Does the PBR Theorem Rule out a Statistical Understanding of QM?

Abstract

The PBR theorem gives insight into how quantum mechanics describes a physical system. This paper explores PBRs’ general result and shows that it does not disallow the ensemble interpretation of quantum mechanics and maintains, as it must, the fundamentally statistical character of quantum mechanics. This is illustrated by drawing an analogy with an ideal gas. An ensemble interpretation of the Schrödinger cat experiment that does not violate the PBR conclusion is also given. The ramifications, limits, and weaknesses of the PBR assumptions, especially in light of lessons learned from Bell’s theorem, are elucidated. It is shown that, if valid, PBRs’ conclusion specifies what type of ensemble interpretations are possible. The PBR conclusion would require a more direct correspondence between the quantum state (e.g., $\left| {\psi \rangle } \right.$) and the reality it describes than might otherwise be expected. A simple terminology is introduced to clarify this greater correspondence.

Appendices

Appendix 1: Derivation of Statistics of Ideal Gas (N Particles, Total Energy = E ₁ at Temperature T)

(Note, on a glance, this title and the appendix itself might look like simply a standard derivation. However, it is not. It is written to address the issues raised by the PBR theorem about the nature of the statistical analysis of a physical system. Parts of the calculation are standard; however, it is approached and analyzed from a wholly unique perspective that, as mentioned in the text, affects the understanding, direction and final output of the calculation. Even the standard parts, in addition to being integral to the presentation, are needed to enable reflection on (and reference to) the key aspects of a statistical description of a physical system.)

As is evident from Eq. (1), there are many ways to assign the energy of each particle so as to get the total energy E₁, indeed infinitely many ways. To avoid the infinities, start with energy bins of finite size; later, we can take the continuum limit. Divide the energy axis from zero to E₁ into M bins, and label the energy of a bin by the energy of the middle of the bin. Thus, the ith bin has energy ${{\epsilon}_{i}}$.⁴⁴ We take M to be very large and we take $N \gg M$. Now, taking n_i to be the number of particles in the ith bin, the energy and particle number constraint are:

$$ \sum\limits_{i=1}^{M}{{{n}_{i}}{{\epsilon}_{i}}}=E,\quad \sum\limits_{i=1}^{M}{{{n}_{i}}}=N $$

(7)

Any possible way of arranging the particles that satisfies these constraints is a possible physical state for the gas of temperature T. We call each possible state a microstate. However, we divide these states up into classes of intermediate states. These intermediate states are simply defined by how we bin the particles. We specify the binning by listing the number of particles in each energy state. Thus, each intermediate state is defined by a list that looks like: $\{n_{1},n_{2},n_{3} \ldots \} \equiv \{n_{i} \}$. Clearly, there are many different arrangements of particles that can satisfy any given binning assignment. Indeed, combinatorics shows that the number of microstates, $\varOmega$, in a given intermediate state, $\lambda \equiv \{{{n}_{i}}\}$ is given by $\varOmega \left({\{n_{i} \}} \right) \equiv \frac{N!}{{n_{1} !n_{2} !n_{3} ! \ldots n_{M} !}}$. The differences between the various microstates that make up an intermediate state are ignored. After all, the difference between microstates within one intermediate state just arises by switching the particles with each other. One can change which particles are where without changing the number in each bin. For example, we could switch particle 2400 in $n_{1}$ with particle 22 which is currently in $n_{49}$without affecting the number of particles in each bin. This makes no difference, because, at our level of analysis, a particle with a given energy acts the same as another particle with the same energy. Thus, an ensemble of microstates makes up an intermediate state. We will not attempt to completely specify this ensemble, but we will need the number of states in it, i.e. $\varOmega \left({\{n_{i} \}} \right)$ .

Now, many intermediate states can be assigned to the same macro-state, i.e. something evident on ordinary human scales. That is, the differences between the various intermediate states are not evident on that ordinary scale. Hence, we can consider an ensemble of intermediate states to represent a given macro-state, such as our gas with energy E. When we look in further detail at a given macro-state, we can find any member of this ensemble. This is what triggered our trek to understand the statistics of an ideal gas. But, we are not yet done.

