Why can't we infer the amplitude of the spin vector components from the spin probability when the latter is 50%?

Question

I'm currently reading Quantum Mechanics: The Theoretical Minimum. I'm stuck trying to understand a minor point in the book where the authors seem to suggest that we can know the probability of measuring a spin in a certain direction to be 50% while not knowing the amplitude of the components of the spin states in that same direction.

I would have thought that if the probability is 0.5, then the amplitudes of the components of the spin state must each be 0.5 as well.

Here is the quote from the book:

Here is what we know: when the spin has been prepared in the left configuration, the probabilities for $ \sigma_z $ are again equal to ½. That is not enough to determine the values $ \alpha_{u}^{*}\alpha_{u} $ and $ \alpha_{d}^{*}\alpha_{d} $, but there is another condition that we can infer...

Here is my understanding of what each of the terms mean in the above quote;

$ \sigma_z $ is the probability of +/- spins detected along the z axis. $ \alpha_{u} $ is the component of the state vector in the up direction, which is a positive spin on the z axis. $ \alpha_{d} $ is the same but for a negative spin on the z axis (d is for down). The asterisk denotes the complex conjugate.

Given that $ \sigma_z $ is 0.5 for both spins, doesn't it also follow that the amplitude of each $\alpha$ (ie $\alpha^*\alpha$) is 0.5 as well? Ie. Isn't it true that

$$ \alpha_{u}^{*}\alpha_{u} = \alpha_{d}^{*}\alpha_{d} $$

As I understand, it is also the case that

$$ \alpha_{u}^{*}\alpha_{u} + \alpha_{d}^{*}\alpha_{d} = 1 $$

Therefore

$$ \alpha_{u}^{*}\alpha_{u} = \alpha_{d}^{*}\alpha_{d} = 0.5 $$

The last sentence in the quote seems to state that my understanding is incorrect. Why?

Answer 1

The probability of measuring 'up' is precisely given by the magnitude of the amplitude squared: $P_u = |\alpha_u|^2 = \alpha_u^* \alpha_u$ . If you know $P_u = 1/2$, then you are correct that you also know $\alpha_u^* \alpha_u = 1/2$. That may be a typo in the text.

However, the amplitude of the wavefunction is $\alpha_u$, rather than $\alpha_u^* \alpha_u$.

In particular: the wavefunction is given by $|\psi\rangle = \alpha_u |\uparrow\rangle + \alpha_d |\downarrow\rangle$. We cannot determine $\alpha_u$ and $\alpha_d$ from knowing the values of $\alpha_u^* \alpha_u$ and $\alpha_d^* \alpha_d$. All we know is the magnitude of $\alpha_u$ and $\alpha_d$: in particular, each have magnitude $1/\sqrt{2}$. In addition to the magnitude, they also carry a phase: $\alpha_u = 1/\sqrt{2} e^{i\phi}$. As a simple case, note that the phase could be either +1 or -1 (if $\phi = 0 \text{ or } \pi$). Or it could be imaginary, if $\phi = \pi/2$. This phase is a crucial part of quantum mechanics, and is the source of all types of interference and wave phenomena.

Note that this is what separates the wavefunction from the probability distribution of measurement outcomes.

Answer 2

A state with definite “up” spin along $+x$ can be expanded in terms of states with spin along $\pm z$ as $$ \vert +\rangle_x =\frac{1}{\sqrt{2}}\left(\vert +\rangle_z+\vert -\rangle_z\right) $$ A state with definite “up” spin along $+y$ can likewise be expanded as $$ \vert +\rangle_y =\frac{1}{\sqrt{2}}\left(\vert +\rangle_z+i\vert -\rangle_z\right) $$ In both case the coefficients $\alpha_u\alpha_u^*=\alpha_d\alpha_d^*=\frac{1}{2}$ but clearly the states are different.

To complete the trick you might also care to expand $\vert +\rangle_x$ in terms of $\vert +\rangle_y$ and $\vert -\rangle_y$: the coefficients (call them $\beta_u$ and $\beta_d$ this time) also satisfy $\beta_u\beta_u^*=\beta_d\beta_d^*=\frac{1}{2}$.

Hopefully this makes it quite clear that knowledge of just probabilities of outcomes is actually rather incomplete. In fact the reconstruction of a quantum state is actually not such an easy task.