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Say we have a particle in an infinite deep well that is $V(x)=\left\{\begin{array}{ll}0 & 0 \leq x \leq L \\ \infty & \text { elsewhere }\end{array}\right.$.

The energies corresponding to various states are given as $E_n=\frac{n^{2} \pi^{2} \hbar^{2}}{2 m L^{2}}$.This means that the particle can have different energies upon different measurements. But this goes against the rule that the total energy of a system remains constant,because if I measure the energy now it's something and later on it may be something else, violating conservation of energy! Where am I wrong?

K7PEH
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Kashmiri
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    Take any superposition state you wish, and work out the variation in time of the expectation value of the energy in this state. Is it non-vanishing? – Cosmas Zachos Aug 29 '20 at 14:48
  • if you start in the state $\psi_n(x)$ with energy $E_n$, the system will remain in this state. Energy is conserved. – ZeroTheHero Aug 29 '20 at 14:50
  • @ZeroTheHero thank you. But I've studied that as soon as we stop measuring, the wavefunction evolves again and upon next measurement we might find the particle in some other state. – Kashmiri Aug 29 '20 at 15:00
  • yes but if you start with a state of definite energy the system evolves "only to that state" and energy is conserved. – ZeroTheHero Aug 29 '20 at 15:29
  • (or at least only to another state of the same energy) – ZeroTheHero Aug 29 '20 at 15:36
  • "but if you start with a state of definite energy the system evolves "only to that state""is that a postulate of Quantum mechanics? – Kashmiri Aug 29 '20 at 16:21
  • You are right that this seems like a contradiction. The energy is a special measurement: in an isolated system, once a particular eigenvalue of the Hamiltonian (the energy) is measured, the system evolves in such a way that repeated measurement of the Hamiltonian returns the same eigenvalue indefinitely. – Charlie Aug 29 '20 at 16:38
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    The evolution operator is $\exp(-i t H/\hbar)$ so for an eigenstate of $H$, i.e. a state $\psi_n(x)$ of definite energy, $H\psi_n(x)=E_n\psi(x)$ and the evolution $\exp(-i t H/\hbar)\psi_n(x)= \exp(-i t E_n/\hbar)\psi_n(x)$ without changing $\psi_n(x)$. – ZeroTheHero Aug 29 '20 at 16:48

2 Answers2

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What is a law in physics?

A law is a distillation of observations and measurements that when imposed on the mathematical models of physics, picks up those solutions that fit the data which imposed the law, so that there is a predictive physics theory.

In classical physics theories the laws of conservation of energy, together with the conservation ofmomentum and angular momentum have the strength of axioms, and have always been found true.

In quantum mechanics, there is a different connection between measurements in the lab and and the variables used in the theory, because the predictions are on the probability of observing a four vector $(p_x,p_y,p_z,E)$ for a given particle. The experiment you designed is theoretical , your particle cannot be observed. To be observed there should be an interaction , which cannot happen in an infinite potential well with one particle.

In the comments it is pointed out that if your problem is correctly mathematically formulated there is no violation of energy conservation, because the probable states occupied by the particle are not an observable. There is a probability that the particle is in one of the given energy states, but you have no method of measuring it. Wavefunctions are not observable, only $Ψ^*Ψ$ is observable , and it is a distribution of many measurements.

If it happens to be in that eigenvalue state, it will be stable there as the comments say.

anna v
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If you begin with a state $\psi(x,t)$ and perform an energy measurement, you will find the state to be in one of the energy eigenstates in the decomposition of $\psi$. Since, after a measurement, the state of the system collapses to an energy eigenstate and for such a state, evolution by Hamiltonian just introduces a phase to the wavefunction subsequent measurements will give the same result for the value of energy, consistent with energy conservation.

In quantum mechanics, however, you can think the energy conservation in terms of expectation values when you are not performing any measurements. Consider an operator $A$ and state $\psi$. Then we have

$ \frac{d\langle A\rangle_{\psi}}{dt} = i\langle [H,A]\rangle_{\psi} + \langle \frac{\partial A}{\partial t} \rangle_{\psi} $

For, the case when $A=H$ and $H$ does not depend on time explicitly, we have that

$ \frac{d\langle H\rangle_{\psi}}{dt} =0$

Viraj Meruliya
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