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Given a particle with spin $1$, let its states be $|1,1\rangle$, $|1,0\rangle$, $|1,-1\rangle$.

  • Question:

Does there exist a rotation $D(R)$ such that $D(R)|1,0\rangle=|1,1\rangle$?

  • My solution:

$D(R)$ is given by $\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$ for some unit vector $\hat n$ and angle $\phi$.

Also, we see that $\langle 1,1| \hat{n}\cdot\vec{S} |1,1\rangle=n_z\hbar $ because:

$\langle s',m'|S_{z}|s,m\rangle=m\hbar\delta_{s',s}\delta_{m',m}$

$ \langle s',m'|S_{\pm}|s,m\rangle =\sqrt{(s\mp m)(s\pm m+1)}\hbar\delta_{s',s}\delta_{m',m\pm1}$

$S_{x}=\frac{1}{2}(S_{+}+S_{-})$

$S_{y}=\frac{1}{2i}(S_{+}-S_{-})$ .

Now, let's assume there is such a $D(R)$. So there is also a $D'(R)$ fulfilling:

$D'(R)|1,1\rangle=|1,0\rangle$

Say $D'(R)=\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)$

So we have: $\exp\left(-i\frac{\phi\hat{n}\cdot\vec{S}}{\hbar}\right)|1,1\rangle=|1,0\rangle$.

Multiplying both sides by the bra $\langle 1,1|$, we get:

$\exp\left(-i\phi n_z\right)=0$

Which can never happen. $\blacksquare$

  • Does my proof seem all-right? Any comments on my way of reasoning? Any ideas of how to do it in other ways? What is the general approach to problems of this kind?
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    $\begingroup$ I didn't do any calculations but I'm pretty sure that your proof is wrong and that there is such a rotation. To point out one error in the proof: the exponential $\exp \left(-i \frac{\phi \hat n \cdot \vec S}{\hbar}\right)$ contains all terms like $\left(\hat n \cdot \vec S\right)^n$ and it is not true that $\langle 1,1|\left(\hat n \cdot \vec S\right)^n|1,1\rangle = (n_z \hbar)^n$ $\endgroup$ Dec 12, 2016 at 14:06
  • $\begingroup$ @Herr_Mitesch I was wondering about that, but I saw countless times this change from an exponent of an operator to an exponent of it's corresponding eigenvalue. Is the problem that this is a sum of operators? $\endgroup$
    – Whyka
    Dec 12, 2016 at 14:23
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    $\begingroup$ This change only works if the state you apply the function of the operator on is an eigenstate of the operator: assume $A|a\rangle = a |a\rangle$ (where $A$ is an operator, $|a\rangle$ its eigenstate to the eigenvalue $a$.) Then, assuming $f$ can be expanded in a Taylor series $f(A) = \sum_{j=0}^\infty c_j A^j$ we have $$f(A)|a\rangle = \sum_{j=0}^\infty c_j A^j |a\rangle = \sum_{j=0}^\infty c_j a^j |a\rangle = f(a) |a\rangle.$$ $\endgroup$ Dec 12, 2016 at 14:28
  • $\begingroup$ In your example consider say $\hat n = \vec e_x$, and calculate $$\langle 1,1|S_x |1,1\rangle = 0 \qquad \text{(this one's correct)}\\ \langle 1,1|S_x^2|1,1\rangle = \hbar/2 \qquad \text{(it stops working here)}$$ $\endgroup$ Dec 12, 2016 at 14:30
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    $\begingroup$ @Herr_Mitesch Oh, I see. Although $|1,1\rangle$ is an eigenstate of $S_z$, it is NOT an eigenstate of $\hat n \cdot \vec S$. That's a good point. I see where I was wrong now. But, how is $D(R)$ found?.. $\endgroup$
    – Whyka
    Dec 12, 2016 at 14:35

3 Answers 3

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The answer is negative.

Suppose there is such rotation $R$ and let us denote its inverse by $R'$. In this case $$S_z D(R') |1,1\rangle = S_z |1,0\rangle=0\:,$$ As a consequence, $$D(R')^\dagger S_z D(R') |1,1\rangle =0$$ that is $$\vec{n} \cdot \vec{S}|1,1 \rangle =0\tag{1}$$ where $\vec{n}= R' \vec{e}_z$.

