2
$\begingroup$

In thermodynamics, we assume that an isolated system has an internal energy $U$, and if there is no heat given, or work done by/to the the system, this energy is constant.

However, in QM, we do know that, as long as we don't do a measurement, the particle do not acquire a definite (up to experimental error) energy, so how can we talk about a closed system having a constant total internal energy $U$ in that case ?

$\endgroup$
  • 1
    $\begingroup$ Is there something wrong with replacing "constant total internal energy" with "constant expectation value of total internal energy" for QM? $\endgroup$ – ACuriousMind Feb 3 at 13:43
  • $\begingroup$ @ACuriousMind I mean it can be done, but will the results change or stay the same ? (I don't know the answer) $\endgroup$ – onurcanbektas Feb 3 at 14:05
2
$\begingroup$

To be specific, consider a closed system of $N$ spinless particles described by the Schrodinger equation $$ i\frac{d}{dt}\psi(t,\mathbf{x}_1,\mathbf{x}_2,...,\mathbf{x}_N) =H\psi(t,\mathbf{x}_1,\mathbf{x}_2,...,\mathbf{x}_N) \tag{1} $$ where the Hamiltonian $H$ is $$ H = \frac{\mathbf{P}_1^2}{2m_1} + \frac{\mathbf{P}_2^2}{2m_1} + \cdots + \frac{\mathbf{P}_N^2}{2m_N} \tag{2} $$ with $\mathbf{P}_n=-i\hbar\nabla_n$ being the momentum operator for the $n$-th particle, and $\nabla_n$ is the gradient with respect to $\mathbf{x}_n$. For simplicity, equation (2) assumes that the particles are non-interacting, but interaction terms $V(\mathbf{x}_j-\mathbf{x}_k)$ may be included if desired.

(To represent a system of identical bosons or fermions, simply set all the masses equal to each other and require that the wavefunction $\psi$ be totally symmetric or antisymmeric, respectively.)

The Hamiltonian is the observable corresponding to the system's total energy. A function $\psi$ that satisfies $$ H\psi(t,\mathbf{x}_1,\mathbf{x}_2,...,\mathbf{x}_N) =E\psi(t,\mathbf{x}_1,\mathbf{x}_2,...,\mathbf{x}_N) \tag{3} $$ would represent a state of definite energy $E$, and equation (1) says that if $\psi$ has definite energy initially, then it has that same definite energy forever.

In general, the state $\psi$ of interest won't have strictly definite energy, but any normalizable state $\psi$ can be written as a linear combination (superposition) of functions that do. Using $\langle\cdots\rangle$ to denote the expectation value of an operator "$\cdots$" in a given state $\psi$, we can define the mean energy $U\equiv\langle H\rangle$ and the standard deviation $$ \Delta U\equiv \sqrt{\langle H^2\rangle-\langle H\rangle^2}. $$ These definitions make sense whether or not any measurements have been made, and equation (1) implies that the quantities $U$ and $\Delta U$ are both independent of time. Even if $\psi$ doesn't have a definite energy, as long as $\Delta U$ is small compared to $U$, we can still reasonably refer to $U$ as "the" internal energy of the system, with the understanding that this is just an (excellent) approximation. If desired, we could think of this as being the result of an imperfect measurement of $H$ (with finite resolution $\Delta U$), but that's not necessary.

Referring to $U$ as "the" internal energy of the system makes sense for the same reason that it makes sense to talk about the location of a book resting on a table, even though we know that the book is a quantum object that doesn't really have a strictly well-defined location. As long as the quantum spread in the book's location is neglible for most practical purposes, we can reasonably talk about "the" location of the book.

Whether or not $\psi$ has a definite total energy, the individual particles typically don't have well-defined energies. In other words, even if $\psi$ is an eigenstate of $H$, it generally won't be an eigenstate of the single-particle operator $\mathbf{P}_n^2/2m_n$. The individual particles generally don't even have wavefunctions of their own, because they are typically entangled with each other. In other words, the multi-particle wavefunction $\psi$ typically cannot be factorized into single-particle wavefunctions. Still, as long as $\Delta U$ is small compared to $U$, we can refer to $U$ as "the" internal energy of the system, as a good approximation.

Now, what if $U$ is not equal to any of the possible eigenvalues $E$ in (3)? As pointed out in a comment by the OP, this can occur if the system is restricted to a finite volume, in which case the particles' momenta (and therefore the total energy) are restricted to a discrete set of values. That's okay, because as long as the volume is large enough, the spacing between the allowed energies will be so fine that it might as well be continuous for all practical purposes. The definition $U=\langle H\rangle$ does not require that $U$ be precisely equal to any eigenvalue of $H$, nor does the idea that $U$ represents the system's total internal energy — with the understanding that this is just an (excellent) approximation.

Altogether, for all practical purposes, we can regard $U=\langle H\rangle$ as "the" total internal energy of the system as long as $\Delta U$ is sufficiently small. Measurement is not required for this to make sense, and equation (1) implies that this $U$ is indeed constant in time.

$\endgroup$
  • 1
    $\begingroup$ In this case then what would constitute a measurement of the entire energy of the system in order to determine $\langle U\rangle$ and $\Delta U$? Is this even possible, or is this where statistical mechanics/thermodynamics comes into play? $\endgroup$ – Aaron Stevens Feb 4 at 4:35
  • $\begingroup$ @AaronStevens That's a really good question, and I'm afraid I can't fit a good answer in a comment-box. In principle, you could determine the energy by measuring the observable $H$, and if this measurement were perfectly precise, you'd end up with $\Delta U=0$ afterward. In reality, a measurement requires interaction with some other part of the system so that the $N$-particle system itself would no longer be "closed," and then the trick is designing that interaction so that it specifically measures $H$ (necessarily with imperfect resolution) rather than merely, say, exchanging heat. $\endgroup$ – Dan Yand Feb 4 at 4:45
  • $\begingroup$ Thanks for the reply. One thing though, $\Delta U$ doesn't depend on how precise your measurement is, right? You can calculate what it is based on the quantum state your system is in. $\endgroup$ – Aaron Stevens Feb 4 at 4:50
  • 2
    $\begingroup$ @AaronStevens Right. The $\Delta U$ that the state has prior to measuring $H$ is just a calculable property of the wavefunction, and it is independent of (can be greater than or less than) the resolution of whatever measurement is applied later. After the measurement, $\Delta U$ may end up being different, if the resolution of the measurement is finer than the original $\Delta U$. $\endgroup$ – Dan Yand Feb 4 at 4:53

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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