Which member of the ensemble is most likely to be picked. Answer: the one with the most number of microstates associated with it. Thus, we need to maximize $\varOmega \left({\{n_{i} \}} \right)$with respect to $\{n_{i} \}$. Because an additive quantity is easier to work with than a muliplicative one, we define the entropy, S as:

$$ S \equiv k\ln \varOmega \left({\{n_{i} \}} \right) $$

(8)

Now, using Lagrange multipliers to incorporate the constraints from Eq. (7), our maximization condition (δS = $\sum\nolimits_{i = 1}^{M} {\frac{\delta S}{{\delta n_{i}}}\delta n_{i}}$) becomes:

$$ \delta S=k\delta \left(\ln N!-\ln {{n}_{1}}!-\ln {{n}_{2}}! \ldots .-\ln {{n}_{M}}!-\alpha \sum\limits_{i=1}^{M}{{{n}_{i}}}-\beta \sum\limits_{i=1}^{M}{{{n}_{i}}}{{\epsilon}_{i}} \right)=0. $$

(9)

Using Sterling’s approximation ($\ln m! \approx m\ln m$, for large m)⁴⁵ by taking $n_{i}$ large and $\delta N = \delta \sum\limits_{i = 1}^{M} {n_{i}} = \sum\limits_{i = 1}^{M} {\delta n_{i}} = 0$, $\delta E=\delta \sum\limits_{i=1}^{M}{{{n}_{i}}}{{\epsilon}_{i}}=\sum\limits_{i=1}^{M}{{{\epsilon}_{i}}\delta {{n}_{i}}}=0$ gives:

$$ \delta S=k\delta \left(N\ln N-\sum\limits_{i=1}^{M}{{{n}_{i}}\ln {{n}_{i}}}-\alpha \sum\limits_{i=1}^{M}{{{n}_{i}}}-\beta \sum\limits_{i=1}^{M}{{{n}_{i}}}{{\epsilon}_{i}} \right)=0 $$

(10)

so: $\left(-\sum\limits_{i=1}^{M}{\delta {{n}_{i}}\ln {{n}_{i}}}-\sum\limits_{i=1}^{M}{\delta {{n}_{i}}}-\alpha \sum\limits_{i=1}^{M}{\delta {{n}_{i}}}-\beta \sum\limits_{i=1}^{M}{\delta {{n}_{i}}}{{\epsilon}_{i}} \right)=0$.

$$ \left(\sum\limits_{i=1}^{M}{\delta {{n}_{i}}\ln {{n}_{i}}}+\sum\limits_{i=1}^{M}{\delta {{n}_{i}}}+\alpha \sum\limits_{i=1}^{M}{\delta {{n}_{i}}}+\beta \sum\limits_{i=1}^{M}{\delta {{n}_{i}}}{{\epsilon}_{i}} \right)=\sum\limits_{i=1}^{M}{\left(\alpha +\beta {{\epsilon}_{i}}+\ln {{n}_{i}} \right)\delta {{n}_{i}}}=0. $$

So, we get: $\alpha +\beta {{\epsilon}_{i}}+\ln {{n}_{i}}=0$, or ⁴⁶:

$$ {{n}_{i}}={{e}^{-\alpha}}{{e}^{-\beta {{\epsilon}_{i}}}} $$

(11)

(Note: if we took the more accurate version of Sterling’s approximation, i.e. $\ln m! \approx m\ln m - m + \frac{1}{2}\ln 2\pi m$, we would get: ${{n}_{i}}{{e}^{\frac{1}{4\pi {{n}_{i}}}}}={{e}^{-\alpha}}{{e}^{-\beta {{\epsilon}_{i}}}}$. This is actually M equations, one for each of the $n_{i}$’s. We take ${{\epsilon}_{i}}\equiv i\,{{E}_{1}}/M$and we also take M to be fixed. $a \equiv e^{- \alpha}$ is a normalization factor that is fixed by the number constraint: $N=\sum\nolimits_{1}^{M}{{{n}_{i}}}$. We also have the energy constraint: ${{E}_{1}}=\sum\nolimits_{1}^{M}{{{n}_{i}}{{\epsilon}_{i}}}$. So, we have $M + 2$ equations and $M + 2$unknowns (M n_i’s, a, β and E₁). In theory, we could solve the system of equations iteratively as follows. For a given, $n_{i}$, say $n_{1}$, we substituting chosen values for M and E₁ and, we pick a trial a and a trial β and numerically find n_i (say by graphical methods). We, then, do the same to find all the n_i’s. These should satisfy the number and energy constraints. If they do not we iterate our choices for a and β until the result comes out correctly.

The important point here is that for fixed E₁, we have a single β. That is, there is, even for low N, a one to one correspondence the total system energy and β (note: β characterizes the highest probability energy-binning state and will become 1/kT in the large N limit.)

