I Orthogonal state with m = 0 carries s = 0 ... explanation?

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Do not understand statement: orthogonal state with m= 0 carries s = 0
Hello
I could use some help understanding a statement / sentence within my Griffiths Quantum Mechanics book. The same statement is made within video lecture I found surfing to understand the Griffiths text.
I have the 2nd edition. (On page 185)
Discussing addition of angular momenta – 2 spin ½ particles.
The end result is the “triplet” and the “singlet”

The singlet is my issue. I think I understand how the triplet was built / derived.
After the triplet discussion, the sentence reads: “Meanwhile, the orthogonal state with m = 0 carries s = 0.” And he shows the state | 0 0 >
I do not understand that sentence. It is apparently obvious; I have found 2 sources that both just make that statement.

Can you help explain, “The orthogonal state with m= 0 carries s = 0.”

Thanks
Sparky_
 

DrClaude

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I don't have the book in front of me, but I guess that the meaning is the following.

For a given s, the possible values of m are ##m = -s, -s+1, ..., s##. Therefore, if there is a lone state with ##m=0##, it must belong to ##s=0##.
 
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I am not with you yet. Can you explain further?

I see the key statements are : Griffiths "Meanwhile, the orthogonal state with m = 0 carries s = 0"
and your : "if there is a lone state with m=0m=0 , it must belong to s=0s=0 ."

I'm not there yet

Help?

Thanks
 

DrClaude

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Looking at the book, I am not sure why Griffiths chose the verb to carry. He uses it also on the next page:
I claim, then, that the cobintation of two spin-1/2 particles can carry a total spin of 1 or 0
"Meanwhile, the orthogonal state with m = 0 carries s = 0"
I think you could use "corresponds to" instead of "carries" here.

My point is the following. You consider first the four combinations of spin-up and spin-down in [4.175], and calculate the possible values of ##m##. It looks like a spin-1 state with an extra ##m=0## state. Figuring out what the actual three eigenstates with ##s=1## are, you are left with one state for which ##m=0##. Since you have only this one state, the only possible value of spin it can correspond to is ##s=0##, because if it were any other value of spin, you would need the extra ##m## states (such as ##m = \pm s##).

The point is that you can assign the value of spin immediately, without an additional check. Griffiths then proves that this is correct by actually calculating the expectation value of ##S^2## for that singlet state.
 

vanhees71

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If you have two independent spins, which we write as ##\hat{s}_{1j} \otimes \hat{1}## and ##\hat{1} \otimes \hat{s}_{2j}## for utmost clarity, their sum is
$$\hat{S}_j=\hat{s}_{1j} \otimes \hat{1} + \hat{1} \otimes \hat{s}_{2j}.$$
The two-spin basis is the product basis
$$|s_1,s_2 \rangle=|s_1 \rangle \otimes |s_2 \rangle,$$
where ##s_{1,2} \in \{-1/2,1/2 \}##. What you want to find is the eigenbasis of the spin-eigenvalue problem for ##\hat{S}_j## in terms of the product basis above. These eigenvectors are simultaneous eigenvectors of ##\hat{\vec{S}}^2## and ##\hat{S}_3## (usual choice by convention), ##|S,M \rangle##,
$$\hat{\vec{S}}^2 |S,M \rangle=S(S+1) |S,M \rangle, \quad \hat{S}_3 |S,M\rangle = M |S,M \rangle.$$
Now you have
$$\hat{S}_3 |s_1,s_2 \rangle=(s_1+s_2) |s_1,s_2 \rangle.$$
The ##|s_1,s_2 \rangle## are already an eigenbasis of ##\hat{S}_3##. Then it's clear, which eigenvalues ##\hat{S}_3## can take, namely ##1/2+1/2=1##, ##1/2-1/2=0##, and ##-1/2-1/2=-1##.

Now we have to find the eigenvalues of ##\hat{\vec{S}}^2##. For each possible representation of the rotation group, i.e., for each possible ##S \in \{0,1/2,1,\ldots, \}## we have ##M \in \{-S,-S+1,\ldots,S-1,S \}##. Since in our case there's no ##M>1##, we must have ##S \leq 1##. There are no half-integer ##M##'s, we can thus only have ##S=0## and ##S=1##. Further we must have ##S=1##, because otherwise there'd be only ##M=0##.