Using $S_z|1,1\rangle=|1,1\rangle$, (1) implies $$n_x S_x|1,1 \rangle + n_y S_y |1,1 \rangle = - n_z |1,1 \rangle \tag{2}\:.$$ Appliyng $S_z$ to both sides, we also have $$n_x S_zS_x|1,1 \rangle + n_y S_zS_y |1,1 \rangle =- n_z |1,1 \rangle\tag{3}\:,$$ which, together with $$n_x S_xS_z|1,1 \rangle + n_y S_yS_z |1,1 \rangle =- n_z |1,1 \rangle \tag{4}\:,$$ implies $$n_x [S_z,S_x]|1,1 \rangle + n_y [S_z,S_y] |1,1 \rangle =0 \tag{5}\:.$$ Using the commutation relations of the $S_k$ operators we have $$n_x S_y|1,1 \rangle - n_y S_x |1,1 \rangle =0 \tag{6}\:.$$ Now focus on (2) and (6). Computing the determinant $d$ of the linear system made of this pair of equations (the unknowns being the two vectors $S_x |1,0 \rangle $ and $S_y |1,1 \rangle$) we find $d= n_x^2 +n_y^2$. If $d=0$ that is $n_x= n_y=0$, we have that $\vec{n}= \vec{e}_z$ and thus $R$ is a rotation around $\vec{e}_z$. This is not possible because $$D(R_z(\theta))|1,0\rangle = e^0 |1,0\rangle = |1,0\rangle \neq |1,1 \rangle\:.$$ If $d\neq 0$, the unique solution of the afore-mentioned system leads to either $$S_x |1,1 \rangle =-\frac{n_xn_z}{1-n_z^2}|1,1 \rangle$$ or $$S_y |1,1 \rangle =-\frac{n_yn_z}{1-n_z^2} |1,1 \rangle$$ where $n^2_z \neq 1$ otherwise $d=0$, which are false by direct inspection, using the explicit expressions of the matrices $S_x$ and $S_y$ because $|1,1\rangle$ is not an eigenvector of $S_x$ and $S_y$.

We conclude that there are no chances to find the wanted $R$.

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  • $\begingroup$ Thank you for the detailed answer, I appreciate it. But how did you go from the first equation to $D(R')^\dagger S_z D(R') |1,0\rangle =0$ ? It seems to me like it should be $|1,1\rangle$ $\endgroup$
    – Whyka
    Dec 12, 2016 at 16:11
  • $\begingroup$ Sorry I wrote $|1,0\rangle$ in place of $|1,1\rangle$, I am correcting... $\endgroup$ Dec 12, 2016 at 16:15
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    $\begingroup$ I have corrected my answer now... $\endgroup$ Dec 12, 2016 at 16:52
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I don't know if this follows from some general theorem, but I've found a simple proof.

Let me denote with $\vert 0,\pm1\rangle$ the base of the $j=1$ representation, with $\vert \uparrow \downarrow\rangle$ that of the $j=\frac{1}{2}$ representation. If there exists such an $R$: $$D^{(1)}(R)\vert 0\rangle =\vert 1\rangle,$$ then we must have, in the standard phase convention: $$D^{(1)}(R)\dfrac{\vert \uparrow \rangle\vert \downarrow \rangle +\vert \downarrow \rangle\vert \uparrow \rangle}{\sqrt 2}=\dfrac{D^{(\frac{1}{2})}(R)\vert \uparrow \rangle \otimes D^{(\frac{1}{2})}(R) \vert \downarrow \rangle +D^{(\frac{1}{2})}(R) \vert \downarrow \rangle \otimes D^{(\frac{1}{2})}(R) \vert \uparrow \rangle}{\sqrt 2}=\vert \uparrow \rangle\vert \uparrow \rangle. $$ Now, if we put $$D^{(\frac{1}{2})}(R)\vert \uparrow\rangle = \alpha \vert \uparrow\rangle +\beta \vert \downarrow \rangle \\D^{(\frac{1}{2})}(R)\vert \downarrow\rangle = \gamma \vert \uparrow\rangle +\delta \vert \downarrow \rangle, \\$$ we obtain the conditions: $$\alpha \gamma \neq 0\\ \alpha \delta+\gamma \beta =0\\ \beta \delta =0.$$ Taken together, these three equations imply $\beta = \delta =0$. So $D^{(\frac{1}{2})}(R)\vert \uparrow\rangle$ and $D^{(\frac{1}{2})}(R)\vert \downarrow\rangle$ are proportional to each other, which is impossible since $D(R)$ is invertible.

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    $\begingroup$ Here I'm using the fact that the $j=1$ representation is a subrepresentation of the $\frac{1}{2}\otimes \frac{1}{2}$ representation. $\endgroup$
    – pppqqq
    Dec 12, 2016 at 17:13
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Consider the inverse rotation R'. Compute the expectation value $1 = \langle U^\dagger(R') L_z U(R') \rangle_0 = \langle R_{3,i} \cdot L_i \rangle_0 = 0$ The first equality is because the rotation sends you to the "highest" eigenvector, the second one is because L transform as a vector under rotation (it is a result of representation theory), the last one is because all of the three components of L have zero mean.

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