with
$$\langle L_z \rangle = \hbar [m_1 \frac{1}{3}+ m_2\frac{2}{3}]$$
The coefficients in front of ψ matter here.

Brilliant thanks Doc.

So to continue the question for ##J_z + L_z + S_z## If I try to make it more general I think I get the following, is it correct? Do we just use additivity?

$$J_z f^l_m = (L_z + S_z) f^l_m \implies J_z f^l_m = (\hbar m + \hbar m) f^l_m$$

Then for ##\psi_a## and ##\psi_b## I get ##2\hbar## as an eigenvalue for each.

##\psi_c## though is not an eigenstate again as ##m_1 \neq m_2##. Is that correct?The last question asks for ##S_x## and I'm not totally sure what to do with this one.

Here's what I know

The normalised eigenspinors for S_x are respectively spin and down.

$$\frac{1}{\sqrt2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \,\,\, , \frac{1}{\sqrt2} \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$

with eigenvalues ## \pm \frac{\hbar}{2}##

As ##\psi_a## is a down state I assume this gives us an eigenvalue of ##-\frac{\hbar}{2}## but I'm not sure how to justify it without deriving the normalised eigenspinors. Also we have the Pauli matrix which could be applied to the state but again I'm not sure what to do here. If I apply the operator to the state i.e ## S_x f^l_m## do I represent the state as the normalised eigenspinor for spin down? And so get the following?

$$S_x = \frac{\hbar}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \psi_{321\downarrow} = \frac{\hbar}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix} 1\\ -1 \end{bmatrix} = \frac{\hbar}{2\sqrt2}$$

pogs said:
Brilliant thanks Doc.

So to continue the question for ##J_z + L_z + S_z## If I try to make it more general I think I get the following, is it correct? Do we just use additivity?

$$J_z f^l_m = (L_z + S_z) f^l_m \implies J_z f^l_m = (\hbar m + \hbar m) f^l_m$$
Not that kind of additivity. Note that your wavefunctions have four subscripts, ##n,~l,~m_l,~m_s.##
So,
##L_z\psi_{nlm_lm_s}=m_l\hbar \psi_{nlm_lm_s}## and ##S_z\psi_{nlm_lm_s}=m_s\hbar \psi_{nlm_lm_s}##.
pogs said:
$$S_x = \frac{\hbar}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \psi_{321\downarrow} = \frac{\hbar}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix} 1\\ -1 \end{bmatrix} = \frac{\hbar}{2\sqrt2}$$
This doesn't make sense. How can a vector be equal to a scalar? It should be (in general) a linear combination of ##|\uparrow\rangle## and ##|\downarrow\rangle## times ##\psi_{321}.## Hint: Review the rules of matrix multiplication.

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Ah yes, thanks for the help, its a typo but also a wrong calculation. Also the second part has a missing vector. It should be $$S_x = \frac{\hbar}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \psi_{321\downarrow} = \frac{\hbar}{2} \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt2} \begin{bmatrix} 1\\ -1 \end{bmatrix} = \frac{\hbar}{2\sqrt2} \begin{bmatrix} -1\\ 1 \end{bmatrix}$$

but I'm not really sure what I'm doing here, let's take it at the end. In regards to the first part.

$$J_z f^l_m = (L_z + S_z) f^l_m \implies J_z f^l_m = (\hbar m_l + \hbar m_s) f^l_m$$

So for ##\psi_{321\downarrow}## I get ##\hbar m_l + \hbar m_s = - \frac{\hbar}{2}##

And for ##\psi_b##

$$(L_z + S_z)(\psi_{321\downarrow} + \psi_{321\uparrow}) = 2 \hbar$$

With ## \psi_c## not being an eigenstate.

$$\langle J_z \rangle = (L_z + S_z)\frac{1}{3} + (L_z + S_z)\frac{2}{3} = \frac{\hbar}{2}$$

Is that better?

Then the last part about ##S_x## could you maybe just say some words about the process here because I'm totally lost now on how to tackle this.$$S_x | \downarrow \rangle= \frac{\hbar}{2} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

It would have been better if in post #8 you wrote equations
$$|\uparrow\rangle = \frac{1}{\sqrt2}\begin{bmatrix}1 \\1\end{bmatrix} \,\,\, , |\downarrow\rangle=\frac{1}{\sqrt2} \begin{bmatrix}1 \\-1\end{bmatrix}.$$
You need to find ##\text{const}## and what to put in place of the question mark in $$S_x = \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \psi_{321\downarrow} =(\text{const})\psi_{321?}$$Do that by taking into account what you got already.
You wrote
$$\langle J_z \rangle = (L_z + S_z)\frac{1}{3} + (L_z + S_z)\frac{2}{3} = \frac{\hbar}{2}$$I don't see how you got that. Can you be more specific? I suggest that you find ##(L_z+S_z)|\psi_c\rangle## first. Call that ##|\varphi\rangle## and then find ##\langle \psi_c|\varphi\rangle##.

