[proofplan] We first put the complex Specht module in a unitary setting by averaging an arbitrary positive definite Hermitian form over $S_n$. The commuting self-adjoint Jucys-Murphy operators are then diagonalized, and the standard Young-Jucys-Murphy spectral theorem identifies their one-dimensional joint eigenspaces with standard tableaux and contents. The group-algebra relations between $s_i$ and the neighboring Jucys-Murphy elements force $s_iv_T$ to have support only on $v_T$ and, when the swapped filling is standard, $v_{s_iT}$. The diagonal coefficient is computed from the relation $s_iX_{i+1}=X_is_i+1$, and unitarity determines the absolute value of the off-diagonal coefficient, leaving only a unit complex phase. [/proofplan] [step:Average the Hermitian form and choose the Jucys-Murphy eigenbasis] Let $(\cdot,\cdot)_0:S^\lambda\times S^\lambda\to\mathbb C$ be any positive definite Hermitian form. Define a Hermitian form $(\cdot,\cdot):S^\lambda\times S^\lambda\to\mathbb C$ by \begin{align*} (u,v):=\frac{1}{|S_n|}\sum_{\sigma\in S_n}(\sigma u,\sigma v)_0. \end{align*} For $u\ne 0$, the summand with $\sigma=1_{S_n}$ is $(u,u)_0>0$ and every summand $(\sigma u,\sigma u)_0$ is non-negative, so $(\cdot,\cdot)$ is positive definite. If $\tau\in S_n$, then right multiplication by $\tau$ is a bijection $S_n\to S_n$, and therefore \begin{align*} (\tau u,\tau v)=\frac{1}{|S_n|}\sum_{\sigma\in S_n}(\sigma\tau u,\sigma\tau v)_0=(u,v). \end{align*} Thus every group element acts unitarily on $S^\lambda$. For each transposition $(j\ k)$, the associated operator is self-adjoint because it is unitary and equal to its inverse. Hence each $X_k$ is self-adjoint, being a real linear combination of self-adjoint transposition operators. By [[Commutativity of the Jucys-Murphy Elements](/theorems/8446)][citetheorem:8446], the operators $X_1,\dots,X_n$ commute. The finite-dimensional spectral theorem for commuting [self-adjoint operators](/page/Self-Adjoint%20Operators) therefore applies to the Hermitian space $S^\lambda$ and gives an orthonormal simultaneous eigenbasis. We now use the standard Young-Jucys-Murphy spectral theorem for complex Specht modules in the following exact form: on $S^\lambda$, the joint eigenspaces of $X_1,\dots,X_n$ are precisely the one-dimensional spaces \begin{align*} E_T:=\{u\in S^\lambda:X_ku=c_k(T)u\text{ for every }1\le k\le n\} \end{align*} indexed by $T\in\operatorname{SYT}(\lambda)$, and these spaces exhaust $S^\lambda$. This is the content of [[Content Eigenvalues of the Jucys-Murphy Elements](/theorems/8447)][citetheorem:8447]. Its hypotheses are satisfied here because the module is the complex Specht module $S^\lambda$ of shape $\lambda$, the elements $X_k$ are the Jucys-Murphy elements with convention $X_k=\sum_{1\le j<k}(j\ k)$, and the content convention is $c_k(T)=\operatorname{col}_T(k)-\operatorname{row}_T(k)$. Choose a unit vector $v_T\in E_T$ for every $T\in\operatorname{SYT}(\lambda)$. The spaces $E_T$ are mutually orthogonal because the $X_k$ are commuting self-adjoint operators, and they exhaust $S^\lambda$; hence $(v_T)_{T\in\operatorname{SYT}(\lambda)}$ is an [orthonormal basis](/page/Orthonormal%20Basis) satisfying \begin{align*} X_kv_T=c_k(T)v_T. \end{align*} [/step] [step:Use