Now, we’d like to take the continuum limit of the bins, which involves $M \to \infty$. To make this easier, shift the sums above to sums from $i = 0$ to $i = M - 1$. This means, applying our system of binning (which could have been anything up till now) that we gave above, we get: ${{\epsilon}_{i}}=i{{E}_{1}}/M$. We thus have:

$$ n_{i} = {e^{- \alpha}} e^{{- \beta iE_{1}/M}},so \, N = e^{- \alpha} \sum\nolimits_{i = 0}^{M - 1} {e^{{- \beta iE_{1}/M}}}. $$

(12)

where we take $a={{e}^{-\alpha}}$. Then, noting that the geometric series partial sum is: $\sum\nolimits_{i = 0}^{M - 1} {ar^{i}} = a\left({\frac{{1 - r^{M}}}{1 - r}} \right)$, taking $r = e^{{- \beta E_{1}/M}}$, we get: $N = a\left({\frac{{1 - e^{{- \beta E_{1}}}}}{{1 - e^{{- \beta E_{1}/M}}}}} \right)$; this implies: $a = N\left({\frac{{1 - e^{{- \beta E_{1}/M}}}}{{1 - e^{{- \beta E_{1}}}}}} \right)$. Note how:

$$ n_{i} = \mathop {{\rm lim}}\limits_{M \to \infty} ae^{{- \beta iE_{1}/M}} \sim \mathop {{\rm lim}}\limits_{M \to \infty} N\left({\frac{{1 - \left({1 - \beta E_{1}/M} \right)}}{{1 - e^{{- \beta E_{1}}}}}} \right)e^{{- \beta iE_{1}/M}} = \mathop {{\rm lim}}\limits_{M \to \infty} \frac{N}{M}\left({\frac{{\beta E_{1}}}{{1 - e^{{- \beta E_{1}}}}}} \right) \to 0. $$

As the bin gets arbitrarily small (i.e., $M \to \infty$), this is to be expected as this is the only way that the total number of particles can be finite as the number of bins approaches infinity. But, recalling our earlier requirement, we want $N \gg M$, so to maintain this (and thus also maintain our approximation that $n_{i}$is large), we would have to let N also go to infinity. We want to go to the continuum limit in the energy-binning, but, of course, we do not want to have an infinite number of particles. This leads us to back to the core of the matter; the bin occupation number n_i is a way of asking how many particles are in a given energy range (with a fixed number of particles). This fact is implicit in the case of a finite number (M) of bins. We need to be more explicit in the limiting case.

Thus, to take the $M \to \infty$limit, we consider, instead of the number of particles in a bin, the number of particles with a given energy in an arbitrarily small region $\Delta \epsilon$. That is, we define:

$$ \rho \equiv \underset{\Delta \epsilon \to \,0}{\mathop{{\rm lim}}}\,\frac{{{n}_{i}}}{\Delta \epsilon}=\underset{M\to \,\infty}{\mathop{{\rm lim}}}\,\frac{a{{e}^{-\beta i{{E}_{1}}/M}}}{{{E}_{1}}/M}. $$

(13)

Now, calculating the total number of particles, we get, substituting for a and using the Riemann definition of an integral:

$$ \begin{aligned} {\text{Total number of particles }} & = \mathop {\lim }\limits_{M \to \,\infty } \frac{M}{{E_{1} }}\sum\limits_{i = 0}^{M - 1} {ae^{{ - \beta iE_{1} /M}} } \frac{{E_{1} }}{M} \\ & = \mathop {\lim }\limits_{M \to \,\infty } \frac{M}{{E_{1} }}\sum\limits_{i = 0}^{M} {\frac{N}{M}\left( {\frac{{\beta E_{1} }}{{1 - e^{{ - \beta E_{1} }} }}} \right)e^{{ - \beta \left( {iE_{1} /M} \right)}} } \frac{{E_{1} }}{M} \\ & = N\left( {\frac{\beta }{{1 - e^{{ - \beta E_{1} }} }}} \right)\int_{0}^{{E_{1} }} {e^{{ - \beta\epsilon \text{}}} d\epsilon\text{}} = N\left( {\frac{\beta }{{1 - e^{{ - \beta E_{1} }} }}} \right)\left( {\frac{{1 - e^{{ - \beta E_{1} }} }}{\beta }} \right) = N \\ \end{aligned} $$

(14)

which confirms our calculation by giving a result of N.