The rest of the stuff looks OK.

Thanks again Kuruman

In general, when we are asked to find the effect of the ##S_x## operator on some wave function am I to assume we use the eigenspinors for ##S_x## applied to the wave function spin state? and if so why exactly? Why not, use the Pauli matrix applied to the spin down vector? Is it because in general we only use this one for z-component operators? I'm assuming its something to do with the basis in which we are measuring?

Just to be clear this is what I mean,

$$S_x | \downarrow \rangle= \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}0 \\1\end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix}1 \\0\end{bmatrix}$$

Or this below, I'm not sure which is correct.

$$S_x | \downarrow \rangle = \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \frac{1}{\sqrt2}\begin{bmatrix}1 \\-1\end{bmatrix} = \frac{\hbar}{2\sqrt2} \begin{bmatrix} -1\\1 \end{bmatrix}$$The second part you wrote doesn't make sense to me.

##(L_z+S_z)|\psi_c\rangle ## call that ##|\varphi\rangle## and then find ##\langle \psi_c|\varphi\rangle##

I need the expectation of ##J_z## which is this.

$$\langle J_z \rangle = \psi_{c}^* J_z \psi_c$$

but I don't see how this ##\langle \psi_c|\varphi\rangle## applies ##L_z + S_z## ?

Taking a leaf from Dr Claudes book

DrClaude said:
When you integrate, ##\psi_{211\downarrow}^* \psi_{211\downarrow}## and ##\psi_{210\uparrow}^*\psi_{210\uparrow}## equal 1, so you are left with
$$\langle L_z \rangle = \hbar [m_1 \frac{1}{3}+ m_2\frac{2}{3}]$$
The coefficients in front of ψ matter here.

I get

$$= \frac{1}{3} \psi_{321\downarrow}^* \psi_{321\downarrow}(m_1 + s_{\downarrow}) + \frac{2}{3}\psi_{321\uparrow}^* \psi_{321\uparrow}(m_1 + s_{\uparrow})$$

$$= \frac{1}{3} \psi_{321\downarrow}^* \psi_{321\downarrow}\hbar(1 + (-\frac{1}{2})) + \frac{2}{3}\psi_{321\uparrow}^* \psi_{321\uparrow}\hbar(1 + \frac{1}{2}) = \frac{7\hbar}{6}$$

So I get a totally different result anyhow... Actually just thinking about another question, if the spin values are different for each term in a wave function (like with the principal quantum numbers ##m_l##) does it mean we can't add them and so we don't have an eigenstate?

I'm not sure now why we only have an eigenstate if ##m_{1l} = m_{2l}## like for the wave function

$$\psi_c = \psi_{321\downarrow} + \psi_{321\uparrow}$$

but it doesn't seem to matter if ##m_{s1} \neq m_{s2}##

pogs said:
Thanks again Kuruman

In general, when we are asked to find the effect of the ##S_x## operator on some wave function am I to assume we use the eigenspinors for ##S_x## applied to the wave function spin state? and if so why exactly? Why not, use the Pauli matrix applied to the spin down vector? Is it because in general we only use this one for z-component operators? I'm assuming its something to do with the basis in which we are measuring?

Just to be clear this is what I mean,

$$S_x | \downarrow \rangle= \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}0 \\1\end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix}1 \\0\end{bmatrix}$$

Or this below, I'm not sure which is correct.

$$S_x | \downarrow \rangle = \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \frac{1}{\sqrt2}\begin{bmatrix}1 \\-1\end{bmatrix} = \frac{\hbar}{2\sqrt2} \begin{bmatrix} -1\\1 \end{bmatrix}$$
Both are correct depending on how you define ##|\downarrow\rangle## and what you want to do. In the first equation that you have written it is used as the eigenstate of ##S_z## for eigenvalue ##-\frac{\hbar}{2}##, i.e. $$|\downarrow\rangle = \begin{bmatrix}0 \\1\end{bmatrix}$$ In the second equation it is used as the eigenstate of ##S_x## for eigenvalue ##-\frac{\hbar}{2}##, i.e. $$|\downarrow\rangle = \frac{1}{\sqrt2}\begin{bmatrix}1 \\-1\end{bmatrix}.$$

What you have written above are answers in search of a question. What is the question that makes you consider one or the other?

pogs said:
The second part you wrote doesn't make sense to me.

