Why does it matter that Schmidt number is invariant under unitary transformations?

Question

I am reading Nielsen & Chuang and they say this:

"The bases $|i_A\rangle$ and $|i_B\rangle$ are called the Schmidt bases for A and B, respectively, and the number of non-zero values $\lambda_i$ is called the Schmidt number for the state $|\psi\rangle$. The Schmidt number is an important property of a composite quantum system, which in some sense quantifies the `amount' of entanglement between systems A and B. To get some idea of why this is the case, consider the following obvious but important property: the Schmidt number is preserved under unitary transformations on system A or system B alone. To see this, notice that if $\sum_i \lambda_i |i_A\rangle |i_B\rangle$ is the Schmidt decomposition for $|\psi\rangle$, then $\sum_i \lambda_i(U |i_A\rangle)|i_B\rangle$ is the Schmidt decomposition for $ U|\psi\rangle$, where $U$ is a unitary operator acting on system A alone. Algebraic invariance properties of this type make the Schmidt number a very useful tool."

My questions:

• How does the invariance of Schmidt number under unitary transformations help it quantify the 'amount' of entanglement between systems A and B?
• Intuitively, why does applying a unitary not change the Schmidt number? (can't a unitary bring us from a maximally entangled state such as |00> + |11> to an unentangled state such as |10> + |11>)?

Thank you!

Answer 1

Crucially, "the Schmidt number is preserved under unitary transformations on system A or system B alone". So Schmidt numbers are invariant under local operations $U_A\otimes U_B$, but not under global unitary operations: as you pointed out, a global unitary can bring any state to any other state, thus not preserving entanglement properties.

As for the example you listed, $|00\rangle+|11\rangle\mapsto|10\rangle+|11\rangle$ is of course a global but not a local operation because if the a local $U$ on the first qubit sends $|0\rangle$ to $|1\rangle$, then it has to send $|1\rangle$ (from the second summand) to $e^{i\varphi}|0\rangle$ (i.e. it cannot stay $|1\rangle$)

Answer 2

Regarding your first question. Consider having the following bipartite state:

$$ |\psi\rangle_{AB} = \frac{1}{2}|00\rangle_A|00\rangle_B+ \frac{\sqrt{3}}{6}|01\rangle_A|01\rangle_B+ \frac{\sqrt{6}}{6}|01\rangle_A|10\rangle_B+ \frac{\sqrt{3}}{6}|10\rangle_A|01\rangle_B− \frac{\sqrt{6}}{6}|10\rangle_A|10\rangle_B+ \frac{1}{2}|11\rangle_A|00\rangle_B $$

At first glance, it is not trivial to figure out if subsystem $A$ and subsystem $B$ are entangled, and if they are, how is it that they correlated to each other.

Expressing this state in its Schmidt decomposition allows you to see these correlations a bit more clearly:

$$ \begin{aligned} |\psi\rangle_{AB} &= \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}|00\rangle_A+\frac{1}{\sqrt{2}}|11\rangle_A \right)|00\rangle_B \\ &+ \frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{2}}|01\rangle_A-\frac{1}{\sqrt{2}}|10\rangle_A \right)|10\rangle_B \\ &+ \frac{1}{\sqrt{6}}\left(\frac{1}{\sqrt{2}}|01\rangle_A+\frac{1}{\sqrt{2}}|10\rangle_A \right)|01\rangle_B . \end{aligned} $$

Notice this state is of the form:

$$ |\psi\rangle_{AB} = \lambda_0 |\Phi^+\rangle_A |00\rangle_B + \lambda_1|\Psi^-\rangle_A |10\rangle_B + \lambda_2|\Psi^+\rangle_A|01\rangle_B, $$

with Schmidt coefficients: $\lambda_0^2 = \frac{1}{2}$, $\lambda_1^2 = \frac{1}{3}$, $\lambda_2^2 = \frac{1}{6}$.

What this implies is that, if you were to measure subsystem $B$ in the computational basis, with probability of $\frac{1}{2}$ you will project this subsystem to state $|00\rangle_B$, and you are guaranteed the state of subsystem $A$ will remain in the Bell State $|\Phi^+\rangle_A$. Similarly, with probability $\frac{1}{3}$, project to $|10\rangle_B$ and $A$ remains in $|\Psi^-\rangle_A$, etc.

But what if you want to measure subsystem $A$ and you can only do so in the computational basis? Well, performing a unitary transformation $U_A$ only on subsystem $A$ guarantees that the Schmidt number will be preserved, so you could select:

$$ U_A = CX \, \left(H \otimes I \right) $$

to transform the Bell states on subsystem $A$ to the computational basis and get:

$$ |\psi\rangle_{AB} = \lambda_0 |00\rangle_A |00\rangle_B + \lambda_1|11\rangle_A |10\rangle_B + \lambda_2|01\rangle_A|01\rangle_B, $$

And this way, you could know after performing measurements on subsystem $A$ how the results are correlated with subsystem $B$.

As for your second question, notice that this is only true when the unitary is applied to only one of the two subsystems, not when applying a unitary that affects both of them. Intuitively, the idea is that local operations on a any system cannot affect the state of a second system, even if they are entangled.