Example states YouTube

Non-entangled states

Recall, a general state is a linear combination of the basis vectors, $$\ket{A} = \alpha \ket{uu} + \beta \ket{ud} + \gamma \ket{du} + \delta \ket{dd}$$ such that $$\braket{A}{A} = \alpha^*\alpha + \beta^*\beta + \gamma^*\gamma + \delta^*\delta = 1$$

However, we have already considered two electron states, called product states, which can be written as a simple product of two single-electron states. That is, if $$\begin{align*}\ket{a} &= \alpha_u \ket{u} + \alpha_d \ket{d} \\ \ket{b} &= \beta_u \ket{u} + \beta_d \ket{d} \end{align*}$$ where $$\braket{a}{a} = \braket{b}{b} = 1$$ then we can construct the product state, $$\ket{A} = \ket{a} \otimes \ket{b}$$

$\ket{A}$ is unit-length, as required, $$\braket{A}{A} = \braket{a}{a}\braket{b}{b} = 1 \cdot 1 = 1$$

We can write the state in component form, $$\begin{align*}\ket{A} &= \ket{a} \otimes \ket{b}\\ &= \left(\alpha_u \ket{u} + \alpha_d \ket{d}\right) \otimes \left(\beta_u \ket{u} + \beta_d \ket{d}\right)\\ &= \alpha_u \beta_u \ket{uu} + \alpha_u \beta_d \ket{ud} + \alpha_d \beta_u \ket{du} + \alpha_d \beta_d \ket{dd}\end{align*}$$

We only need four real numbers to describe a product state, as opposed to six for the general state, hence a product state is not entangled. This is because the two single-electron states are both phase-independent and unit-length, hence we can reduce the total of real numbers needed by two for each state.

Product states correspond to the situation in which each electron is prepared independently, and can be measured independently. We can see this by recalling that, given $\ket{a}, \ket{b}$, there exist unique spatial directions, $\hat{m}, \hat{n}$, such that $$\begin{align*}\bra{a}\sigma_m\ket{a} &= \pm 1\\\bra{b}\sigma_n\ket{b} &= \pm 1\end{align*}$$

But then, $$\begin{align*}\bra{A} \boldsymbol{\sigma}_m \ket{A} &= \left(\bra{a} \otimes \bra{b}\right) \left(\sigma_m \otimes \mathbf{I}\right) \left(\ket{a} \otimes \ket{b}\right)\\&= \bra{a} \sigma_m \ket{a} \braket{b}{b}\\&= \pm 1\\\bra{A} \boldsymbol{\tau}_n \ket{A} &= \left(\bra{a} \otimes \bra{b}\right) \left(\mathbf{I} \otimes \tau_n\right) \left(\ket{a} \otimes \ket{b}\right)\\&= \braket{a}{a} \bra{b} \tau_n \ket{b}\\&= \pm 1\end{align*}$$

This means that, for a product state, there is always a direction along which you will measure the spin of the first electron to be $\pm 1$ with 100% certainty, and there is always a direction along which you will measure the spin of the second electron to be $\pm 1$ with 100% certainty. This is equivalent to saying that each electron can be measured independently.

Entangled states

We now consider examples of entangled states, namely the states $$\ket{A_\pm} = \tfrac{1}{\sqrt{2}}\left(\ket{ud} \pm \ket{du}\right)$$

By definition, these states are systems where the second electron is in the opposite configuration to the first. If one of electrons is measured to be up along a given direction, then the other will definitely be down without the need for measurement. A pair of electrons in this state are said to be entangled.

We can show that neither of these states are product states. We have just seen that for every product state there is a direction along which you will measure the spin of the first electron to be $\pm 1$ with 100% certainty and a direction along which you will measure the spin of the second electron to be $\pm 1$ with 100% certainty. This does not hold for $\ket{A_\pm}$.

For any direction $\hat{n}$, we can show that the averages of the associated spins, $\boldsymbol{\sigma}_n, \boldsymbol{\tau}_n$, in this state, are zero, meaning that measurements along this direction are equally likely to be $+1$ as $-1$. That is, $$\begin{align*}\bra{A_\pm} \boldsymbol{\sigma}_n \ket{A_\pm} &= \tfrac{1}{2}\left(\bra{ud} \pm \bra{du}\right)\boldsymbol{\sigma}_n\left(\ket{ud} \pm \ket{du}\right)\\&= \tfrac{1}{2}\left(\bra{ud} \pm \bra{du}\right)\left(\boldsymbol{\sigma}_n\ket{ud} \pm \boldsymbol{\sigma}_n\ket{du}\right)\\&= \tfrac{1}{2}\left(\bra{ud} \pm \bra{du}\right)\left(n_z\ket{ud} + n_+\ket{dd} \pm \left(n_-\ket{uu} - n_z\ket{du}\right)\right)\\&= \tfrac{1}{2}\left(n_z\braket{ud}{ud} - n_z\braket{du}{du}\right)\\&= \tfrac{1}{2}\left(n_z - n_z\right)\\&= 0\end{align*}$$