##(L_z+S_z)|\psi_c\rangle ## call that ##|\varphi\rangle## and then find ##\langle \psi_c|\varphi\rangle##

I need the expectation of ##J_z## which is this.

$$\langle J_z \rangle = \psi_{c}^* J_z \psi_c$$

but I don't see how this ##\langle \psi_c|\varphi\rangle## applies ##L_z + S_z## ?
This is not hrd to understand. I am suggesting that you break sown the calculation in two steps.
Step 1. Operate with ##(L_z+S_z)## on ##|\psi_c\rangle ##. The result is some kind of linear combination of eigenfunctions of ##L,~L_z,~S_z##. Call that ##|\varphi \rangle##. Mathematically,
\begin{align}(L_z+S_z)|\psi_c\rangle=|\varphi \rangle.\end{align}Step 2. The expectation value you are looking for is

##\langle \psi_c|(L_z+S_z)|\psi_c\rangle ##

In view of equation (1), the expectation value becomes
##\langle \psi_c|(L_z+S_z)|\psi_c\rangle=\langle \psi_c|\varphi \rangle.##
pogs said:
Actually just thinking about another question, if the spin values are different for each term in a wave function (like with the principal quantum numbers ##m_l##) does it mean we can't add them and so we don't have an eigenstate?
Why can you not add them? Look at $$\psi_c = \sqrt{\frac{1}{3}}\psi_{211↓} + \sqrt{\frac{2}{3}}\psi_{210↑}.$$ It's the addition of two different eigenstates of ##S_z##. As such,##\psi_c## cannot be an eigenstate of ##S_z## but in general it could be an eigenstate of some other operator.

Thanks again Kuruman

Maybe you can help me interpret the question then. How am I supposed to define spin down here? As far as I can tell there are two different possible ways.

Part two. Whilst I'm grateful for your help, please don't tell me its not hard to understand, its an unnecessary comment that doesn't contribute anything positive to my predicament, quite the opposite.

I assumed ##(L_z + S_z)## would operate on both the Bra and the Ket, but now its clear.

Part three, was not as clearly written as it could have been. I should have added the spin operator.

$$S_z |\psi_c \rangle = \hbar(s_1 \psi_{321\downarrow} + s_2 \psi_{321\uparrow})$$

If ##s_1 \neq s_2## then its not an eigenstate for S_z correct?

And I suppose in general, if something is an eigenstate it has to be for that particular operator I assume?

pogs said:
If ##s_1 \neq s_2## then its not an eigenstate for S_z correct?
Correct.

pogs said:
And I suppose in general, if something is an eigenstate it has to be for that particular operator I assume?
The qualifier "eigenstate" applies to a state with respect to a given operator.

Note that even if a state is not an eigenstate of ##S_z##, it can still be an eigenstate of ##J_z = L_z + S_z##.

pogs said:
Thanks again Kuruman

In general, when we are asked to find the effect of the ##S_x## operator on some wave function am I to assume we use the eigenspinors for ##S_x## applied to the wave function spin state? and if so why exactly? Why not, use the Pauli matrix applied to the spin down vector? Is it because in general we only use this one for z-component operators? I'm assuming its something to do with the basis in which we are measuring?

Just to be clear this is what I mean,

$$S_x | \downarrow \rangle= \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}0 \\1\end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix}1 \\0\end{bmatrix}$$

Or this below, I'm not sure which is correct.

$$S_x | \downarrow \rangle = \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \frac{1}{\sqrt2}\begin{bmatrix}1 \\-1\end{bmatrix} = \frac{\hbar}{2\sqrt2} \begin{bmatrix} -1\\1 \end{bmatrix}$$The second part you wrote doesn't make sense to me.

##(L_z+S_z)|\psi_c\rangle ## call that ##|\varphi\rangle## and then find ##\langle \psi_c|\varphi\rangle##

I need the expectation of ##J_z## which is this.

$$\langle J_z \rangle = \psi_{c}^* J_z \psi_c$$

but I don't see how this ##\langle \psi_c|\varphi\rangle## applies ##L_z + S_z## ?
Where I come from, ##| \downarrow \rangle ## is assumed to refer to z axis and is equivalent to Pauli notation $$\begin{bmatrix}0 \\1\end{bmatrix}$$
See https://en.wikipedia.org/wiki/Pauli_matrices for the usual definitions
So how to evaluate ## \langle S_x \rangle## ? You really have it
$$S_x | \downarrow \rangle= \frac{\hbar}{2}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} \begin{bmatrix}0 \\1\end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix}1 \\0\end{bmatrix}$$

Notice $$\langle \downarrow | S_x | \downarrow \rangle=0$$ and $$\langle \uparrow | S_x | \uparrow \rangle=0$$ but for the mixed state (## \psi_{c} ## ) the cross terms will contribute $$\langle \uparrow | S_x | \downarrow \rangle \neq 0$$

Thanks both.

$$S_x \psi_{c} = S_x (\frac{1}{\sqrt3}\psi_{321\downarrow} + \sqrt{\frac{2}{3}}\psi_{321\uparrow})$$

So to me its really not clear which spinor I should be using and why.

