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For two spin 1/2 particles I understand that the triplet states ($S = 1$) are: $\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $

\begin{align} \ket{1,1} &= \ket{\up\up} \\ \ket{1,0} &= \frac{\ket{\up\dn} + \ket{\dn\up}}{\sqrt2} \\ \ket{1,-1} &= \ket{\dn\dn} \end{align}

And that the singlet state ($S = 0$) is:

$$ \ket{0,0} = \frac{\ket{\up\dn} - \ket{\dn\up}}{\sqrt2} $$

What I'm not too sure about is why the singlet state cannot be $\ket{0,0}=(\ket{↑↓} + \ket{↓↑})/\sqrt2$ while one of the triplet states can then be $(\ket{↑↓} - \ket{↓↑})/\sqrt2$. I know they must be orthogonal, but why are they defined the way they are?

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  • $\begingroup$ The triplet states are degenerate, while the singlet state has a different energy. $\endgroup$ Mar 3, 2018 at 22:03
  • $\begingroup$ What does this have to do with symmetry? $\endgroup$
    – Matt Hill
    Mar 3, 2018 at 22:05
  • $\begingroup$ Anti-symmetry is the only way to get a total spin of 0. The singlet and triplet are classified according to total spin. $\endgroup$ Mar 3, 2018 at 22:16
  • $\begingroup$ No that is not right. $\endgroup$
    – Matt Hill
    Mar 3, 2018 at 23:07
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    $\begingroup$ The singlet, i.e. the antisymmetric combination, is killed by all angular momentum operator and is thus an eigenstate of $L^2$ with eigenvalue $0$. The action of the angular momentum raising and lowering operators on the triplet will give the $L=1, M=\pm 1$. $\endgroup$ Mar 4, 2018 at 0:16

5 Answers 5

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Let's temporarily forget that the two $m=0$ states exist, and consider just the two completely aligned triplet states, $\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $ $\ket{\up\up}$ and $\ket{\dn\dn}$. There's not any physical difference between these: you can "transform" your state from one to the other by changing your coordinate system, or by standing on your head. So any physical observable between them must also be the same.

Either of the single-particle states are eigenstates of the spin operator on the $z$-axis, $$\sigma_z = \frac\hbar 2\left(\begin{array}{cc}1&\\&-1\end{array}\right),$$ and "standing on your head," or reversing the $z$-axis, is just the same as disagreeing about the sign of this operator.

But let's suppose that, on your way to reversing the $z$-axis, you get interrupted midway. Now I have a system which I think has two spins along the $z$-axis, but you are lying on your side and think that my spins are aligned along the $x$-axis. The $x$-axis spin operator is usually $$\sigma_x = \frac\hbar 2\left(\begin{array}{cc}&1\\1\end{array}\right).$$

Where I see my single-particle spins are the eigenstates of $\sigma_z$, $$\ket\up = {1\choose0} \quad\text{and}\quad \ket\dn = {0\choose1},$$ you see those single-particle states as eigenstates of $\sigma_x$, \begin{align} \ket\rt &= \frac1{\sqrt2}{1\choose1} = \frac{\ket\up + \ket\dn}{\sqrt2} \\ \ket\lf &= \frac1{\sqrt2}{1\choose-1} \end{align}

If you are a $z$-axis chauvinist and insist on analyzing my carefully prepared $\ket{\rt\rt}$ state in your $\up\dn$ basis, you'll find this mess:

\begin{align} \ket{\rt\rt} = \ket\rt \otimes \ket\rt &= \frac{\ket\up + \ket\dn}{\sqrt2} \otimes \frac{\ket\up + \ket\dn}{\sqrt2} \\ &= \frac{\ket{\up\up}}2 + \frac{\ket{\up\dn} + \ket{\dn\up}}2 + \frac{\ket{\dn\dn}}2 \end{align}

This state, which has a clearly defined $m=1$ in my coordinate system, does not have a well-defined $m$ in your coordinate system: by turning your head and disagreeing about which way is up, you've introduced both $\ket{\up\up}$ and $\ket{\dn\dn}$ into your model. You've also introduced the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$.