Thus we must have 3 linear eigenvectors with ##S=1##. Now the largest value ##M=1## must be the state
$$|S=1,M=1 \rangle=|1/2,1/2 \rangle$$
and the state with ##M=-1## must be
$$|S=1,M=-1 \rangle = |-1/2,-1/2 \rangle.$$
The ##M=0## state is given using the ladder operator ##\hat{S}_+## to the ##M=-1## state, which is (with the appropriate normalization)
$$\hat{S}_+ |-1/2,-1/2 \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle + |-1/2,1/2\rangle).$$
The ##S=1## states are thus simply the symmetrized product states. These must be the three orthonormal simultaneous eigenvectors of ##\hat{\vec{S}}^2## and ##\hat{S}_3##. It's easy to check that this is really the case.

Now it's very easy to also get the one remaining ##S=0## state. It must have ##M=0## and thus it's a linear combination
$$|S=0,M=0 \rangle=C_1 |1/2,-1/2 \rangle + C_2 |-1/2,1/2 \rangle.$$
It must be orthogonal to all three ##S=1## states. The only non-trivial condition is
$$\langle S=1,M=0|S=0,M=0 \rangle=\frac{1}{\sqrt{2}} (C_1+C_2) \stackrel{!}{=}0 \; \Rightarrow \; C_2=-C_1.$$
With the usual convention for the phases you thus get, together with the normalization condition ##C_1=-C_2=-1/\sqrt{2}##. Thus the only possibility is
$$|S=0,M=0 \rangle = \frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle - |-1/2,1/2 \rangle),$$
i.e., the singlet state is the antisymmetrized combination of the product states.
 
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Looking at the book, I am not sure why Griffiths chose the verb to carry. He uses it also on the next page:



I think you could use "corresponds to" instead of "carries" here.

My point is the following. You consider first the four combinations of spin-up and spin-down in [4.175], and calculate the possible values of ##m##. It looks like a spin-1 state with an extra ##m=0## state. Figuring out what the actual three eigenstates with ##s=1## are, you are left with one state for which ##m=0##. Since you have only this one state, the only possible value of spin it can correspond to is ##s=0##, because if it were any other value of spin, you would need the extra ##m## states (such as ##m = \pm s##).

The point is that you can assign the value of spin immediately, without an additional check. Griffiths then proves that this is correct by actually calculating the expectation value of ##S^2## for that singlet state.
Dr. Claude,

I am closer to the light bulb going off.

Here is what I do (think) I understand:
I followed the text showing the 4 possible states.
I see how the lowering operator applied starting at the "up-up" can get me to the 3 states.
I agree at this point it looks like a spin 1

I hope / think I am with you when you say: "

Since you have only this one state, the only possible value of spin it can correspond to is s=0, because if it were any other value of spin, you would need the extra m states"

Spin 0 is the only possible for s=0 ?? that gives you m = 0 and s =0 (m going from -s to +s)

The word I am hung up on in my Griffiths text (and the lecture on Youtube - which I'm sure is just using Griffiths) is "orthogonal"

Meaning " the orthogonal state with m = 0 carries (or as you say corresponds) s = 0"

I have thumb back through the first three chapters for clarity on what I assume should be obvious based on how he states it. - no luck

what does he mean by "the orthogonal state with m= 0"?

(Unfortunately when I first hear / read "orthogonal" I think "at-right-angle" I am an engineer - this is just a bucket list item to learn some QM) I know there is more to it from chapters 2 and 3. I followed the delta functions topic when the orthogonal property of the wave function was introduced.


Thanks so much for the help!
 
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Vanhees71 -

I am currently studying your reply, it is a little (or perhaps more) ahead of me. I have been going through Griffiths book, Chaps 1-4

I am reviewing your reply and googling the topic to help with your reply.

First dumb question what is this : ⊗ operation?


Thanks
 

vanhees71

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Well, Griffiths's QM book seems not to be as good as its popularity suggests, given that we have many questions by confused readers of this book in the forum. I've tried to give a complete explanation, how to add two spins with ##s=1/2##. Don't hesitate to ask for clarification. The issue of "addition of angular momenta" is not an easy task, and it takes a while to fully understand it. For me the best explanation is in

J. J. Sakurai, Modern Quantum Mechanics.
 
194
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Well, Griffiths's QM book seems not to be as good as its popularity suggests, given that we have many questions by confused readers of this book in the forum. I've tried to give a complete explanation, how to add two spins with ##s=1/2##. Don't hesitate to ask for clarification. The issue of "addition of angular momenta" is not an easy task, and it takes a while to fully understand it. For me the best explanation is in

J. J. Sakurai, Modern Quantum Mechanics.
Thanks,
fortunately I am working at my own pace and have committed to "keep going" and so forth even when frustration sets in.

what is the operation

all through your explanation?

