I Tensor product expansion

Alex Dingo

Hi,
I'm currently working through a tensor product example for a two qubit system.
For the expression:

$$\rho_A = \sum_{J=0}^{1}\langle J | \Psi \rangle \langle \Psi | J \rangle$$
Which has been defined as from going to a global state to a local state.
Here
$$|\Psi \rangle = |\Psi^+ \rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle )$$

I'm not sure how to get from here to here: $$\rho_A = \frac{1}{2} |1\rangle\langle 1| + \frac{1}{2} |0\rangle\langle 0|$$

I'm finding that the terms in the summation reduce to dot products rather than the form of the answer. i.e
$$\rho_A = \frac{1}{2}(\langle 0 | 01 \rangle + \langle 0 | 10 \rangle + \langle 1| 01 \rangle + \langle 1 | 10 \rangle + \langle 01 | 0 \rangle + \langle 10 | 0 \rangle + \langle 10 | 1 \rangle + \langle 01 | 1 \rangle )$$
Which is where I'm thinking of going wrong?

Thanks for any help.

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Truecrimson

You either multiplying out $|\psi^+ \rangle \langle \psi^+|$ wrong or writing + in the result where there shouldn't be. (Each term in $\rho_A$ should be a ket-bra $|\dots \rangle \langle \dots |$) Other than that, use the orthonormality $\langle x|y \rangle = \delta_{xy}$ of subsystem B. Make sure you understand the meaning of the expression you wrote.

vanhees71

Gold Member
Let's first get clear about the math. Your first equation makes no sense to me. What you need is the "partial trace" of a statistical operator.

To describe two Q-bits you need the tensor-product space $\mathcal{H}=\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$. Since $A$ and $B$ are Q-bits, and each of the Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ is two-dimensional with bases $|J_1 J_2 \rangle$ (where $J_1,J_2 \in \{0,1 \}$), the resulting space is four-dimensional, and $|J_1 \rangle \otimes |J_2 \rangle \equiv |J_1,J_2 \rangle$ is a basis.

Now we have the two Q bits prepared in a state described by the statistical operator $\hat{\rho}$. The probability that $A$ takes the state $J_1$ and $B$ the state $J_2$ is then given by
$$W(J_1,J_2) = \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle.$$
Now suppose you are not interested at all in part $B$ of the two-bit system. You only want to describe the part $A$. Then all you are interested in are the probabilities to find $A$ in state $J_1$. Obviously that probability is given by
$$W_A(J_1)=\sum_{J_2=0}^2 W(J_1,J_2).$$
The corresponding density matrix which describes system $A$ only is then given by
$$\hat{\rho}_A=\sum_{J_1=0}^1 W_A(J_1) |J_1 \rangle \langle J_1|.$$
Note that this operator is defined on system $A$'s Hilbertspace $\mathcal{H}_A$, as it should be since that's the Hilbert space used to describe this very system $A$ alone.

Now one can show that $\hat{\rho}_A$ doesn't depend on the bases used for $\mathcal{H}_A$ and $\mathcal{H}_B$ and that's why you can write it in a basis-independent form, as what's called the "partial trace", and it's precisely defined in the way explained above, i.e.,
$$\hat{\rho}_A = \mathrm{Tr}_{B} \hat{\rho} = \sum_{J_1,J_2=0}^1 |J_1 \rangle \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle \langle J_1|.$$

Now let's check your example. Here the two-q-bit system is in a pure state represented by the normalized ket $|\Psi \rangle$, of more precisely formulated, the state of the system is represented by the corresponding statstical operator
$$\hat{\rho} =|\Psi \rangle \langle \Psi \rangle.$$
It's a pure state, but since $|\Psi \rangle$ is not a tensor product but a sum of tensor products, it's what's called an entangled state, and that's why we'll find out now that the subsystem $A$ is not in a pure state although the state as a whole is. We get the statistical operator for the subsystem $A$ by the just defined partial trace, i.e., by "tracing out system $B$"
$$\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho}=\sum_{J_1,J_2} |J_1 \rangle \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle.$$
So first we need the matrix elements:
$$\langle J_1,J_2 |\hat{\rho}|J_1,J_2 \rangle =\frac{1}{2} |\langle J_1,J_2|01+10 \rangle|^2 = \frac{1}{2} (\delta_{J_1 0} \delta_{J_2 1} + \delta_{J_1 1} \delta_{J_2 0}.$$
Plugging this into the previous formula and doing the sums you just find
$$\hat{\rho}_A=\frac{1}{2} (|0 \rangle \langle 0| + |1 \rangle \langle 1|) = \frac{1}{2} \mathbb{1}_A.$$
This is amazing! Although your composite system is in a pure state, i.e., you have complete knowledge about it (i.e., as complete as it can get in quantum theory), but for part $A$ of the system you have the least complete knowledge you can have. This is in the sense of the usual Shannon-Jaynes-von Neumann entropy, which takes its maximum for this state.