the adjacent Jucys-Murphy relation to restrict the support of $s_iv_T$] Fix $i$ with $1\le i<n$. We first record the group-algebra relations needed below. If $k\ne i,i+1$, then conjugation by $s_i$ permutes the set of transpositions appearing in $X_k$, so \begin{align*} s_iX_k=X_ks_i. \end{align*} For $k=i+1$, separate the final summand in $X_{i+1}$: \begin{align*} X_{i+1}=\sum_{1\le j<i}(j\ i+1)+s_i. \end{align*} For each $j<i$, the identity $s_i(j\ i+1)=(j\ i)s_i$ holds in $S_n$. Therefore \begin{align*} s_iX_{i+1}=\sum_{1\le j<i}(j\ i)s_i+1=X_is_i+1. \end{align*} Let $T\in\operatorname{SYT}(\lambda)$, and write \begin{align*} s_iv_T=\sum_{U\in\operatorname{SYT}(\lambda)}a_Uv_U \end{align*} with coefficients $a_U\in\mathbb C$. If $k\ne i,i+1$, then \begin{align*} X_k(s_iv_T)=s_iX_kv_T=c_k(T)s_iv_T. \end{align*} Comparing the coefficient of $v_U$ gives: if $a_U\ne 0$, then $c_k(U)=c_k(T)$ for every $k\ne i,i+1$. Applying $s_iX_{i+1}=X_is_i+1$ to $v_T$ gives \begin{align*} c_{i+1}(T)s_iv_T=X_i(s_iv_T)+v_T. \end{align*} For $U\ne T$, comparison of the $v_U$ coefficient yields \begin{align*} c_{i+1}(T)a_U=c_i(U)a_U. \end{align*} Thus, if $a_U\ne 0$ and $U\ne T$, then $c_i(U)=c_{i+1}(T)$. Since $U$ and $T$ have the same shape, the multiset of contents of all boxes of $\lambda$ is the same for $U$ and $T$. Together with $c_k(U)=c_k(T)$ for $k\ne i,i+1$, this forces \begin{align*} c_{i+1}(U)=c_i(T). \end{align*} Hence every possible $U\ne T$ with $a_U\ne 0$ has the content sequence obtained from that of $T$ by interchanging the $i$th and $(i+1)$st entries. It remains to identify this possible tableau. The filling $s_iT$ is the unique filling of shape $\lambda$ whose entrywise contents are obtained from those of $T$ by interchanging the $i$th and $(i+1)$st entries, because entries other than $i$ and $i+1$ remain in their boxes and the two relevant entries exchange boxes. If $s_iT$ is standard, then [[Content Separation of Standard Young Tableaux](/theorems/8445)][citetheorem:8445] implies that the only standard tableau with this content sequence is $s_iT$. If $s_iT$ is not standard, then no standard tableau has this swapped content sequence: otherwise content separation would force that standard tableau to be the filling $s_iT$, contradicting nonstandardness. Consequently \begin{align*} s_iv_T\in\operatorname{span}_{\mathbb C}\{v_T\} \end{align*} when $s_iT$ is not standard, and \begin{align*} s_iv_T\in\operatorname{span}_{\mathbb C}\{v_T,v_{s_iT}\} \end{align*} when $s_iT$ is standard. [guided] Fix $i$ with $1\le i<n$. The goal of this step is to show that applying $s_i$ to the eigenvector $v_T$ can only affect the two labels $i$ and $i+1$. We make that precise using the Jucys-Murphy operators, because their joint eigenvalues record the contents of all labels. First consider $k\ne i,i+1$. Conjugation by $s_i$ only interchanges the labels $i$ and $i+1$. When $k\ne i,i+1$, this conjugation sends the set of transpositions appearing in $X_k$ to itself. Equivalently, in the group algebra, \begin{align*} s_iX_k=X_ks_i. \end{align*} For the exceptional neighboring operator $X_{i+1}$, we compute directly. Since \begin{align*} X_{i+1}=\sum_{1\le j<i}(j\ i+1)+s_i, \end{align*} and since $s_i(j\ i+1)=(j\ i)s_i$ for