Thanks
 

vanhees71

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This is the tensor product. If you have a composite system made of two parts (in your case two spins) you build the Hilbert space of this composite system by the tensor product of the two parts. For details, see, e.g.,


This looks like a great lecture on introductory QM (although I've clicked only randomly at some of the chapters of the manuscript):

https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/index.htm
 
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PeroK

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This is the tensor product. If you have a composite system made of two parts (in your case two spins) you build the Hilbert space of this composite system by the tensor product of the two parts. For details, see, e.g.,
@Sparky_

Griffiths actually manages to present spin without the tensor product. It's definitely a weakness of the book. I took some time out when I was studying from it to learn about the tensor product and put the theory of spin on a more solid mathematical footing.

I guess you have a decision on whether to do the same.

Regarding your question, the issue is resolved on the next page, 186, where Griffiths shows that these states are indeed eigenstates of ##S^2##.

PS what @vanhees71 has done in post #5 is a more elegant presentation of what Griffiths does on page 186.
 

PeroK

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The word I am hung up on in my Griffiths text (and the lecture on Youtube - which I'm sure is just using Griffiths) is "orthogonal"

Meaning " the orthogonal state with m = 0 carries (or as you say corresponds) s = 0"

I have thumb back through the first three chapters for clarity on what I assume should be obvious based on how he states it. - no luck

what does he mean by "the orthogonal state with m= 0"?

(Unfortunately when I first hear / read "orthogonal" I think "at-right-angle" I am an engineer - this is just a bucket list item to learn some QM) I know there is more to it from chapters 2 and 3. I followed the delta functions topic when the orthogonal property of the wave function was introduced.


Thanks so much for the help!
I'm not sure how you missed this if you've reached chapter 4, but "orthogonal" is the Hilbert space generalisation of the concept for ordinary 3D vectors. Two vectors are orthogonal if their inner product is zero.
 
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I'm not sure how you missed this if you've reached chapter 4, but "orthogonal" is the Hilbert space generalisation of the concept for ordinary 3D vectors. Two vectors are orthogonal if their inner product is zero.
PeroK, I think this is a (or at least my) problem doing a self-study and not having the structure of classroom lecture and a professor to test if I am getting "it". I did get through Chap 3 and could duplicate the work and worked the examples.

I obviously didn't get it all - in the big picture, meaning in the big picture, I didn't fully own and understand chapter 3 with me. I am going incredibly slow through this and seems like I am staying frustrated; it is cool when things click though.

Thank you all!!
I will review the Hilbert space discussion again
 
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A follow-up question,

For my own benefit, I would like to see how the inner product of the (singlet m=0 and s=0) and the triplet is zero?

I assume matrix multiplication is used and the product will be zero.

Can you show me the first few steps setting up the inner product that would show the orthogonal result (zero)?

Thanks
 

PeroK

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A follow-up question,

For my own benefit, I would like to see how the inner product of the (singlet m=0 and s=0) and the triplet is zero?

I assume matrix multiplication is used and the product will be zero.

Can you show me the first few steps setting up the inner product that would show the orthogonal result (zero)?

Thanks
First, you need to define the inner product on the composite Hilbert space of the two particle system. See here, for example:


Note that Griffiths uses the simplified notation ##\chi_1 \chi_2## instead of ##\chi_1 \otimes \chi_2##.

In any case, we have:

##\langle \chi_1 \chi_2, \phi_1 \phi_2 \rangle = \langle \chi_1 , \phi_1 \rangle \langle \chi_2 , \phi_2 \rangle##

So, you can see that two composite spin states are orthogonal if and only if either pair of single spin states is orthogonal.

That explains why all four states on page 185 (##\uparrow \uparrow## etc.) are othogonal.

In general, combining orthonormal bases for the constituent Hilbert spaces in this way gives you a ready-made orthonormal basis for the composite (tensor product of) Hilbert spaces.
 

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