Measurement YouTube

Measurement in practise

We first have to remember that spin states of an electron cannot be represented as points in sets. They are complex vectors that are, in principle, determined by experiment.

By measuring the spin of the electron. we mean finding the Hermitian operator with the electron state as an eigenvector with eigenvalue $+1$.

For example, suppose some process leaves an electron is in some state $\ket{a}$. What we really mean here is that we run an experiment many times, and hone in on the state that occurs 100% of the time - we might do many more experiments to confirm this.

Having found $$\ket{a} = \begin{bmatrix}\alpha \\ \beta \end{bmatrix}$$ then the spin operator associated with this state should satisfy $$\sigma_n \ket{a} = \ket{a}$$

To find out what this operator is we note that since $\ket{a}$ is a unit vector then the average of the operator in this state to be one. That is, $$\bra{a} \sigma_n \ket{a} = \braket{a}{a} = 1$$

But, from the last section we saw that $$\bra{a} \sigma_n \ket{a} = \left( \beta^*\alpha + \alpha^* \beta \right)n_x + i\left( \beta^*\alpha - \alpha^* \beta \right)n_y + \left( \alpha^*\alpha - \beta^* \beta \right)n_z$$

Recall that we can insist that $$\hat{n} = \begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix}$$ is a unit vector, so that $$n_x^2 + n_y^2 + n_z^2 = 1$$

So, we can choose $$\begin{align*}n_x &= \beta^*\alpha + \alpha^* \beta\\n_y &= \mathrm{i}\left( \beta^*\alpha - \alpha^* \beta \right)\end{align*}$$ from which it also follows that $$n_z = \alpha^*\alpha - \beta^* \beta$$

If we write $$\begin{align*}n_+ &= n_x + in_y\\&= \left( \beta^*\alpha + \alpha^* \beta \right) + \mathrm{i}\left( \mathrm{i}\left( \beta^*\alpha - \alpha^* \beta \right) \right)\\&= \beta^*\alpha + \alpha^* \beta - \beta^*\alpha + \alpha^* \beta\\&= 2\alpha^* \beta\\n_- &= n_+^*\\&= 2\beta^* \alpha\end{align*}$$ then the spin operator is $$\sigma_n = \begin{bmatrix}n_z & n_-\\n_+ & -n_z\end{bmatrix} = \begin{bmatrix}\alpha^*\alpha - \beta^* \beta & 2\beta^*\alpha\\2\alpha^*\beta & -\left( \alpha^*\alpha - \beta^* \beta \right)\end{bmatrix}$$

Measuring in arbitrary directions

Figure 4.1 - Arbitrary directions

Figure 4.1 - Arbitrary directions

Suppose we prepare the spin of an electron by putting it in a large magnetic field, waiting long enough to allow a photon to be emitted or not thus forcing it down a particular axis, $\hat{n}$, say.

Then, we measure the component of spin along a different direction, $\hat{m}$, say. We want to know the probability that, given the prepared state, $\hat{n}$, that the measured value along the $\hat{m}$ direction is $+1$ (and the probability that it is $-1$). That is we want to find $$\left| \braket{+m}{+n} \right|^2 = \braket{+m}{+n}^* \braket{+m}{+n}$$

We could look at the situation where we prepare in the $\hat{m}$ direction and measure along $\hat{n}$. We would expect the same answer, due to symmetry. In fact, we should expect the answer to depend only on the angle $\theta_{mn}$ between them.

We first note the following identities $$\begin{align*}\hat{m} \cdot \hat{n} &= \cos \theta_{mn}\\\bra{+m} &= \sqrt{\frac{1 + m_z}{2}} \begin{bmatrix}1 & \frac{1 - m_z}{m_+}\end{bmatrix}\\m_+ n_- + m_- n_+ &= 2 \left( m_x n_x + m_y n_y \right)\\m_+ m_- &= 1 - m_z^2\\n_+ n_- &= 1 - n_z^2\end{align*}$$

