# Divergence of probabilities for a position eigenstate in an energy basis

If we have an electron in a one-dimensional infinite potential well, and we have measured its position and found it to be let's say at $$x=0$$ at the center of the well. The state vector after measurement becomes $$\ket\{x=0\}.$$ If we calculate the probability then to get any eigen-value of the energy we will find it to be $$\frac{1}{a}$$ where $$a$$ is the width of the well. But that means the total probability (the sum of infinite terms each equals $$\frac{1}{a}$$) is infinite which is absurd. So, what exactly is happening here please?

• The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.) Oct 21, 2018 at 14:06
• @Arthur Your state $|x=0\rangle$ is not normalized. Normalized states have unit probability. But unnormalized states don't have unit probability. There's no mathematical theory that says that an unnormalized states have probability add to one. The ONLY rigorous mathematical statement is that normalized states will have probabilities that add to one. Oct 21, 2018 at 15:57
• @Arthur $\langle x|n\rangle\langle n|x\rangle$ is NOT a probability, so you writing is at $P(n)$ is simply incorrect. The postulates of quantum mechanics say that for a normalized state you have $P(n)=\langle \psi|n\rangle\langle n|\psi\rangle$. If your state $|\psi\rangle$ is not normalized, then the formula you've written is meaningless. The formula for "probability" REQUIRES you to first normalize your state. Oct 21, 2018 at 23:37
• @Arthur Some authors will write $P(n)=\frac{|\langle n|\psi\rangle|^2}{\langle\psi|\psi\rangle}$, and then this formula holds for any state whether or not it's normalized. Of course, if you try to put in $|\psi\rangle=|x\rangle$, you'll simply get $\frac{\infty}{\infty}$, which is a little hard to interpret. But at least then you can see directly why considering $|\psi\rangle=|x\rangle$ is problematic for QM, and why we never consider a particle to actually be in the state $|x\rangle$. Oct 21, 2018 at 23:39
• @Arthur You cannot normalize the position eigenstate. It is not a normalizable function. That is precisely why you can never have a particle in a position eigenstate. Actual, physical wavefunctions are always normalizable. The position eigenstates are useful for EXPANDING physical wavefunctions and doing FORMAL MANIPULATIONS, but are not physical states a particle can be in. Oct 21, 2018 at 23:41

When the distribution of probabilities is continuous, the probability of finding an object exactly at one specific position is zero. You can calculate what is the probability of finding the object at a certain interval, you have a probability density. The right way to treat the probability is $$dP(x)=dx/a$$, and the total probability now becomes 1, because $$\int dP(x)=\int^a_0 dx/a=1$$. For more details see https://en.wikipedia.org/wiki/Particle_in_a_box

• But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a. Oct 21, 2018 at 14:08
• The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– user65081
Oct 21, 2018 at 14:11
• you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– user65081
Oct 21, 2018 at 14:16
• The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend. Oct 21, 2018 at 14:25