If I express it like this: $$ \psi = \frac{1}{\sqrt 2} \lvert +z \rangle + \frac{1}{\sqrt 2} \lvert -z \rangle $$

that will give a $50\%$ spin up / $50\%$ spin down measurement along $z$ which is proper. However, this is also the $\lvert +x \rangle$ eigenstate, which means that if I instead measure spin along $x$ I will always get spin up along $x$. What am I forgetting?

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    $\begingroup$ Hi and welcome to the Physics SE! The equations become much easier to read, search and edit when mathjax is used. I've proposed an edit to your post this time, but you should use it yourself in your future posts. $\endgroup$ – Samarth Jan 24 '18 at 16:22

What you're looking for is a density operator. Writing a quantum state as a Dirac state $\lvert \psi \rangle $ implies a well-defined state, that is, that there is a set of Hermitian operators with non-degenerate simultaneous eigenvectors such that $\lvert \psi \rangle$ is one of them.

Dirac kets are used to write states of single particles, while density operators describe ensembles. You can, for instance, write the density operator for an ensemble of particles that are $50\%$ spin up and $50\%$ spin down:

$$ \rho = \frac{1}{2}\left(\lvert +z \rangle \langle +z \rvert + \lvert -z \rangle \langle -z \rvert \right) $$

And you can, with a basis change, verify that the state is $50\%$ spin up and $50\%$ spin down in any direction it is measured

$$ \rho = \frac{1}{2}\left(\lvert +\hat n \rangle \langle +\hat n \rvert + \lvert -\hat n \rangle \langle -\hat n \rvert \right) $$

  • $\begingroup$ Usually density operators have trace equal to one. Then their eigenvalues are probabilities. Your operators have trace equal to $\sqrt{2}$. $\endgroup$ – Gec Jan 24 '18 at 16:35
  • $\begingroup$ @Gec whoops lemme fix that real quick. Thanks for pointing it out. $\endgroup$ – Samarth Jan 24 '18 at 17:08
  • $\begingroup$ Thanks for answer. So |ψ⟩ has to be in just one eigenvector of a Hermition operator for me to write it as a Dirac (ket) state? Can't I write the state of the particle before it is measured by the z-apparatus? $\endgroup$ – Chris Robertson Jan 24 '18 at 18:31
  • $\begingroup$ The point is that any ket in a Hilbert space is an eigenvector of some Hermitian operator. An arbitrary ket $\lvert \psi \rangle$ is an eigenvector of the projection operator $P = \lvert \psi \rangle \langle \psi \rvert$, for example. Thus for a single particle, there is (theoretically) always some measurement you can do that will give you a value with 100% certainty. $\endgroup$ – Samarth Jan 25 '18 at 13:06
  • $\begingroup$ Before z-measurement, the ket is in a superposition of the $\lvert +z \rangle$ and $\lvert -z \rangle$ states. If you have read about rotation operators in the context of spin-1/2 systems, you should try and apply a rotation to $\lvert +z \rangle$, of $\theta$ about the $y$ axis and $\phi$ about the $z$ axis, which results in $\lvert \hat n \rangle$ (where $\hat n$ is the unit vector with polar angle $\theta$ and azimuth $\phi$) and see if you can write an arbitrary superposition of $\lvert +z \rangle$ and $\lvert -z \rangle$ as $\lvert \hat n \rangle$ for some $\hat n$ $\endgroup$ – Samarth Jan 25 '18 at 13:18

Your thinking is correct. If you want to specify a state that will result in random up/down upon measurement along the z-axis, $|\psi> = \frac{1}{\sqrt 2} \lvert +z \rangle + \frac{1}{\sqrt 2} \lvert -z \rangle$ is correct. It will give you a random result: up 50% of the time and down 50% of the time. Note that, before the measurement, there is no way to tell, whether it is going to be up or down.

However, you are right that the same state is also the $|+x>$ state and hence, if you measure them along the x-axis, you will get the + value 100% of the time. So, you are not forgetting anything here.