It's more intuitive if you think of a concrete physics example for your q-bits. One common (and nowadays even easy to realize in the lab) is to use the polarization states of two photons. Say $|0 \rangle$ is realized as a horizontally polarized photon and $1 \rangle$ by a vertically polarized photon. Then doing experiments using two polarizers oriented precisely in the same direction (say horizontally) than, if photon $A$ passes its filter, then necessarily photon $B$ doesn't and vice versa, i.e., there's a 100% correlation between the polarizations of the two photons, and you can place the detectors as far away from each other as you want, i.e., you have long-ranged correlations between the two photons' polarization states, but according to our calculation, photon $A$ is completely unpolarized, i.e., if you use an ensemble of such prepared two photons Alice, observing one of the photons (at place $A$), just finds randomly with probability 50% a horizontally and with 50% probability a vertically polarized photon.

It's clear that also
$$\hat{\rho}_B=\mathrm{Tr}_{A} \hat{\rho} = \sum_{J_1,J_2} |J_2 \rangle \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle \langle J_2 |=\frac{1}{2} \mathbb{1}_B,$$
i.e. also Bob at place $B$ registers just unpolarized photons.

If both Alice and Bob note in the measurement protocols the exact times when registering (or not registering) a photon and than compare their measurement protocols, however they'll find the 100% correlation, i.e., if Alice finds her photon to be horizontally (vertically) polarized, then Bob always, i.e., with 100% probability, finds his photon to be vertically (horizontally) polarized. It never ever happens that Alice and Bob find both photons to be in the same polarization state.

That's a very mind-boggling situation. What's even more interesing is that with such "maximally entangled states" you can disprove Bell's inequality, which he has derived under the assumption that QT is NOT right but that there's a hidden variable making the polarization states of the single photons determined. Then by measuring the polarizations not in the same direction but with the polarizers in an adequate angle to each other, you just find out that this inequality is violated as it is predicted by the above quantum formalism. This means that there's no (local) deterministic hidden-variable theory leading under all circumstances to the same statistics for the polarization states of the two photons as quantum theory predicts, and such Bell tests always reveal that quantum theory is right! So the very strong correlations, which are incompatible with ANY local hidden-variable theory, described by quantum-entanglement are what's realized in nature and not any local hidden-variable theory as famously Einstein (and I guess also Bell) would have expected!

As weird as QT may look to us, it's how nature behaves. If you want back determinism for sure the corresponding hidden-variable theory must be non-local, and that's may be even more quirky than QT itself, particularly in few of the relativistic causal structure of physics.

• odietrich

Alex Dingo

Let's first get clear about the math. Your first equation makes no sense to me. What you need is the "partial trace" of a statistical operator.

To describe two Q-bits you need the tensor-product space $\mathcal{H}=\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$. Since $A$ and $B$ are Q-bits, and each of the Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ is two-dimensional with bases $|J_1 J_2 \rangle$ (where $J_1,J_2 \in \{0,1 \}$), the resulting space is four-dimensional, and $|J_1 \rangle \otimes |J_2 \rangle \equiv |J_1,J_2 \rangle$ is a basis.

Now we have the two Q bits prepared in a state described by the statistical operator $\hat{\rho}$. The probability that $A$ takes the state $J_1$ and $B$ the state $J_2$ is then given by
$$W(J_1,J_2) = \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle.$$
Now suppose you are not interested at all in part $B$ of the two-bit system. You only want to describe the part $A$. Then all you are interested in are the probabilities to find $A$ in state $J_1$. Obviously that probability is given by
$$W_A(J_1)=\sum_{J_2=0}^2 W(J_1,J_2).$$
The corresponding density matrix which describes system $A$ only is then given by
$$\hat{\rho}_A=\sum_{J_1=0}^1 W_A(J_1) |J_1 \rangle \langle J_1|.$$
Note that this operator is defined on system $A$'s Hilbertspace $\mathcal{H}_A$, as it should be since that's the Hilbert space used to describe this very system $A$ alone.

