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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 $E$?

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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.

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  • $\begingroup$ 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

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