However, note that, $\rho = \frac{1}{2}\left(\lvert +z \rangle \langle +z \rvert + \lvert -z \rangle \langle -z \rvert \right)$ means that 50% of the time you are sending |+z> state and 50% of the time you are sending |-z> state. So, "on average", over time, you will still get up-down result randomly if measured along the z-axis; "but" for each case, the source actually knows, beforehand, what it has prepared and hence also knows what will be the measurement outcome. So, technically, it's not a random spin up/down in the z-basis.


I will assume that considered system is in a pure quantum state, i.e. in a spin eigenstate, so the best approach to use are spinors. If the spin is directed in the direction $\bf{n}$, the corresponding eigen-functions are

$\frac{1}{2}({\bf n}\vec{\sigma})w^{(\lambda)} = \lambda w^{(\lambda)}$

where $w^{(\lambda)}$ is the eigen-function to the eigenvalue $\lambda$. As the Pauli-matrices $\vec{\sigma}=(\sigma_x, \sigma_y, \sigma_z)$ operate on it, it is actually a Weyl-spinor. Is $\bf{n} = ( \sin(\theta)\cos\phi, \sin(\theta)\sin\phi, \cos(\theta))$, ( the z-direction corresponds to ${\bf n} = ( 0,0,1)$ or equivalently to $\phi=0=\theta$) the eigen-function which solves the above eigenvalue equation is:

$w^{(\lambda=1/2)} =\left( \begin{array}{c} e^{-\frac{i}{2}\phi}\cos\frac{\theta}{2} \\ e^{\frac{i}{2}\phi}\sin\frac{\theta}{2} \end{array} \right)$

We will now use this eigenfunction for your case. According to your writing you have your system always in a $|+x>$ state, as always spin up is measured along the x-axis. So your spin is directed in the x-direction so $\phi=0$ and $\theta =\frac{\pi}{2}$. Then we get:

$w^{(\lambda=1/2)} =\left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) = |+x>$

We can decompose this spinor in eigen-functions of the spin directed in z-direction (and opposite to z-direction):

$w^{(\lambda=1/2)} =\left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) = \frac{1}{\sqrt{2}}\left( \begin{array}{c} 1 \\ 0 \end{array} \right) + \frac{1}{\sqrt{2}}\left( \begin{array}{c} 0 \\ 1 \end{array} \right) =\frac{1}{\sqrt{2}} |+z> +\frac{1}{\sqrt{2}} |-z> $ (1)

So you get the decomposition as you proposed for spin states directed in z-direction and opposite to z-direction.

As well you can check that the constructed state (eigen-function) is an eigen-function of the spin directed in x-direction.

$\frac{1}{2}(\bf {n_x}\sigma)w^{(1/2)} = \frac{1}{2}(\sigma_x)w^{(1/2)} = \frac{1}{2}\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) = \frac{1}{2} \left( \begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array} \right) = \frac{1}{2} w^{(1/2)} $

So our spinor $w^{(\lambda=1/2)}$ is indeed the eigen-function of the spin directed in x-direction and as shown in (1) will produce with equal probability upon spin measurement in z-direction spin-up and spin-down results.


Thank you for answers. I was interested in more the philosophy of quantum mechanics - I assumed my question was simpler. Now I realize that in chapters 1-2 of Sakurai he is talking about particles that have already been measured. Its not till chapter 3 that he gets into pure and mixed ensembles.

I realize now that if I want to write down a definite state vector for a particle with 50/50 spin in z-direction, and I specify no other information, I have four choices. Particle that is completely spin up in x, particle that is completely spin down in x, particle that is completely spin up in y, particle that is completely spin down in y. There is a problem like this in Sakurai chapter 3 - given a pure ensemble where expectation values of spin x and spin z and only the sign of spin y, write down the state vector for the ensemble. You only need the sign in y because the magnitude is fixed.

Anyone care to comment or illuminate? Thanks!


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