Just as an exercise, we show the same is true for $\boldsymbol{\tau}_n$ using matrix notation. Thus, $$\begin{align*}\bra{A_\pm} \boldsymbol{\tau}_n \ket{A_\pm} &= \tfrac{1}{\sqrt{2}}\begin{bmatrix} 0 & 1 & \pm 1 & 0 \end{bmatrix}\begin{bmatrix} n_z & n_- & 0 & 0 \\ n_+ & -n_z & 0 & 0 \\ 0 & 0 & n_z & n_- \\ 0 & 0 & n_+ & -n_z \end{bmatrix}\tfrac{1}{\sqrt{2}}\begin{bmatrix}0 \\ 1 \\ \pm 1 \\ 0 \end{bmatrix}\\&= \tfrac{1}{2}\begin{bmatrix} 0 & 1 & \pm 1 & 0 \end{bmatrix} \begin{bmatrix}n_- \\ -n_z \\ \pm n_z \\ \pm n_+ \end{bmatrix}\\&= \tfrac{1}{2}\left(-n_z + n_z \right)\\&= 0\end{align*}$$

All we need to do to produce an entangled state is to bring two electrons in close enough proximity so that their magnetic fields interact and then leave them alone. Eventually, the electrons will be entangled, after maybe having emitted a photon. This is due to the fact that the electrons in opposite configuration is the lowest energy state of a two electron system. We can also think about as the first electron being the magnet that prepares the second (or vice versa).

We might ask ourselves how odd this entangled behaviour is. We can think of classical systems that are entangled. For example, suppose you have two counters in your hands, one red and one blue. You cup your hands together, give them a shake and take one counter in each hand without looking to see which one went into which hand. Now, if you look in one hand and see a red counter, you immediately know, without looking, that the counter in the other hand is blue. This isn't so different to the quantum case (but it isn't exactly the same, either).

The singlet state

Figure 5.1 - Vector sum

Figure 5.1 - Vector sum

There is a significant difference between the two states, $$\begin{align*}\ket{S} &= \ket{A_-} = \tfrac{1}{\sqrt{2}}\left(\ket{ud} - \ket{du}\right)\\\ket{T} &= \ket{A_+} = \tfrac{1}{\sqrt{2}}\left(\ket{ud} + \ket{du}\right)\end{align*}$$

One way to distinguish them is to compute the result of the sum of the two spin operators, $\boldsymbol{\sigma}_n + \boldsymbol{\tau}_n$, for any direction $\hat{n}$, on each state.

This is analagous to the vector sum of a pair of directions, $\hat{m}, \hat{n}$ in normal space.

In matrix form, $$\begin{align*}\boldsymbol{\sigma}_n + \boldsymbol{\tau}_n &= \begin{bmatrix} n_z & 0 & n_- & 0 \\ 0 & n_z & 0 & n_-\\ n_+ & 0 &-n_z & 0 \\ 0 & n_+ & 0 &-n_z \end{bmatrix} + \begin{bmatrix} n_z & n_- & 0 & 0 \\ n_+ & -n_z & 0 & 0 \\ 0 & 0 & n_z & n_- \\ 0 & 0 & n_+ & -n_z \end{bmatrix}\\&= \begin{bmatrix} 2n_z & n_- & n_- & 0 \\ n_+ & 0 & 0 & n_- \\ n_+ & 0 & 0 & n_- \\ 0 & n_+ & n_+ & -2n_z \end{bmatrix}\end{align*}$$

The state $\ket{S}$, called the singlet state, has the property of being in the null-space of the operator $\boldsymbol{\sigma}_n + \boldsymbol{\tau}_n$. That is, for any direction $\hat{n}$, the singlet state is an eigenvector of $\boldsymbol{\sigma}_n + \boldsymbol{\tau}_n$, with eigenvalue $0$, $$\begin{align*}\left(\boldsymbol{\sigma}_n + \boldsymbol{\tau}_n\right)\ket{S} &= \begin{bmatrix} 2n_z & n_- & n_- & 0 \\ n_+ & 0 & 0 & n_- \\ n_+ & 0 & 0 & n_- \\ 0 & n_+ & n_+ & -2n_z \end{bmatrix} \tfrac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix}\\&= \tfrac{1}{\sqrt{2}}\begin{bmatrix} n_- - n_- \\ 0 \\ 0 \\ n_+ - n_+ \end{bmatrix}\\&= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}\end{align*}$$

This is not the case for the other state $\ket{T}$, known as the triplet state. In general, $$\begin{align*}\left(\boldsymbol{\sigma}_n + \boldsymbol{\tau}_n\right)\ket{T} &= \begin{bmatrix} 2n_z & n_- & n_- & 0 \\ n_+ & 0 & 0 & n_- \\ n_+ & 0 & 0 & n_- \\ 0 & n_+ & n_+ & -2n_z \end{bmatrix} \tfrac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ 1 \\ +1 \\ 0 \end{bmatrix}\\&= \tfrac{1}{\sqrt{2}}\begin{bmatrix} n_- + n_- \\ 0 \\ 0 \\ n_+ + n_+ \end{bmatrix}\\&= \sqrt{2} \begin{bmatrix} n_- \\ 0 \\ 0 \\ n_+ \end{bmatrix}\end{align*}$$

It turns out that, experimentally, the singlet state, $$\ket{S} = \ket{A_-} = \tfrac{1}{\sqrt{2}}\left(\ket{ud} - \ket{du}\right)$$ is the state that a pair of electrons will end up in, if brought together so that their magnetic fields interact.

Nb. We shall see later where the names singlet state and triplet state come from.