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Nielsen's pure-state conversion theorem concerns exact deterministic transformations of bipartite pure states by local operations and classical communication. Let $\lambda (\psi )$ be the decreasing vector of squared Schmidt coefficients, equivalently the eigenvalues of either party's reduced density matrix, padded with zeros when necessary. ThenThe majorization relation means ${\sum }_{j=1}^k{\lambda }_j(\psi )\le {\sum }_{j=1}^k{\lambda }_j(\chi )$ for every $k$, with equality for the total sum. The direction allows a maximally entangled state to become a product state, not the reverse. Local-unitary invariance of reduced-state spectra alone would require equality of all the weights; measurements and communicated outcomes permit this larger class of conversions. A comparison of entanglement entropies alone is not a sufficient test in higher dimension.^[ref]

For the supplied states, the weight vectors and cumulative sums areAt $k=2$, $0.8>0.75$, violating the required inequality. The desired conversion is impossible with certainty using LOCC alone. The reverse conversion also fails, already at $k=1$, so these states are incomparable under this deterministic ordering.

Taking a tensor product with the catalyst multiplies every original weight by each catalyst weight. Regrouping the two local systems on Alice's side and the two on Bob's side leaves a Schmidt decomposition across the Alice–Bob partition. After sorting, the resulting weight vectors areThe actual nonnegative Schmidt amplitudes are $\sqrt{x_j}$ and $\sqrt{y_j}$; the displayed numbers are their squares. The cumulative sums areEvery component obeys $X_k\le Y_k$, so $x\prec y$. HenceThe final catalyst is exactly its initial state and is uncorrelated with the desired output. It can therefore be returned and reused. This is entanglement catalysis; it is stronger than merely retaining the same catalyst marginal while leaving correlations with the output.^[ref]

For the maximally entangled two-level auxiliary state, the sorted input weights are $(0.2,0.2,0.2,0.2,0.05,0.05,0.05,0.05)$, while the output weights are $(0.25,0.25,0.125,0.125,0.125,0.125,0,0)$. Their first four sums are $0.8$ and $0.75$. This maximally entangled auxiliary state cannot catalyze the conversion while being returned unchanged.

Nevertheless, $(0.5,0.5)\prec (0.6,0.4)$, so it can first be converted deterministically into the nonmaximal catalyst. The desired conversion can then proceed, after which Alice and Bob can return $|\varphi ⟩$, rather than the borrowed maximally entangled state. Turning that returned pair alone back into a maximally entangled pair would increase its entanglement entropy and is forbidden by LOCC. No joint restoration procedure can both keep the exact desired $|{\psi }_2⟩$ and return the original maximally entangled auxiliary state deterministically: its composition with the preceding steps would contradict the failed four-weight inequality above.

Use the thin singular value decomposition $a=Ud{V}^{†}$, with $U^{†}U=I_N$, $V$ unitary and $d=\mathrm{diag}(d_k)$ nonnegative. Expanding its entries givesThe columns of $U$ and of $V^{\ast }$ are orthonormal, establishing the Schmidt decomposition. Bob's basis uses the complex-conjugated columns of $V$. Omitting that conjugation would not reproduce the coefficient matrix. The squared Schmidt coefficients are the eigenvalues of $a^{†}a$; equivalently $d^2=V^{†}{a}^{†}aV$ and $d=U^{†}aV$. The two reduced density matrices are $a{a}^{†}$ and $(a^{†}a{)}^T$, which have the same nonzero eigenvalues.

The printed coefficient $0.41231$ slightly rounds a normalized state: each squared norm is $0.9999995361$. Normalize before calculating an entropy. For the two states, the resulting Gram matrices are approximatelyFor a trace-one two-dimensional Hermitian matrix with diagonal entry $u$ and off-diagonal entry $c$, its eigenvalues areSort them decreasingly and take square roots:The Von Neumann entropy is $S=-{\lambda }_0{\log }_2{\lambda }_0-{\lambda }_1{\log }_2{\lambda }_1$. $\Psi$ is more entangled: its Schmidt weights are closer to equal and its reduced-state entropy is larger. Neither state is a product state or maximally entangled on its two-dimensional support.

By Nielsen's pure-state conversion theorem, deterministic LOCC conversion requires the input vector of squared Schmidt coefficients to be majorized by the output vector. Here $0.699941<0.894238$ and both sums are one, so $\lambda (\Psi )\prec \lambda (\Phi )$. Therefore $\Psi \to \Phi$ is possible deterministically; $\Phi \to \Psi$ is not. Entanglement can be discarded by local operations but cannot increase deterministically for these pure inputs. Probabilistic concentration is a separate question and is allowed below.^[ref]

There is a dimensional qualification in the proposed measurement. Write $P=|0_A^{\prime }⟩⟨0_A^{\prime }|+|1_A^{\prime }⟩⟨1_A^{\prime }|$. The two supplied measurement operators obeyThus their effects form a POVM on the occupied support, where $P$ is the identity. Alice's full space is three-dimensional, so these two operators alone do not sum to $I_3$. A full-space completion is $M_3=|2_A^{\prime }⟩⟨2_A^{\prime }|$. It has zero probability on either input, so it changes none of the requested outcomes. Strictly, the POVM consists of the effects $E_r=M_r^{†}{M}_r$; the specified $M_r$ additionally fix a quantum instrument and therefore the conditional states.

In the appropriate Schmidt basis for either input, write $|\psi ⟩=\sqrt{{\lambda }_0}|00⟩+\sqrt{{\lambda }_1}|11⟩$. The two outcomes have probabilitiesand normalized conditional statesThese formulas give both outcomes for each state by substituting its row of the table. For example,The two inputs have different Schmidt bases; the primed measurement basis is adapted separately to the state, as the question specifies, not one common unexamined laboratory basis.

For the $M_1$ outcome to be a maximally entangled state, its two amplitudes must be equal. HenceThe requested numerical results areThus $\Psi$ has the larger success probability, consistent with its greater initial entanglement. The $M_2$ branch cannot be balanced in the allowed angle range for either nonmaximal input: it would require $\tan \theta =\sqrt{{\lambda }_0/{\lambda }_1}>1$.

The success probabilities belong to this particular binary Schmidt filter. They are not the optimum over every possible quantum instrument. The diagonal success operator $\mathrm{diag}(\sqrt{{\lambda }_1/{\lambda }_0},1)$ instead gives the standard single-copy entanglement concentration probability $2{\lambda }_1$. Postselection can increase entanglement in the retained branch, while the outcome-averaged entanglement cannot increase under LOCC.