# Quantum Entanglement

My book is generally being quite unclear about something.

So firstly I know that if the system is not entangled, we can write its state as $|ab\rangle=|a\rangle|b\rangle$ (if we understand the product is actually a tensor product). If it is entangled, we cannot do this.

It states that in general the composite system has Hamiltonian $H_{ab}=H_a+H_b+H_{int}$. Then it does some maths and works out that if $H_{int}=0$, the systems will remain unentangled provided they begin unentangled, otherwise the systems have to be entangled.

Somewhere else it says to consider two non-interacting subsystems (which I interpret as $H_{int}=0$) so that $|ab\rangle=|a\rangle|b\rangle$ - this disagrees with the above in that we haven't ruled out the prospect of the system being initially entangled, which would mean we couldn't write $|ab\rangle=|a\rangle|b\rangle$.

So my question is, which of the books two statements are correct? I.e, does $H_{int}=0$ imply the system is not entangled, or does it imply that the system is not entangled only if it is not entangled at time zero?

• I would guess that when your book says consider two non-interacting subsystems it actually means consider two unentangled non-interacting subsystems. Does the context suggest otherwise? – John Rennie Apr 4 '15 at 12:11
• Could you please say what book you're referring to? – Phonon Apr 4 '15 at 12:16

## 1 Answer

You might be misunderstanding entangled states. An entangled state, is a state which we cannot express as a factorised tensor product, i.e it cannot be cast in this form:

$$|\psi_A\rangle \otimes |\psi_B\rangle \equiv |\psi_A\psi_B\rangle.$$

Consider the state

$$|\psi\rangle = \frac{1}{2}\Big (|0\rangle_A|0\rangle_B + |0\rangle_A|1\rangle_B + |1\rangle_A|0\rangle_B + |1\rangle_A|1\rangle_B \Big).$$

This state can be factorised, i.e it can be expressed into a tensor product of two other states:

$$\to |\psi\rangle_A \otimes |\psi\rangle_B = \frac{1}{\sqrt 2}\Big (|0\rangle_A + |1\rangle_A \Big ) \otimes \Big (|0\rangle_B + |1\rangle_B \Big).$$

Now consider the Bell state:

$$|\Phi ^{+}\rangle = \frac {1}{\sqrt 2} \Big (|0\rangle_A|0\rangle_B + |1\rangle_A |1\rangle_B \Big)$$

There is no way to decompose this state now into the tensor product of two states (try it!), hence $|\Phi ^{+}\rangle$ is said to be entangled.

Bear in mind that I have suppressed tensor notation when I wrote down the basis states, i.e technically I should have written for example:

$$|\psi\rangle = \frac{1}{2}\Big (|0\rangle_A \otimes |0\rangle_B + |0\rangle_A \otimes |1\rangle_B + |1\rangle_A \otimes |0\rangle_B + |1\rangle_A \otimes |1\rangle_B \Big),$$

but do not confuse products of states with entanglement. Entanglement exists when a system cannot be separated into two subsystems, therefore you can only consider them together and describe them using an entangled state. In your case, as you yourself has said, $|ab\rangle$ can be decomposed into $|a\rangle \otimes |b\rangle$ so the states are not entangled and the system can be described in terms of two independent subsystems.

• So if $a$ and $b$ are entangled, does this mean not only that I can't write the composite system state as a product of the states of the two subsystems, but that I am also not permitted to write $|\psi_a\rangle=\displaystyle\sum_{i}a_i|a_i\rangle$ and $|\psi_b\rangle=\displaystyle\sum_{i}b_i|b_i\rangle$, because doing so would be assuming the two systems could be treated separately? – Watw Apr 4 '15 at 17:09
• @Watw Yes, the state vector formalism isn't enough in this case. One needs the density matrix formalism to describe entanglement. You can then calculate the reduced density matrix of the subsystem you want to isolate, because a classical probability gets introduced. – Constandinos Damalas Apr 4 '15 at 17:22
• But say these kets are the energy eigenkets. Then even if $a$ and $b$ are entangled, surely the two systems alone have various probabilities to have each possible energy, and so then they should be able to be expanded in terms of energy eigenstates (however, the coefficients would depend on the other system). Why is this not the case? – Watw Apr 4 '15 at 18:07