Generalized Uncertainty Principle

Recall the generalized uncertainty principle $\sigma_a \sigma_b \geq (1/2)<\psi|[A,B]|\psi>$. Note that the quantity $\sigma_a$ is known as the uncertainty in quantity $A$ (and similarly for $B$).

a) Suppose you knew quantity $A$ precisely for a quantum mechanical particle. Would the uncertainty be greater than, less than or equal to zero? Explain.

b) Suppose that you were to precisely know all three components of the angular momentum vector for a particle with $l \neq 0$. Prove that this is inconsistent with the uncertainty principle. 

c) Suppose a quantum mechanical particle has a well defined value of $L_z$ only. Starting from the left hand side of the uncertainty principle, determine the expectation value of $L_y$, $<\psi|L_y|\psi>$. Does your answer depend on the value of $L^2$ for the particle? Explain.

d) With which of the following statements, if either, do you agree? Explain.

Student 1. “If one component of the angular momentum is well defined, then the expectation value of the perpendicular components will be zero”.

Student 2. “I agree. that means that the probability of measuring zero for those other components will always be the largest, no matter which value we started with.”

The uncertainty principle for two observables $A$ and $B$ is ($σ$ are uncertainties)

$σ_A*σ_B≥1/2*|(<ψ|[A,B] |ψ>)|$

If one knows exactly the value of one of the observable $A$ , it means the uncertainty in its measurement is exactly zero: $σ_A=0$

One has the commutation relation

$[L_x,L_y ]=-i\hbar*L_z$   then $σ_{Lx}*σ_{Ly}≥\hbar/2*L_z$

so that if $L_z$ is exactly known (and nonzero), then always the uncertainties in $L_x$  and $L_y$  will be greater than zero. (Because $l>0$ at least one component that is exactly known need to be nonzero)

The uncertainty of an observable is defined as

$σ_A=√(<A^2>-<A>^2 )$

We begin by making the observation that

$<L^2>=<L_x^2>+<L_y^2>+<L_z^2>$  because angular momentum is a vector quantity

We know that

$<L^2>=l(l+1)*\hbar^2$   and  $<L_z^2>=m^2*\hbar^2$

From the symmetry of the problem one has also

$<L_x^2>=<L_y^2>$  so that  $<L_x^2>=(\hbar^2)/2*[(l(l+1)-m^2]$

Again, because the problem is symmetric one has

$σ_{L x}*σ_{Ly}=σ_{L x}^2=\hbar/2*|<ψ| L_z |ψ>|=\hbar/2*<L_z>=m\hbar*\hbar/2$

So that

$<L_x^2>-<L_x >^2=(m\hbar^2)/2$   that is $<L_x >^2=(\hbar^2)/2*[l(l+1)-m^2-m]$


It seems that $<L_x>$ depends on the value of $L^2=\hbar^2*l(l+1)$ but in reality this is not true. Explanation follows: if one know exactly the value of one component of $L$, for example $Lz$, then if you also are fixing the values of the two other perpendicular components $Lx$ and $Ly$ this means the electron needs to rotate in a fixed plane (all angular momentum components are exactly know). This is in contradiction with the uncertainty principle and quantum mechanics, and in reality if $Lz$ is known the electron is moving in the space of an inverted cone having the height along the z axis. So, always if you fix $Lz$, the expectation values of the other two components will be zero $<Lx>=<Ly> =0$

From the above discussion it follows that first student is right. (If you know one component of angular momentum the other two components will have an expectation value of zero). The second student is wrong since the probability to measure a value of an observable is one thing and the expectation value of that observable is another thing (it is the average of all measured values). Again, the electron is moving in the space of an inverted cone (positioned along z axis if $lz$ is fixed) so one particular measurement can yield a nonzero value of $Lx$ or $Ly$.