You should use $$|\uparrow\rangle = \begin{bmatrix}1 \\0\end{bmatrix}~~~|\downarrow\rangle = \begin{bmatrix}0\\1\end{bmatrix}.$$ That's because the convention is that "up" and "down" have been defined in terms of an already existing magnetic field that chooses the axis of quantization ##z##. That is the implicit convention when one writes something like ##\psi_{321\downarrow}##.

I understand what the convention is, I'm trying to get clarity on when one is supposed to use the S_x eigenspinors.

What you seem to be saying is the fact that I'm measuring ##S_x## has no bearing on the spinors I use. So when would you be expected to use the ##S_x## eigenspinors? Is it only when explicitly stated in the question?

pogs said:
What you seem to be saying is the fact that I'm measuring ##S_x## has no bearing on the spinors I use. So when would you be expected to use the ##S_x## eigenspinors? Is it only when explicitly stated in the question?
That is what I'm saying.

If the problem is asking you to find the expectation value ##\langle\psi_{321\uparrow} |S_z|\psi_{321\downarrow}\rangle##, you use the conventional eigenstates of ##S_z##.

If the problem is asking you to find the expectation value ##\langle \psi_{321\uparrow}|S_x|\psi_{321\downarrow}\rangle##, you use the conventional eigenstates of ##S_z##.

If the problem is asking you to find the expectation value ##\langle \psi_{321\beta}|S_x|\psi_{321\alpha}\rangle##, where ##|\alpha\rangle## and ##|\beta\rangle## are the eigenstates of ##S_x##, then you use those.

The first part of the question only asks whether we have an eigenstate and what the eigenvalue is aka.

pogs said:
$$S_x \psi_{c} = S_x (\frac{1}{\sqrt3}\psi_{321\downarrow} + \sqrt{\frac{2}{3}}\psi_{321\uparrow})$$

So to me its really not clear which spinor I should be using and why.

If its not an eigenstate THEN you calculate the expectation value. I'm asking when I apply the operator to the wavefunction how am I supposed to know which spinor I use.

pogs said:
The first part of the question only asks whether we have an eigenstate and what the eigenvalue is aka.
If its not an eigenstate THEN you calculate the expectation value. I'm asking when I apply the operator to the wavefunction how am I supposed to know which spinor I use.
$$S_x \psi_{c} = S_x (\frac{1}{\sqrt3}\psi_{321\downarrow} + \sqrt{\frac{2}{3}}\psi_{321\uparrow})= \frac{1}{\sqrt3}S_x\psi_{321\downarrow} + \sqrt{\frac{2}{3}}S_x\psi_{321\uparrow}.$$Find what each term gives you separately and then add them. As I said before, use the spinors that are eigenstates of ##S_z##.

Ok, but you're not really answering my question.

What about the eigenspinors of ##S_x## then? Under what conditions would I be expected to use them?

pogs said:
Ok, but you're not really answering my question.

What about the eigenspinors of ##S_x## then? Under what conditions would I be expected to use them?
A corollary of my statements in my post #25 is

If you need to find what you get from ##S_x|\psi_{321\downarrow}\rangle## then you use the eigentate of ##S_z##.
If you need to find what you get from ##S_x|\psi_{321\alpha}\rangle##, where ##|\alpha\rangle## is an eigenstate of ##S_x##, then you use that.

What you need to find depends on what you are asked to find, i.e. the statement of the problem.

Ok thanks.

1. What is a quantum exam practice?

A quantum exam practice is a set of questions and problems designed to test a student's understanding and knowledge of quantum mechanics. It typically covers topics such as operators, eigenstates, and other fundamental concepts in quantum physics.

2. What are operators in quantum mechanics?

In quantum mechanics, operators are mathematical objects that represent physical quantities, such as position, momentum, or energy. They act on quantum states to produce new states, and their eigenvalues correspond to the possible outcomes of a measurement.

3. What are eigenstates in quantum mechanics?

Eigenstates are special quantum states that are associated with specific eigenvalues of an operator. These states are the only ones that can be measured with certainty, and they form a complete basis for describing the quantum system.

4. How can I prepare for a quantum exam?

To prepare for a quantum exam, it is important to review the fundamental concepts and equations of quantum mechanics, practice solving problems and applying operators, and familiarize yourself with the mathematical formalism used in the subject. It can also be helpful to work through past exams or practice problems to get a sense of the types of questions that may be asked.

5. What is the significance of eigenstates and operators in quantum mechanics?

Eigenstates and operators are fundamental concepts in quantum mechanics that allow us to describe and understand the behavior of quantum systems. They play a crucial role in predicting the outcomes of measurements and in the mathematical formalism of quantum mechanics.

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