How to understand combination states vs pure/mixed states?

Question

I've learned that representing a combination of two states, I simply need to take the tensor product of the states. For example:

$$\left|\Psi\right>=\alpha_0\left|0\right>+\beta_0\left|1\right>$$

$$\left|\Phi\right>=\alpha_1\left|0\right>+\beta_1\left|1\right>$$

The combination state is given by: $$ \left|\Psi\right>\otimes\left|\Phi\right>=\alpha_0\alpha_1\left|00\right>+\alpha_0\beta_1\left|01\right>+\beta_0\alpha_1\left|10\right>+\beta_0\beta_1\left|11\right> $$

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"Mixing" them in a classical sense can be done by simply finding their density matrices, and adding them together weighed by their probabilities respectively:

$$\rho=p_1|\Psi\rangle\langle\Psi|+p_2|\Phi\rangle\langle\Phi|$$

Which gives us a mixed state.

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Can I understand it as the combination state is EITHER $\left|\Psi\right>$ OR $\left|\Phi\right>$ and the mixed state is BOTH $\left|\Psi\right>$ AND $\left|\Phi\right>$?

If so, is it possible to calculate the probability of finding $\left|\Psi\right>$ in $(\left|\Psi\right>\otimes\left|\Phi\right>)$? (By my intuition, if the combination state is OR, then the probability should be just $1$?)

Answer 1

It seems that your intution is wrong. Logical operators OR and AND takes two inputs (bits) and return only one output (bit).

In case of your mixed states, you have two one-qubit inputs and two-qubits output, so it is not AND operation. Sometimes AND is refered to as a logical product, while combination is produced via tensor product. Maybe this is a reson why you are missled.

In case of mixed states, XOR operator seems reasonable, as only one state is returned. However, I do not see any reason while to look for such analogs.

Concerning finding $|\Phi\rangle$ in $|\Phi\rangle \otimes |\Psi\rangle$. State $|\Phi\rangle$ has diffrent number of qubits than $|\Phi\rangle \otimes |\Psi\rangle$ - at least one qubit is added if $|\Psi\rangle$ is composed of one qubit. So, you cannot find $|\Phi\rangle$ in $|\Phi\rangle \otimes |\Psi\rangle$.

Answer 2

I think you are 'mixing' the terms a bit!

I shall take the example of single-qubits to make the analogy simpler but $|\Psi \rangle$ and $| \Phi \rangle$ can be n-qubit states too.

Let us first clarify what the combination state i.e. $|\Psi \rangle \otimes |\Phi \rangle$ means. It is a state which describes a 2-qubit system. We can say that the first qubit is in the state $|\Psi \rangle$ and the second qubit is in the state $|\Phi \rangle$. Now, the combined system of qubit 1 and qubit 2 is what you are calling the combination state and note that this 4-dimensional state describes a 2-qubit system.

Now, coming to the mixed state. Mixed state is a representation; it states that given that we have a system which can be in any one of the states, say $| \Phi_i \rangle$ with a probability $p_i$, then the system can be represented as a density matrix which is just - $$ \rho = \sum_i p_i |\Phi_i \rangle \langle \Phi_i|$$ This is only one system's uncertain state representation.

So, the combination state is either $|\Psi \rangle$ or $ |\Phi \rangle$ and the mixed state is both $| \Psi \rangle $ and $| \Phi \rangle $ is not a valid question to ask as $|\Psi \rangle \otimes |\Phi \rangle$ is a combined state representing two single qubit systems whereas $\rho = p_1 |\Psi \rangle \langle \Psi| + p_2 |\Phi \rangle \langle \Phi|$ is the representation of the uncertain state of one single-qubit system.

This is my interpretation and I hope it helps. Any additions to this are more than welcome!