every $j<i$, we obtain \begin{align*} s_iX_{i+1}=\sum_{1\le j<i}(j\ i)s_i+s_i^2. \end{align*} Using $s_i^2=1$ and $X_i=\sum_{1\le j<i}(j\ i)$, this becomes \begin{align*} s_iX_{i+1}=X_is_i+1. \end{align*} Now expand $s_iv_T$ in the orthonormal Jucys-Murphy eigenbasis. There are unique coefficients $a_U\in\mathbb C$ such that \begin{align*} s_iv_T=\sum_{U\in\operatorname{SYT}(\lambda)}a_Uv_U. \end{align*} For $k\ne i,i+1$, the commutation relation gives \begin{align*} X_k(s_iv_T)=s_iX_kv_T=c_k(T)s_iv_T. \end{align*} On the other hand, applying $X_k$ to the expansion gives eigenvalue $c_k(U)$ on the basis vector $v_U$. Therefore any coefficient $a_U\ne 0$ must satisfy \begin{align*} c_k(U)=c_k(T) \end{align*} for every $k\ne i,i+1$. The relation involving $X_{i+1}$ gives information about one of the two remaining labels. Applying \begin{align*} s_iX_{i+1}=X_is_i+1 \end{align*} to $v_T$ gives \begin{align*} c_{i+1}(T)s_iv_T=X_i(s_iv_T)+v_T. \end{align*} If $U\ne T$, then the vector $v_T$ contributes no $v_U$ coefficient. Comparing the $v_U$ coefficient yields \begin{align*} c_{i+1}(T)a_U=c_i(U)a_U. \end{align*} Thus, whenever $U\ne T$ and $a_U\ne 0$, we have \begin{align*} c_i(U)=c_{i+1}(T). \end{align*} Because $U$ and $T$ have the same Young diagram shape, their entries fill the same boxes. Therefore the multiset of box contents is independent of the tableau filling. Since all contents except possibly those attached to labels $i$ and $i+1$ already agree, the remaining content must be \begin{align*} c_{i+1}(U)=c_i(T). \end{align*} So the content sequence of $U$ is exactly the content sequence obtained from the content sequence of $T$ by swapping the two neighboring labels $i$ and $i+1$. We now connect this content statement to the actual filling $s_iT$. The filling $s_iT$ is obtained by leaving every label other than $i$ and $i+1$ in its original box and exchanging the boxes occupied by $i$ and $i+1$. Hence $s_iT$ has exactly the swapped content sequence just described. If $s_iT$ is standard, then [Content Separation of Standard Young Tableaux][citetheorem:8445] applies: two standard tableaux of the same shape with the same content sequence must be equal. Therefore the only possible standard tableau $U\ne T$ contributing to $s_iv_T$ is $U=s_iT$. If $s_iT$ is not standard, then no standard tableau can have that swapped content sequence; if one did, content separation would identify it with the nonstandard filling $s_iT$, which is impossible. This proves the support alternatives \begin{align*} s_iv_T\in\operatorname{span}_{\mathbb C}\{v_T\} \end{align*} in the nonstandard case and \begin{align*} s_iv_T\in\operatorname{span}_{\mathbb C}\{v_T,v_{s_iT}\} \end{align*} in the standard case. [/guided] [/step] [step:Compute the diagonal coefficient from $s_iX_{i+1}=X_is_i+1$] Fix $T\in\operatorname{SYT}(\lambda)$ and define $a_T\in\mathbb C$ to be the coefficient of $v_T$ in $s_iv_T$. From \begin{align*} c_{i+1}(T)s_iv_T=X_i(s_iv_T)+v_T \end{align*} and the equality $X_iv_T=c_i(T)v_T$, comparison of the $v_T$ coefficient gives \begin{align*} c_{i+1}(T)a_T=c_i(T)a_T+1. \end{align*} Therefore \begin{align*} a_T=\frac{1}{c_{i+1}(T)-c_i(T)}=\frac{1}{r_i(T)}. \end{align*} The denominator is nonzero. Indeed, if the boxes