# Our choice of basis surely cannot effect possible outcomes of a measurement?

Common sense says that, of course, the outcome of a measurement on a quantum system cannot be affected by what base we choose to represent it in. However, while studying QM text, it seems like they sometimes just almost suggest it... My question is quite vague, so let me instead give you an example which I think contains most of my confusion.

Consider a system in a basis of simultaneous eigenstates of $L^2$ and $L_z$, with respective eigenvalues $l(l+1)\hbar^2$ and $m\hbar$. Let $\mathbf{\hat{n}}$ be a unit vector in a direction specified by polar angles ($\theta,\phi$).

Clearly $L_n = \sin\theta\cos\phi L_x+\sin\theta\sin\phi L_y +\cos\theta L_z$

Now I have two slightly different questions:

1. What are the possible results of a (precise) measurement of $L_n$, $L_n^2$?
2. What are the possible expectation values of $L_n$, $L_n^2$

I would tell you

1. $m\hbar$, $l(l+1)\hbar^2$ or maybe $m \hbar \cos\theta$, but this would mean that my choice of direction in space affected the outcome of the measurement?
2. No idea and makes me question my answer to 1) again...
• If you have a state that is an eigenstate of $L_z$, then it will not be an eigenstate of $L_n$ that you define above, since $L_x$, $L_y$, and $L_z$ don't commute, and therefore the measured value could be anything. I think what you mean to do is to do a coordinate change, but then you also need to transform the state in the same way you transformed the operators, such that if $\phi$ was an eigenvector of $L_z$ the new $\phi'$ is an eigenvector of $L_n$. After this you'll get exactly the same results.
– Stan
May 26, 2016 at 8:46

Phenomena in quantum mechanics may be expressed using any basis (that's the English word for the set of vectors, not a "base"). It doesn't mean that all bases are equally useful for a given situation. In particular, a fundamental postulate of quantum mechanics says that right after every measurement, the system is found in one of the eigenstates of the observable that was just measured.

That's why the basis of the eigenstates of $K$ is obviously more useful to describe the measurement of $K$ than other bases. Note that which basis is useful – or which basis describes possible post-measurement state – does depend on the kind of a measurement we decide to perform. This dependence of the "right analysis of the physical system" on the chosen way of observing it is really a main point of quantum mechanics.

If the physical analysis could be made independently of the nature of observations, the theory would be by definition classical physics, not quantum mechanics. Such independence on the observations could be called "common sense" by someone – but that changes nothing about the fact that Nature contradicts this assumption.

1. $L_n$ always has eigenvalues $m\hbar$ where $m$ is an integer. $L_n^2$ has eigenvalues $m^2\hbar^2$ – because it's just the square of the operator from the previous sentence. This shouldn't be confused with $L^2$ which has the eigenvalues $\ell(\ell+1)\hbar^2$ where $\ell=0,1,2,3,\dots$ $L^2$ may always be measured simultaneously with any $L_n$ and in that case, $m\in\{-\ell, -\ell+1,\dots, \ell-1,\ell\}$.

2. The expectation value of any operator may be any number from the interval between the lowest and highest eigenvalue. So if we measure $L^2$ and $L_n$ at the same moment, then $\langle L_n\rangle$ may be any real number between $-\ell$ and $+\ell$.

Note that for every $\vec n$, the spectrum of $L_n$ is the same. For all choices of $\vec n$, the operators $L_n$ are conjugate to each other i.e. $$\exists U\in U({\mathcal H}):\quad L_{n'} = U L_n U^\dagger$$ Here, the operator $U$ is an operator on the Hilbert space that represents a rotation that turns the $\vec n$ axis to $\vec n'$ (passively or actively, one would have to be careful).

The corresponding eigenstates of $L_n$ and $L_{n'}$ are also conjugate to each other – but the detailed sets of eigenvectors are different. So when we measure $L_n$, we bring the system to one of the basis vectors of $L_n$ by the measurement, and if we measure $L_{n'}$, the candidate post-measurement states are elements of the basis of different eigenvectors.

• One question about 2 though, if we are talking about the expectation value of $L_n$ in a simultaneous eigenstate of $L_z$ and $L^2$, so $\langle l m_l \vert L_n \vert l m_l \rangle$, does your answer to 2) remain the same? May 26, 2016 at 8:53
• I was and had to be talking about simultaneous eigenstates of $L^2$ and $L_n$, otherwise the symbol $\ell$ wouldn't have meant anything. On the contrary, if you didn't guarantee that the state was an eigenstate of $L^2$, I could only say that the expectation value of $L_n$ is any real number because $\ell$ would be unbounded. May 26, 2016 at 8:56
• Very clarifying answer +1! May 26, 2016 at 8:57