Given an infinite potential square well with $0<x<L$, I need to calculate the uncertainties of position and momentum. The eigenstates in the position basis are $$\lvert E_n\rangle\to \psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right), n = 1,2,3,...$$

I know that the position operator is defined by $x$, and the momentum operator is defined by $-i\hbar\frac{\partial }{\partial x}$, but the inequality formula for uncertainty, which is $$\Delta A\Delta B\ge \frac{1}{2}\lvert \langle [A,B] \rangle\rvert$$ requires not operators for position and momentum, but corresponding observables. How can I get these observables?

  • $\begingroup$ I'm not sure what the problem is, observables are given by (Hermitian) operators. That is, you can just plug in the operators for position and momentum for $A$ and $B$. $\endgroup$ Jul 8, 2016 at 7:47
  • $\begingroup$ The confusion might come from that the right-hand-side should be applied to a given state. Right? Which is basically what the triangle brackets mean. Is that right? $\endgroup$
    – fffred
    Jul 8, 2016 at 7:49
  • $\begingroup$ @KennethGoodenough that should probably be an answer. $\endgroup$
    – David Z
    Jul 8, 2016 at 8:04
  • $\begingroup$ An example is: $$ <x^{2}> - (<x>)^{2} = \triangle x $$ $\endgroup$
    – user97261
    Jul 8, 2016 at 8:37

2 Answers 2


You can work in the position representation. It is not difficult:

$$\langle \psi \mid[x,p] \mid \psi \rangle = \langle \psi \mid (xp-px)\mid \psi \rangle =\\=\int dx \ \psi^*(x) \left(-i \hbar x \ \partial_x \psi(x) + i \hbar \ \partial_x (x \ \psi(x)) \right) = \\ =-i \hbar \int dx \ \psi^* \left(x \ \psi'-(\psi+x \ \psi')\right) = \\ =i \hbar \int dx \mid \psi(x)\mid^2 = i \hbar$$

What you have to realize is that the identities $\hat x = x$ and $\hat p = -i \hbar \partial_x$ are meaningful only if you express the operators in the position basis, that is to say

$$\langle x' \mid \hat p \mid \psi \rangle = -i \hbar \partial_{x'} \ \psi(x') \\ \langle x' \mid \hat x \mid \psi \rangle = x' \ \psi(x') $$


$$\langle x' \mid \psi \rangle = \psi(x')$$

is the wave function in the position basis. You could also choose the momentum representation:

$$\langle p' \mid \hat p \mid \psi \rangle = p' \tilde \psi(p')\\\langle p' \mid \hat x \mid \psi \rangle = i \hbar \partial_{p'} \tilde \psi(p')\\\langle p'\mid \psi \rangle =\tilde \psi(p')$$

The result will be the same.

In general, every time you have an hermitian operator $A$ with a complete set of eigenvectors $\{\mid a \rangle\}$, you can in principle express your state vectors (kets) and operators in the $\{\mid a \rangle\}$ basis, but the mathematical form of the operators can become cumbersome if you choose the "wrong" representation. For example, the momentum operator is a nice differential operator in the position basis $\{\mid x \rangle\}$, but could become ugly if expressed in another basis.

Going back to what you wanted to compute, we obtain

$$\Delta x \Delta p \geq \frac 1 2 \mid i \hbar \mid = \frac \hbar 2$$

which is the well-known Heisenberg uncertainty principle for position and momentum.

  • $\begingroup$ Do you mean the uncertainty is $i\hbar$ in this case? But shouldn't it be a real number? $\endgroup$
    – sequence
    Jul 8, 2016 at 16:08
  • $\begingroup$ @sequence You have to take the absolute value and divide by $2$, so it becomes $\hbar/2$ (look at the formula you wrote). $\endgroup$
    – valerio
    Jul 8, 2016 at 16:10
  • $\begingroup$ @sequence The result is $\Delta x \Delta p \geq \hbar/2$, which is the well-known Heisenberg uncertainty principle. I've included the calculation in my answer. $\endgroup$
    – valerio
    Jul 8, 2016 at 16:11

Observables are operators, in particular they are of the self-adjoint type (with discrete spectrum spanning the entire Hilbert space). Given a normalised state $|\psi\rangle$, the expectation value of an operator $A$ thereupon is defined as $\langle A \rangle = \langle\psi |\, A\, |\psi\rangle$; equivalently, one can prove that the uncertainty on the measurement of the operator $A$ onto the state $|\psi\rangle$ can be expressed as $$ \Delta A_{|\psi\rangle} = {\langle A^2 \rangle}_{|\psi\rangle} - {(\langle A\rangle)^2}_{|\psi\rangle}. $$

You are given the expression of the initial state and its wave function, hence you can calculate the expectation values of (any power) of the operators $\hat{x}$ and $\hat{p}$ by inserting the identity operator $1 = \int \textrm{d}x’ |x’ \rangle \langle x’ |$ on the left (respectively on the right) and perform the integrations. You should end up with the usual integrals of the wave function against its variable minus the derivative.

I know that the momentum operator is defined as $-i\hbar \partial_x$...

That is wrong. The momentum operator is $\hat{p}$ and one has $\langle x| \hat{p} |\psi\rangle = -i\hbar \partial_x \psi(x)$.


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