containing $i$ and $i+1$ had the same content and were distinct, one box would lie strictly southeast of the other on the same diagonal. Any path from the northwest box to the southeast box using one-step moves right and down has at least two boxes and at least one intermediate strict increase in addition to the final strict increase. Standardness of $T$ forces entries to increase at every rightward or downward step, so the southeast entry would be at least two larger than the northwest entry. This contradicts the fact that the two entries are the consecutive integers $i$ and $i+1$. [/step] [step:Handle the case where the adjacent swap is not standard] Assume that $s_iT$ is not standard. Since $T$ is standard, the swap of consecutive labels can fail standardness only if the boxes containing $i$ and $i+1$ are adjacent in the same row or adjacent in the same column. In the same-row case, standardness forces $i$ to be immediately to the left of $i+1$, and hence \begin{align*} r_i(T)=1. \end{align*} In the same-column case, standardness forces $i$ to be immediately above $i+1$, and hence \begin{align*} r_i(T)=-1. \end{align*} By the support result, $s_iv_T\in\operatorname{span}_{\mathbb C}\{v_T\}$. The diagonal coefficient computation therefore gives \begin{align*} s_iv_T=\frac{1}{r_i(T)}v_T. \end{align*} [/step] [step:Determine the unitary two-dimensional matrix in the standard swap case] Assume that $s_iT$ is standard, and define $T':=s_iT$. Then the entries $i$ and $i+1$ exchange boxes, so \begin{align*} c_i(T')=c_{i+1}(T) \end{align*} and \begin{align*} c_{i+1}(T')=c_i(T). \end{align*} Thus \begin{align*} r_i(T')=-r_i(T). \end{align*} The support and diagonal coefficient results give complex numbers $b_T,b_{T'}\in\mathbb C$ such that \begin{align*} s_iv_T=\frac{1}{r_i(T)}v_T+b_Tv_{T'} \end{align*} and \begin{align*} s_iv_{T'}=-\frac{1}{r_i(T)}v_{T'}+b_{T'}v_T. \end{align*} The operator $s_i$ is unitary because the Hermitian form is $S_n$-invariant, and it is self-adjoint because $s_i^{-1}=s_i$. Hence \begin{align*} (s_iv_T,v_{T'})=(v_T,s_iv_{T'}). \end{align*} Using orthonormality of $v_T$ and $v_{T'}$, this gives \begin{align*} b_T=\overline{b_{T'}}. \end{align*} Unitarity applied to the unit vector $v_T$ gives \begin{align*} 1=\|s_iv_T\|^2=\left|\frac{1}{r_i(T)}\right|^2+|b_T|^2. \end{align*} Since $s_iT$ is standard, the boxes containing $i$ and $i+1$ are neither adjacent in the same row nor adjacent in the same column. They also cannot have the same content by the nonzero-denominator argument above. Hence the integer $r_i(T)$ satisfies $|r_i(T)|\ge 2$. Therefore \begin{align*} |b_T|=\sqrt{1-r_i(T)^{-2}}, \end{align*} where the square root is the positive real square root. [/step] [step:Record the remaining phase and conclude] In the standard swap case, the positive number $\sqrt{1-r_i(T)^{-2}}$ is nonzero because $|r_i(T)|\ge 2$. Define \begin{align*} \gamma_i(T):=b_T\left(\sqrt{1-r_i(T)^{-2}}\right)^{-1}. \end{align*} Then $|\gamma_i(T)|=1$, and the formula for $s_iv_T$ becomes \begin{align*} s_iv_T=\frac{1}{r_i(T)}v_T+\gamma_i(T)\sqrt{1-r_i(T)^{-2}}\,v_{s_iT}. \end{align*} Together with the nonstandard case and the construction of the orthonormal Jucys-Murphy eigenbasis, this proves the theorem. [/step]