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Bounded self-adjoint operators on separable infinite-dimensional complex Hilbert spaces are unitarily equivalent modulo compact operators exactly when they have the same essential spectrum in the Calkin algebra. The Weyl-von Neumann theorem diagonalizes each modulo a compact error. Their diagonal cluster sets are the essential spectra; cluster-set matching of bounded sequences pairs the two diagonal lists with differences tending to zero, giving a compact diagonal difference. The separability assumption is essential: projections with countable and uncountable range dimensions can have the same essential spectrum but cannot differ by a compact operator after unitary conjugation.

The Weyl-von Neumann theorem states that on a separable Hilbert space a self-adjoint operator is a diagonal self-adjoint operator plus a compact self-adjoint operator. For a bounded operator $T$, the perturbation can in fact be chosen Hilbert-Schmidt with arbitrarily small Hilbert-Schmidt norm. The separable-space formulation is also recorded in Kadison's Corollary B.

Here is a constructive proof. The bounded self-adjoint spectral theorem decomposes a separable $H$ into countably many cyclic reducing spaces, on each of which $T$ is multiplication by $x$ in $L^2(\mu )$ for a finite measure supported in a bounded real interval. First work on one such space. Partition the interval into $N$ intervals of length at most $\delta$, and repeatedly bisect them. Let $E_j$ be the finite-dimensional space of functions constant on the level-$j$ intervals. These spaces are nested and their union is dense, since dyadic step functions approximate continuous functions and continuous functions are dense in $L^2(\mu )$.

Choose normalized interval indicators as a basis of $E_0$, and choose orthonormal bases of $E_j\ominus E_{j-1}$ within each parent interval. This is possible because the refinement splits into orthogonal parent blocks. Each new basis vector at level $j\ge 1$ is supported in an interval of length at most $\delta 2^{-(j-1)}$, and there are at most $N{2}^{j-1}$ such vectors. For every basis vector $e$ choose a real ${\lambda }_e$ in its supporting interval and define the diagonal operator $De={\lambda }_{e}e$. ThensoOne may take $N=O({\delta }^{-1})$, so this bound tends to zero as $\delta \to 0$. The error is Hilbert-Schmidt, hence compact; it is self-adjoint because both $T$ and $D$ are.

On the countably many cyclic summands choose error bounds ${\epsilon }_j$ with ${\sum }_j{\epsilon }_j^2<{\epsilon }^2$. Combining their orthonormal bases and diagonal operators gives $T=D+K$ with $\Vert K{\Vert }_{\mathrm{HS}}<\epsilon$, proving the assertion. For an unbounded self-adjoint operator, first split by spectral intervals $[j,j+1)$, apply the bounded construction on each, and choose summable error bounds. The resulting total $K$ is bounded compact, and $D=T-K$ has the same domain as $T$; the diagonal domain condition follows from the boundedness of $K$.

Separability matters. Take an uncountable orthogonal sum of copies of multiplication by $x$ on $L^2[0,1]$. The range closure of a compact operator is separable, and a separable subspace of this direct sum is supported in only countably many summands: choose a countable dense set and unite its countable supports. A self-adjoint compact error therefore vanishes on every remaining summand, each of which reduces the proposed diagonal operator $D=T-K$. The restriction of a diagonal self-adjoint operator to a reducing subspace has pure point spectrum, whereas multiplication by $x$ on $L^2[0,1]$ has no eigenvectors. This is a contradiction, qualifying the theorem if the unspoken separability convention is dropped.

Work with bounded self-adjoint operators on separable infinite-dimensional Hilbert spaces, the setting in which the classification holds. If $UT{U}^{\ast }-S$ is compact for a unitary $U$, their images in the Calkin algebra have identical spectra, by compact perturbation invariance and unitary conjugation. Thus their essential spectra coincide.

Conversely apply the Weyl-von Neumann theorem to write $T=D_a+K_a$ and $S=D_b+K_b$ in orthonormal bases, with bounded real diagonal sequences and compact errors. The essential spectrum of a diagonal operator is exactly its sequence cluster set. Indeed outside that cluster set all but finitely many entries stay a positive distance from $\lambda$, so the diagonal has an inverse on a finite-codimensional subspace and is Fredholm. At a cluster point, infinitely many distinct basis vectors have $(D-\lambda )e_j\to 0$; these are weakly null unit vectors, contradicting any Fredholm parametrix. This also handles a value with infinite multiplicity.

The two diagonal sequences therefore have the same cluster set. Part (c) provides permutations with $a_{m_k}-b_{n_k}\to 0$. Send the $m_k$th basis vector for $T$ to the $n_k$th basis vector for $S$. Under this unitary, the diagonal difference has entries tending to zero and is compact, by finite-rank truncation. Adding the two conjugated compact errors gives $\boxed{UT{U}^{\ast }-S\text{ compact}}$.

Without separability, equality of essential spectra need not suffice, even on the same Hilbert space. Let $H$ have uncountable dimension and let $T,S$ be projections with infinite-dimensional kernels, but with range dimensions respectively countable and uncountable. Both essential spectra are $\{0,1\}$. After any unitary conjugation the range of $T$ is still separable. If its difference from $S$ were compact, the range of $S$ would lie in the sum of that separable range and the separable closure of the compact error's range, which is impossible. Finite-dimensional spaces also need a separate dimension convention, since essential spectra are then empty. These are genuine qualifications to a statement made without the standard separable infinite-dimensional setting.