This question arises from a discussion I recently had with my friend. We were talking about a particle in an infinite potential well. The particle is in an arbitrary wavefunction $\Psi$. When one measures the energy of the particle, then he gets a specific value $E_1$, which is an eigen value corresponding to an eigen-state of the particle $\psi_1$, to which the system has been forced to be in by making the measurement. The part that tricked us is: How can anyone know with definite certainty what the value of energy is for a quantum particle? Shouldn't there exist some uncertainty in the value of $E$?


Do you refer to the energy-time principle involving the product $\Delta E \Delta t?$ Here, $\Delta E$ is the resulting standard deviation you obtain on measuring $E$ in an infinite number of identically prepared quantum systems. In the case of the system in an eigenstate of energy $E$ (through e.g a collapse as per the Copenhagen interpretation), you will always measure $E$ and so here the standard deviation $\Delta E = 0$.

Now, to satisfy the principle constraint it means $\Delta t$ is infinite. As emphasised within this thread,

What is $\Delta t$ in the time-energy uncertainty principle?

this is a reflection that there is no finite time in which one will measure a deviation in the energy of the state.

  • $\begingroup$ I was not referring to the energy-time uncertainty principle. I was just confused about the definiteness in the measurement of Energy. Another similar confusion I had was with the uncertainty in the measurement of x. If I were to locate the particle's position inside the potential well, it means $\Delta$x is $0$. But what does it mean to truly know where the particle exactly is? And if that does make sense, what does it mean for uncertainty in momentum to be infinite? I am very sure I am missing something quite fundamental here. $\endgroup$ – Ufomammut Dec 7 '18 at 23:20
  • $\begingroup$ The standard deviations in the uncertainty principles are indicative of the inherent randomness of the system, they are not a reflection of the capabilities of the measurement apparatus one uses. You can compute these standard deviations through the difference of expectation values and you’ll find that no eigenstate of the infinite well can saturate the Heisenberg uncertainty bound. $\endgroup$ – CAF Dec 8 '18 at 7:44

My advice is to remember at all times that quantum mechanics is just a set of rules for building mathematical models of reality, so some of its predictions may be features of the model rather than features of reality.

The QM model of a particle in an infinite potential well predicts a set of exact allowed energy values, corresponding to a set of eigenfunctions. The Copenhagen interpretation says that a measurement of the energy of the particle will cause its wave function to be one of the allowed eigenfunctions, and its energy to be the corresponding one of the exact energy values- no uncertainty about that whatsoever. Of course your experiment might not be exact so the reading you see on a dial somewhere might be slightly off the exact allowed value. And, of course, when you set-up your potential well, you might have made the width of the well slightly more or less than the value you plugged into the formula to calculate the allowed energy values. So the theory predicts exact answers, but reality is something else.


It may not be immediately apparent, but the answer does have to do with the energy-time principle. But it's a good question, because it exposes some implicit assumptions that are often made when working out solutions to the Schrodinger Equation.

The infinite potential well problem is well-known to students in introductory quantum mechanics, since when you plug that potential into the Schrodinger Equation you get something that is very easy to solve. But the question you're asking is, essentially, what is the Schrodinger Equation itself? What do its solutions mean?

To answer that question, you have to go back to some of the fundamental postulates of quantum mechanics. One of those postulates is that there is a wavefunction that describes everything about a particle, which has a couple properties:

  1. Just like a vector exists independently of whatever coordinate system you use to describe the vector, the wavefunction exists independently of whatever basis vectors you use to describe the wavefunction. Also, just like a vector can be described using different coordinate systems, the wavefunction can be expressed using different basis vectors, and there is a well-defined process for transforming between different basis vectors.
  2. When you choose a basis set, the square of the wavefunction describes a probability density function in that basis. So if, for example, you're using the position basis, the square of the wavefunction describes the probability for the position of the particle. This implies that the integral of the square of the wavefunction must be 1 (because the probability of it it being somewhere must be exactly 100%). In other words, the wavefunction must be square-integrable.

So going back to the Schrodinger Equation, that equation plays the role of the Hamiltonian: it represents the energy of a non-relativistic particle, and it also describes the time-evolution of the particle's wavefunction. The fact that it does both (energy and time-evolution) tells you that there is a fundamental connection between energy and time.

That fundamental connection is formally expressed by the statement that energy and time are canonically conjugate variables. (https://en.wikipedia.org/wiki/Conjugate_variables). That means to transform between the energy and the time basis, you perform a Fourier transform. Position and momentum are the same: they're also canonically conjugate variables.

However, a pure position state, or pure momentum state, or pure energy state, etc., is not allowable under the postulates of quantum mechanics. Think of if this way: if the energy were known with 100% accuracy to be some value $E$, it would be described by a (Dirac) delta function in the energy basis. But in the time basis (which is the Fourier transform of the energy basis), it would be a pure wave function $e^{i\omega t}$ (with $\omega = E/\hbar$), which is not square-integrable. You can get asymptotically close to to a pure state, but in principle you can't actually reach it.

So that's the answer: the solution of the Schroedinger equation for a particle in an infinite well is really the asymptotic limit of a definite energy state. But it's a limit; you can't actually achieve such a pure energy state in principle.

  • $\begingroup$ This is incorrect for two reasons. First, a quantum system can exist in an eigenstate of an observable as long as the spectrum of the observable is not purely continuous (which it is not, in this case). Second, and far more importantly, time is not a quantum mechanical observable. There is no time basis. $\endgroup$ – J. Murray yesterday
  • $\begingroup$ Stationary states are an asymptotic solution. They're a solution to the Schrodinger equation, but that's the same as saying ΔE = 0, or that you can Fourier transform a plane wave, or that a system can exist in an eigenstate of the Hamiltonian as long as the spectrum is not purely continuous. Those are all identical statements, and they're all asymptotic solutions. Nobody talks about a time basis because (unlike an energy basis) it's not useful. Still, the mathematics supports it: energy and time are canonically conjugate variables in the same way that position and momentum are. $\endgroup$ – Richter65 yesterday
  • $\begingroup$ I'm sorry, but that's just not right. Nobody talks about a time basis because time is not a quantum mechanical observable. There is no time operator, no time eigenstates, and no time basis. Energy and time are not canonically conjugate quantum variables because time is not a dynamical variable in quantum mechanics. $\endgroup$ – J. Murray yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.