Now one can show that $\hat{\rho}_A$ doesn't depend on the bases used for $\mathcal{H}_A$ and $\mathcal{H}_B$ and that's why you can write it in a basis-independent form, as what's called the "partial trace", and it's precisely defined in the way explained above, i.e.,
$$\hat{\rho}_A = \mathrm{Tr}_{B} \hat{\rho} = \sum_{J_1,J_2=0}^1 |J_1 \rangle \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle \langle J_1|.$$

Now let's check your example. Here the two-q-bit system is in a pure state represented by the normalized ket $|\Psi \rangle$, of more precisely formulated, the state of the system is represented by the corresponding statstical operator
$$\hat{\rho} =|\Psi \rangle \langle \Psi \rangle.$$
It's a pure state, but since $|\Psi \rangle$ is not a tensor product but a sum of tensor products, it's what's called an entangled state, and that's why we'll find out now that the subsystem $A$ is not in a pure state although the state as a whole is. We get the statistical operator for the subsystem $A$ by the just defined partial trace, i.e., by "tracing out system $B$"
$$\hat{\rho}_A=\mathrm{Tr}_B \hat{\rho}=\sum_{J_1,J_2} |J_1 \rangle \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle.$$
So first we need the matrix elements:
$$\langle J_1,J_2 |\hat{\rho}|J_1,J_2 \rangle =\frac{1}{2} |\langle J_1,J_2|01+10 \rangle|^2 = \frac{1}{2} (\delta_{J_1 0} \delta_{J_2 1} + \delta_{J_1 1} \delta_{J_2 0}.$$
Plugging this into the previous formula and doing the sums you just find
$$\hat{\rho}_A=\frac{1}{2} (|0 \rangle \langle 0| + |1 \rangle \langle 1|) = \frac{1}{2} \mathbb{1}_A.$$
This is amazing! Although your composite system is in a pure state, i.e., you have complete knowledge about it (i.e., as complete as it can get in quantum theory), but for part $A$ of the system you have the least complete knowledge you can have. This is in the sense of the usual Shannon-Jaynes-von Neumann entropy, which takes its maximum for this state.

It's more intuitive if you think of a concrete physics example for your q-bits. One common (and nowadays even easy to realize in the lab) is to use the polarization states of two photons. Say $|0 \rangle$ is realized as a horizontally polarized photon and $1 \rangle$ by a vertically polarized photon. Then doing experiments using two polarizers oriented precisely in the same direction (say horizontally) than, if photon $A$ passes its filter, then necessarily photon $B$ doesn't and vice versa, i.e., there's a 100% correlation between the polarizations of the two photons, and you can place the detectors as far away from each other as you want, i.e., you have long-ranged correlations between the two photons' polarization states, but according to our calculation, photon $A$ is completely unpolarized, i.e., if you use an ensemble of such prepared two photons Alice, observing one of the photons (at place $A$), just finds randomly with probability 50% a horizontally and with 50% probability a vertically polarized photon.

It's clear that also
$$\hat{\rho}_B=\mathrm{Tr}_{A} \hat{\rho} = \sum_{J_1,J_2} |J_2 \rangle \langle J_1,J_2|\hat{\rho}|J_1,J_2 \rangle \langle J_2 |=\frac{1}{2} \mathbb{1}_B,$$
i.e. also Bob at place $B$ registers just unpolarized photons.

If both Alice and Bob note in the measurement protocols the exact times when registering (or not registering) a photon and than compare their measurement protocols, however they'll find the 100% correlation, i.e., if Alice finds her photon to be horizontally (vertically) polarized, then Bob always, i.e., with 100% probability, finds his photon to be vertically (horizontally) polarized. It never ever happens that Alice and Bob find both photons to be in the same polarization state.

That's a very mind-boggling situation. What's even more interesing is that with such "maximally entangled states" you can disprove Bell's inequality, which he has derived under the assumption that QT is NOT right but that there's a hidden variable making the polarization states of the single photons determined. Then by measuring the polarizations not in the same direction but with the polarizers in an adequate angle to each other, you just find out that this inequality is violated as it is predicted by the above quantum formalism. This means that there's no (local) deterministic hidden-variable theory leading under all circumstances to the same statistics for the polarization states of the two photons as quantum theory predicts, and such Bell tests always reveal that quantum theory is right! So the very strong correlations, which are incompatible with ANY local hidden-variable theory, described by quantum-entanglement are what's realized in nature and not any local hidden-variable theory as famously Einstein (and I guess also Bell) would have expected!

As weird as QT may look to us, it's how nature behaves. If you want back determinism for sure the corresponding hidden-variable theory must be non-local, and that's may be even more quirky than QT itself, particularly in few of the relativistic causal structure of physics.
Thank you for you're fantastic response! You answered my questions perfectly and I have a much better grasp on the significance of the result - much appreciated.

• vanhees71

"Tensor product expansion"

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