Then, we calculate the probability, $$\begin{align*}\left| \braket{+m}{+n} \right|^2 &= \left| \sqrt{\frac{1 + m_z}{2}} \begin{bmatrix}1 & \frac{1 - m_z}{m_+}\end{bmatrix}\sqrt{\frac{1 + n_z}{2}} \begin{bmatrix}1 \\ \frac{1 - n_z}{n_-}\end{bmatrix} \right|^2\\&= \left| \sqrt{\frac{\left( 1 + m_z \right)\left( 1 + n_z \right)}{4}}\left( 1 + \frac{\left( 1 - m_z \right)\left( 1 - n_z \right)}{m_+ n_-} \right) \right|^2\\&= \frac{\left( 1 + m_z \right)\left( 1 + n_z \right)}{4} \left( 1 + \frac{\left( 1 - m_z \right)\left( 1 - n_z \right)}{m_- n_+} \right) \left( 1 + \frac{\left( 1 - m_z \right)\left( 1 - n_z \right)}{m_+ n_-} \right)\\&= \frac{\left( 1 + m_z \right)\left( 1 + n_z \right)}{4}\left( \frac{\left(m_- n_+ + \left(1 - m_z \right)\left(1 - n_z \right) \right)\left(m_+ n_- + \left(1 - m_z \right)\left(1 - n_z \right) \right)}{\left(m_+m_-\right)\left(n_+n_-\right)}\right)\\&= \frac{\left( 1 + m_z \right)\left( 1 + n_z \right)}{4}\left(\frac{\left(m_+m_-\right)\left(n_+n_-\right) + \left(m_- n_+ + m_+ n_-\right)\left(1 - m_z \right)\left(1 - n_z \right) + \left(1 - m_z \right)^2\left(1 - n_z \right)^2}{\left(m_+m_-\right)\left(n_+n_-\right)} \right)\\&= \frac{\left( 1 + m_z \right)\left( 1 + n_z \right)}{4} \left(\frac{\left(1 - m_z^2\right)\left(1 - n_z^2\right) + 2\left(m_x n_x + m_y n_y\right)\left(1 - m_z \right)\left(1 - n_z \right) + \left(1 - m_z \right)^2\left(1 - n_z \right)^2}{\left(1 - m_z^2\right)\left(1 - n_z^2\right)}\right)\\&= \frac{\left(1 + m_z\right)\left(1 + n_z\right) + 2\left(m_x n_x + m_y n_y\right) + \left(1 - m_z\right)\left(1 - n_z\right)}{4}\\&= \frac{\left(1 + m_z + n_z + m_z n_z\right) + 2\left(m_x n_x + m_y n_y\right) + \left(1 - m_z - n_z + m_z n_z\right)}{4}\\&= \frac{2 + 2\left(m_x n_x + m_y n_y + m_z n_z\right)}{4}\\&= \frac{1 + \hat{m} \cdot \hat{n}}{2}\\&= \frac{1 + \cos\theta_{mn}}{2}\end{align*}$$

Figure 4.2 - Probability function

Figure 4.2 - Probability function

So the probability is a continuous function of $\theta_{mn}$ over the interval $\left [0, \pi \right]$

We can check that this corresponds to our experimental knowledge.

We would expect that the probability that the spin of an electron would be in the same direction in which it was prepared should be one, $$\begin{align*}\hat{m} = \hat{n} &\quad \Rightarrow \quad \theta_{mn} = 0\\&\quad \Rightarrow \quad \left| \braket{+m}{+n} \right|^2 = \frac{1 + \cos(0)}{2} = \frac{1 + 1}{2} = 1\end{align*}$$

We would also expect that the probability that it would be in the opposite direction in which it was prepared would be zero, $$\begin{align*}\hat{m} = -\hat{n} &\quad \Rightarrow \quad \theta_{mn} = \pi\\&\quad \Rightarrow \quad \left| \braket{+m}{+n} \right|^2 = \frac{1 + \cos(\pi)}{2} = \frac{1 - 1}{2} = 0\end{align*}$$ and that the probability that it would be perpendicular to the direction in which it was prepared would be $1/2$, $$\begin{align*}\hat{m} \perp \hat{n} &\quad \Rightarrow \quad \theta_{mn} = \frac{\pi}{2}\\&\quad \Rightarrow \quad \left| \braket{+m}{+n} \right|^2 = \frac{1 + \cos(\frac{\pi}{2})}{2} = \frac{1 + 0}{2} = \tfrac{1}{2}\end{align*}$$

As discussed before, in classical mechanics, we would have expected that the result of such an experiment would result in a continous range of results, depending on the angle between the two directions. In quantum mechanics, it is not the result that is continuous - only one of two things can occur in a measurement of the spin of an electron - it is the probability that is continuous.

Simultaneous measurements

In quantum mechanics, measuring the spin of an electron necessarily means leaving the electron in a new eigenstate, along the axis used to make the measurement, in either the positive or negative direction. This is called polarising the electron along a particular direction.

This means that, in general, you cannot measure two different observables simultaneously. However, there are cases when you can measure two different observables simultaneously, namely when the two observables commute with each other. We shall see why this is true in detail later, but it means that the order in which you measured the observables wouldn't matter. So it is useful to know when two observables commute.

Observables with identical eigenvectors commute

Recall that, any square matrices, $A,B$, commute if $$AB = BA$$

Suppose that $A,B$ had identical eigenvectors, $\ket{a_i}$, such that, for some eigenvalues, $\lambda_i, \mu_i$, we have $$\left\{\begin{matrix}A\ket{a_i} &= \lambda_i \ket{a_i}\\B\ket{a_i} &= \mu_i \ket{a_i}\end{matrix}\right.$$

Then, we can apply both operators, $$\begin{matrix}\left(BA\right)\ket{a_i} & B\left(A\ket{a_i}\right)\\& B\left(\lambda_i\ket{a_i}\right)\\& \lambda_i\left(B\ket{a_i}\right)\\& \lambda_i\left(\mu_i\ket{a_i}\right)\\& \mu_i\left(\lambda_i\ket{a_i}\right)\\& \mu_i\left(A\ket{a_i}\right)\\& A\left(\mu_i\ket{a_i}\right)\\& A\left(B\ket{a_i}\right)\\& \left(AB\right)\ket{a_i}\end{matrix}$$

So, $$\left(BA - AB\right)\ket{a_i} = 0$$ where, in general, $$\ket{a_i} \neq 0$$ so we must have $$BA - AB = 0$$ or $$BA = AB$$