Now, by comparing $N=\int_{0}^{{{E}_{1}}}{\rho \left(\epsilon \right)}d\epsilon$ with the last line above, we can write the continuum limit of our binning result:

$$ \rho \left(\epsilon \right)=N\left(\frac{\beta}{1-{{e}^{-\beta {{E}_{1}}}}} \right){{e}^{-\beta \epsilon}} $$

(15)

Connecting this to the empirically verified Maxwell Boltzmann distribution validates our assumptions and gives us $\beta = 1/kT$, so that we have:

$$ \rho \left(\epsilon \right)=\frac{N}{kT}\frac{1}{\left(1-{{e}^{-{{E}_{1}}/kT}} \right)}{{e}^{-\epsilon/kT}} $$

(16)

Note the total energy is:

$$ {{E}_{1}}=\int_{0}^{{{E}_{1}}}{\epsilon \,\rho \left(\epsilon \right)d\epsilon}=\frac{N}{kT}\frac{1}{\left(1-{{e}^{-{{E}_{1}}/kT}} \right)}\int_{0}^{{{E}_{1}}}{\epsilon \,{{e}^{-\epsilon/kT}}d\epsilon}=\frac{N{{E}_{1}}}{\left(1-{{e}^{{{E}_{1}}/kT}} \right)}+NkT $$

(17)

For fixed T, graphing, $E_{1}$and $\frac{{NE_{1}}}{{\left({1 - e^{{E_{1}/kT}}} \right)}} + NkT$ reveals that there is a solution at $E_{1} = 0$, which is physically invalid and there is a solution near⁴⁷

$$ E_{1} = NkT\, {\text{for large}}\,N. $$

(18)

At E₁=NkT, $E_{1} - \left({\frac{{NE_{1}}}{{\left({1 - e^{{E_{1}/kT}}} \right)}} + NkT} \right) = \frac{{N^{2} kT}}{{e^{N} - 1}}$. So, for small N, this solution is far off but for our case of large N, it is a good approximation; in particular, $\frac{N^{2} kT}{e^{N}-1} \ll {E_{1}}$ implies, for $N \gg 1, N^{2} \ll \frac{{e^{N} - 1}}{kT}E_{1} \approx e^{N} \left({\frac{{E_{1}}}{kT}} \right)$ or ${{N}^{2}}{{e}^{-N}} \ll \left(\frac{{{E}_{1}}}{kT} \right)$. For macroscopic values, i.e. $N \approx 6.02 \times 10^{23} \approx 10^{24}$, the log of the left hand side is: $2\ln N - N \approx - 10^{24}$, so $N^{2} e^{- N} \approx 10^{{- 10^{24}}}$, an incredibly small number. Hence, only when the temperature is extremely large or the total system energy is extremely small is this solution not approximately valid.

As mentioned in the text of the article, the important general point is that: for an isolated idea gas (as described in text) at given a temperature, there is a single value for the total energy, establishing the correspondence between temperature and total energy.

Appendix 2 (Imposing a Different Binning Requirement)

Let’s see what happens if we let our system of total energy E₁ be such that no single particle can have an energy less than $\varepsilon_{0}$. We proceed as above.

To take the limit of $M \to \infty$, we first apply this different binning to get: ${{\epsilon}_{i}}={{\varepsilon}_{0}}+i\left({{E}_{1}}-{{\varepsilon}_{0}} \right)/M$. We thus have, using equation ${{n}_{i}}={{e}^{-\alpha}}{{e}^{-\beta {{\epsilon}_{i}}}}$:

$$ n_{i} = e^{- \alpha} e^{{- \beta \left({\varepsilon_{0} + i\left({E_{1} - \varepsilon_{0}} \right)/M} \right)}} = e^{{- \left({\alpha + \beta \varepsilon_{0}} \right)}} e^{{- \beta i\left({E_{1} - \varepsilon_{0}} \right)/M}}, $$

(19)

so $N=a\sum\nolimits_{i=0}^{M-1}{{{e}^{-i\beta \left({{E}_{1}}-{{\varepsilon}_{0}} \right)/M}}}$, where we take $a = e^{{- \left({\alpha + \beta \varepsilon_{0}} \right)}}$.

Then, noting that the geometric series partial sum is: $\sum\nolimits_{i=0}^{M-1}{a{{r}^{i}}}=a\left(\frac{1-{{r}^{M}}}{1-r} \right)$, taking $r = e^{{- \beta \left({E_{1} - \varepsilon_{0}} \right)/M}}$and, we get: $N = a\left({\frac{{1 - e^{{- \beta \left({E_{1} - \varepsilon_{0}} \right)}}}}{{1 - e^{{- \beta \left({E_{1} - \varepsilon_{0}} \right)/M}}}}} \right)$; this implies: $a = N\left({\frac{{1 - e^{{- \beta \left({E_{1} - \varepsilon_{0}} \right)/M}}}}{{1 - e^{{- \beta \left({E_{1} - \varepsilon_{0}} \right)}}}}} \right)$.