And this is where the symmetry argument comes in. The triplet and singlet states are distinguishable because they have different energies. If you propose that the symmetric combination $\ket{\up\dn} + \ket{\dn\up}$ is the singlet state, then you and I will predict different energies for the system based only on how we have chosen to tilt our heads. Any model that says the energy of a system should depend on how I tilt my head when I look at it is wrong. So the $m=0$ projection of the triplet state must be symmetric, in order to have the same symmetry under exchange as the $m=\pm1$ projections.

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    $\begingroup$ Slight correction to your statement "So any physical observable between them must also be the same": that's only true of physical observables (like energy) that are invariant under the operation of standing on your head. Physical observables like $S^z_\text{tot}$ can change. $\endgroup$
    – tparker
    Mar 3, 2018 at 23:21
  • $\begingroup$ All three triplet states are "completely aligned". $\endgroup$
    – my2cts
    Feb 21, 2020 at 21:20
  • $\begingroup$ @my2cts That's more or less the point I was making, yes. Take a state where the alignment is obvious, examine it in a messier representation, and interpret the messy representation based on what we know from the neater one. $\endgroup$
    – rob
    Feb 22, 2020 at 12:57
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There are at least 2 approaches. One is just show that it is symmetric by applying the lower operator for total spin to the maximal $S_z$ state which should satisfy ($\hbar=1$):

$$ S_-|1,1\rangle=\sqrt 2 |1, 0\rangle$$

so

$$|1, 0\rangle =\frac{1}{\sqrt 2}S_-|1,1\rangle=\frac{1}{\sqrt 2}(S_{1-}+S_{2-})\uparrow_1\uparrow_2$$

$$\frac{1}{\sqrt 2}[(S_{1-}\uparrow_1)\uparrow_2+\uparrow_1(S_{2-}\uparrow_2)] $$

$$ =\frac 1 {\sqrt 2}(\uparrow_1\downarrow_2 + \downarrow_1\uparrow_2)$$

That gives a nice demonstration of how to work with the ladder operators, but there is a much deeper reason it must be symmetric.

To find the rotationally invariant subspaces of a tensor product of $N$ states with dimension $d$ you do the following (this just a sketch of the procedure):

Find $N$. It is $N=2$, now we partition $2$ in every way possible:

$$ 2 = 2 $$

and

$$ 2 = 1 + 1 $$

For each of these partitions we draw the Young diagrams and connect those with irreducible representations of the permutation group on $N=2$ letters. This is called the Robinson-Schensted Correspondence.

Take the $2=2$ diagram an make a normal Young Tableau and then compute the Young symmetrizer. In this case, you get the purely symmetric operator: $S=(1 + e_{2,1})/\sqrt 2$

For $2=1+1$, you do the same an get the antisymmetric operator: $A=(1 - e_{2,1})/\sqrt 2$.

Schur-Weyl Duality tells us that applying these to the indices (here, particle labels), will tell us the rotationally invariant subspaces of this tensor product space; moreover, the remarkable Hook-Length Formula tells us the dimensions of the subspace, and the result for $d=2$ is the symmetric one has ${\bf 3}$ dimensions, and antisymmetric is ${\bf 1}$.

This is written:

$$ {\bf 2} \otimes {\bf 2} = {\bf 3}_S + {\bf 1}_A $$

So it simply has to be that all the states in the triplet have the same exchange symmetry.

Note that can add another spin $\frac 1 2$, and the whole procedure will show you that:

$$ {\bf 2} \otimes {\bf 2} \otimes {\bf 2}= {\bf 4}_S + {\bf 2}_M + {\bf 2}_M$$

which means the four $S=\frac 3 2$ states are symmetric and there are two doublet $S=\frac 1 2$ states with mixed symmetry, corresponding to the partitions:

$$ 3 = 3$$

and

$$ 3 = 2 + 1 $$

Note that the hook length formula for:

$$ 3 = 1 + 1 + 1 $$

yields a subspace of dimension zero: there is no antisymmetric combination of 3 spins.