Note how:

$$ \begin{aligned} n_{i} & = \mathop {\lim }\limits_{M \to \infty } ae^{{ - \beta iE_{1} /M}} \sim \mathop {\lim }\limits_{M \to \infty } N\left( {\frac{{1 - \left( {1 - \beta E_{1} /M} \right)}}{{1 - e^{{ - \beta E_{1} }} }}} \right)e^{{ - \beta iE_{1} /M}}\\ &\quad =\mathop {\lim }\limits_{M \to \infty } \frac{N}{M}\left( {\frac{{\beta E_{1} }}{{1 - e^{{ - \beta E_{1} }} }}} \right)e^{{ - \beta iE_{1} /M}} \to 0 \\ n_{i} & = \mathop {\lim }\limits_{M \to \infty } ae^{{ - \beta i\left( {E_{1} - \varepsilon_{0} } \right)/M}} \sim \mathop {\lim }\limits_{M \to \infty } N\left( {\frac{{1 - \left( {1 - \beta \left( {E_{1} - \varepsilon_{0} } \right)/M} \right)}}{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}} \right)e^{{ - \beta i\left( {E_{1} - \varepsilon_{0} } \right)/M}} \\ & = \mathop {\lim }\limits_{M \to \infty } \frac{N}{M}\left( {\frac{{\beta \left( {E_{1} - \varepsilon_{0} } \right)}}{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}} \right)e^{{ - \beta i\left( {E_{1} - \varepsilon_{0} } \right)/M}} \to 0 \\ \end{aligned} $$

As the bin gets arbitrarily small (i.e., $M \to \infty$), this is to be expected as this is the only way that the total number of particles can be finite as the number of bins approaches infinity. As above, to take the $M \to \infty$limit, we thus consider instead of the number of particles in a bin, the number of particles with a given energy in an arbitrarily small region $\Delta \epsilon$. That is, we define, using $n_{i} = ae^{{- \beta i\left({E_{1} - \varepsilon_{0}} \right)/M}}$:

$$ \rho \equiv \underset{\Delta \epsilon \to \,0}{\mathop{{\rm lim}}}\,\frac{{{n}_{i}}}{\Delta \epsilon}=\underset{M\to \,\infty}{\mathop{{\rm lim}}}\,\frac{a{{e}^{-\beta i\left({{E}_{1}}-{{\varepsilon}_{0}} \right)/M}}}{\left({{E}_{1}}-{{\varepsilon}_{0}} \right)/M}. $$

(20)

Now, calculating the total number of particles, we get, substituting for a and using the Riemann definition of an integral:

$$ \begin{aligned} {\text{Total number of particles }} & = \mathop {\lim }\limits_{M \to \,\infty } \frac{M}{{\left( {E_{1} - \varepsilon_{0} } \right)}}\sum\limits_{i = 0}^{M - 1} {ae^{{ - \beta i\left( {E_{1} - \varepsilon_{0} } \right)/M}} } \frac{{\left( {E_{1} - \varepsilon_{0} } \right)}}{M} \\ & = \mathop {\lim }\limits_{M \to \,\infty } \frac{M}{{\left( {E_{1} - \varepsilon_{0} } \right)}}\sum\limits_{i = 0}^{M - 1} {\frac{N}{M}\left( {\frac{{\beta \left( {E_{1} - \varepsilon_{0} } \right)}}{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}} \right)}\\&\qquad \times \,{e^{{ - \beta i\left( {E_{1} - \varepsilon_{0} } \right)/M}} } \frac{{\left( {E_{1} - \varepsilon_{0} } \right)}}{M} \\ & = N\left( {\frac{\beta }{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}} \right)\int_{0}^{{E_{1} - \varepsilon_{0} }} {e^{{ - \beta\epsilon \text{}}} d\epsilon\text{}}\\ & = N\left( {\frac{\beta }{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}} \right)\int_{{\varepsilon_{0} }}^{{E_{1} }} {e^{{ - \beta \left( {\text{} \epsilon - \varepsilon_{0} } \right)}} d\epsilon\text{}} \\ & = N\left( {\frac{\beta }{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}} \right)\left( {\frac{{1 - e^{{ - \beta \left( {E_{1} - \varepsilon_{0} } \right)}} }}{\beta }} \right) = N \\ \end{aligned} $$

(21)

which confirms our calculation by giving a result of N.