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If $\:\mathsf{H}_{\boldsymbol{\alpha}},\mathsf{H}_{\boldsymbol{\beta}}\: $ are the 2-dimensional Hilbert spaces of two particles $\:\boldsymbol{\alpha},\boldsymbol{\beta}\:$ with spins $\:1/2\:$ then the composite system lives in the product 4-dimensional Hilbert space \begin{equation} \mathsf{H}_{\boldsymbol{f}}\equiv \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}} \tag{01} \end{equation} which is the direct sum of two invariant orthogonal subspaces : the 1-dimensional subspace $\:\mathsf{H}_{\boldsymbol{1}}\:$ of angular momentum $\;j=0\;$ (the antisymmetric singlet) and the 3-dimensional subspace $\:\mathsf{H}_{\boldsymbol{2}}\:$ of angular momentum $\;j=1\;$ (the symmetric triplet): \begin{equation} \mathsf{H}_{\boldsymbol{\alpha}}\boldsymbol{\otimes}\mathsf{H}_{\boldsymbol{\beta}}=\mathsf{H}_{\boldsymbol{1}}\boldsymbol{\oplus}\mathsf{H}_{\boldsymbol{2}} \tag{02} \end{equation} expressed also as \begin{equation} \boldsymbol{2}\boldsymbol{\otimes}\boldsymbol{2}=\boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{3} \tag{03} \end{equation} Invariance means that if we apply the same special unitary transformation \begin{equation} U_{\boldsymbol{\alpha}}=U=U_{\boldsymbol{\beta}} \in SU(2) \tag{04} \end{equation} in each one of the spaces $\:\mathsf{H}_{\boldsymbol{\alpha}},\mathsf{H}_{\boldsymbol{\beta}}\:$ then the subspaces $\:\mathsf{H}_{\boldsymbol{1}},\mathsf{H}_{\boldsymbol{2}}\:$ are invariant under the product special unitary transformation \begin{equation} U_{\boldsymbol{\alpha}}\boldsymbol{\otimes}U_{\boldsymbol{\beta}}=U^{\boldsymbol{\otimes}\boldsymbol{2}}\in SU(4) \tag{05} \end{equation} Note that application of the transformation (04) corresponds to a rotation in the 3-dimensional real space $\:\mathbb{R}^{3}$ wherein the two particles coexist.

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According to your last question, the singlet state

$\newcommand\ket[1]{\left|{#1}\right>} \newcommand\up\uparrow \newcommand\dn\downarrow \newcommand\lf\leftarrow \newcommand\rt\rightarrow $ $ \ket{0,0} = \frac{\ket{\up \dn} + \ket{\dn\up}}{\sqrt2} $

cannot be valid, while one of the triplet states (assume it the $\ket{1,0}$) could be written as

$ \ket{1,0} = \frac{\ket{\up\dn} + \ket{\dn\up}}{\sqrt2} $

as it can be shown below.

Defining the spin-exchange operator as

$P\mid \chi_{\uparrow\downarrow} \rangle = \mid\chi_{\downarrow\uparrow} \rangle , P\mid \chi_{\downarrow\uparrow} \rangle =\mid \chi_{\uparrow\downarrow} \rangle $

which implies

$P\mid \chi_{\text{sym.}} \rangle = \mid \chi_{\text{sym.}} \rangle , P \mid \chi_{\text{asym.}} \rangle = -\mid \chi_{\text{asym.}} \rangle$

The above singlet state becomes,

$ P \ket{0,0} = \frac{\ket{\dn\up} + \ket{\up \dn} }{\sqrt2} \neq -\ket{0,0}. $

whereas, for the second triplet state we write,

$ P \ket{1,0} = \frac{ \ket{\dn\up} + \ket{\up\dn} }{\sqrt2} = \ket{1,0}. $

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The singlet and the triplet states are all antisymmetric. For the triplet state the spatial part is symmetric and the spin part is antisymmetric. For the singlet state it is the other way around. In this case the spatial orbitals can overlap or even coincide. In the triplet m=0 state the spins are aligned but oriented perpendicular to the quantisation axis.

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  • $\begingroup$ I thought the question here was only about the symmetry of the spin part of the wavefunction. You might also consider the deuteron, a spin triplet whose spatial wavefunction is mostly $S$- and a little $D$-wave, both symmetric under exchange. To restore antisymmetry under exchange to the deuteron, we have to discover strong isospin. $\endgroup$
    – rob
    Feb 22, 2020 at 13:18

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