Now, by comparing $N=\int_{0}^{{{E}_{1}}}{\rho \left(\epsilon \right)}d\epsilon$with the last line above, we can write the continuum limit of our binning result, using $\beta = 1/kT$:

$$ \rho \left(\epsilon \right)=\frac{N}{kT}\left(\frac{1}{1-{{e}^{-\left({{E}_{1}}-{{\varepsilon}_{0}} \right)/kT}}} \right){{e}^{-\left(\epsilon -{{\varepsilon}_{0}} \right)/kT}} $$

(22)

Note the total energy is:

$$ E_{1} = \int_{\varepsilon_{0}}^{{E_{1} }} {\epsilon \,\rho \left( \epsilon \right)d\epsilon } = \frac{N}{{kT}}\frac{1}{{\left( {1 - e^{{ {-(E_{1} - \varepsilon_{0} )} /kT}} } \right)}}\int_{\varepsilon_{0}}^{{E_{1} }} {\epsilon e^{{ - (\epsilon-{\varepsilon_{0}} /kT}} d\epsilon } = \frac{{N\left( {E_{1} - \varepsilon_{0} e^{{\left( {E_{1} - \varepsilon_{0} } \right)/kT}} } \right)}}{{\left( {1 - e^{{\left( {E_{1} - \varepsilon_{0} } \right)/kT}} } \right)}} + NkT $$

(23)

For fixed T, graphing, E₁ and $\frac{{N\left( {E_{1} - \varepsilon_{0} e^{{\left( {E_{1} - \varepsilon_{0} } \right)/kT}} } \right)}}{{\left( {1 - e^{{\left( {E_{1} - \varepsilon_{0} } \right)/kT}} } \right)}} + NkT$ reveals that there is a solution at ${{E}_{1}}={{\varepsilon}_{0}}$, which is physically invalid and there is, again, a solution near:

$$ E_{1} = NkT + \varepsilon_{0} \, {\text{for large}}\,N. $$

(24)

At ${{E}_{1}}=NkT+{{\varepsilon}_{0}}$, $E_{1} - \left({\frac{{N\left({E_{1} - \varepsilon_{0} e^{{\left({E_{1} - \varepsilon_{0}} \right)/kT}}} \right)}}{{\left({1 - e^{{\left({E_{1} - \varepsilon_{0}} \right)/kT}}} \right)}} + NkT} \right) = \frac{{N(N(\varepsilon_{0} + kT) - \varepsilon_{0})}}{{e^{{N\left({\varepsilon_{0}/kT + 1} \right) - \varepsilon_{0}/kT}} - 1}}$. So, for small N, this solution is far off but for our case of large N, it is a good approximation; in particular, $\frac{{N(N(\varepsilon_{0} + kT) - \varepsilon_{0})}}{{\,e^{{N\left({\varepsilon_{0}/kT + 1} \right) - \varepsilon_{0}/kT}} - 1}} < < E_{1}$ implies, for $N\gg 1$, ${{N}^{2}}\ll {{E}_{1}}\frac{{{e}^{N\left({{\varepsilon}_{0}}/kT+1 \right)}}-1}{({{\varepsilon}_{0}}+kT)}\approx {{E}_{1}}\frac{{{e}^{N\left({{\varepsilon}_{0}}/kT+1 \right)}}}{({{\varepsilon}_{0}}+kT)}$, or $N^{2} e^{{- N\left({\varepsilon_{0}/kT + 1} \right)}} \ll \frac{{E_{1}}}{{(\varepsilon_{0} + kT)}}$,

Hence, for macroscopic size N, as before, only when the temperature is extremely large or the total system energy is extremely small is this solution not approximately valid. (Note: for small, i.e., $\varepsilon_{0} \ll kT$, the inequality becomes: $N^{2} e^{- N} \ll \frac{{E_{1}}}{kT}$, i.e., the same as before. For $\varepsilon_{0} > > kT$, we have$N^{2} e^{{- N\varepsilon_{0}/kT}} < < \frac{{E_{1}}}{{\varepsilon_{0}}}$; the smallest the right hand side can become is one since $E_{1} \ge \varepsilon_{0}$.)

Again, a key general point is that for a given temperature, there is always a single value of E₁ corresponding to it for any fixed N.