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For more particles we would proceed analogously, with a somewhat higher complication.

First of all let me remark that not fermions alone exert a pressure when confined in a finite volume. Bosons do as well. Radiation pressure is an example, and photons are bosons. So let's compute the pressure exerted by a gas of non-interacting bosons at $0\,$K, when all particles are in the ground state (this isn't forbidden for bosons).

For those who find too abstract the above derivation I'll add a semiclassical one. In our box we have free particles bouncing back an forth between boundaries. Their momentum is $p=h/(2L)$. A particle hits one boundary (e.g. the left one) once in a time $${2L \over v} = {2mL \over p} = {4 m L^2 \over h}$$ and every time it exchanges with the boundary a momentum $2p$. Then the momentum exchanged per unit of time, i.e. the force, is $$f = 2p\, {h \over 4 m L^2} = {h^2 \over 4 m L^3}.$$ This holds for one particle. It's only left to multiply by $N$ to get (3)

For more particles we would proceed analogously, with a somewhat higher complication.

First of all let me remark that not fermions alone exert a pressure when confined in a finite volume. Bosons do as well. Radiation pressure is an example, and photons are bosons. So let's compute the pressure exerted by a gas of non-interacting bosons at $0\,$K, when all particles are in the ground state (this isn't forbidden for bosons).

For those who find too abstract the above derivation I'll add a semiclassical one. In our box we have free particles bouncing back and forth between boundaries. Their momentum is $p=h/(2L)$. A particle hits one boundary (e.g. the left one) once in a time $${2L \over v} = {2mL \over p} = {4 m L^2 \over h}$$ and every time it exchanges with the boundary a momentum $2p$. Then the momentum exchanged per unit of time, i.e. the force, is $$f = 2p\, {h \over 4 m L^2} = {h^2 \over 4 m L^3}.$$ This holds for one particle. It's only left to multiply by $N$ to get (3).

$\def\ket#1{|#1\rangle} \let\up=\uparrow \let\dn=\downarrow \def\PD#1#2{{\partial#1\over\partial#2}}$ There is no repulsion and no unexplained force. I would also add that PEP is an outdated way of describing the matter. In QM you should rather speak of antisymmetry* of fermion states. It's only when we build up a many particle state as a tensor product of one particle states that antisymmetry forces us to keep only different states for each single particle. A simple example with two particles will explain this (I hope).

Assume your particles are non-interacting* spin 1/2 fermions. Then above expression for energy eigenfunction is to be supplemented by specifying the spin state. Then Dirac's ket notation is preferable: $$\ket{n\up} \quad \hbox{or} \quad \ket{n\dn}$$ both belonging to $E_n$ eigenvalue.

If your system consists of just two particles, a set of base kets would be obtained by taking tenso products, which in Dirac's notation are written just putting two kets one after another. E.g. $$\ket{m\up} \ket{n\up} \quad \ket{m\up} \ket{n\dn} \quad \ket{m\dn} \ket{n\up} \quad \ket{m\dn} \ket{n\dn}$$ for all positive integers $m$, $n$. A shorthand may be used: $$\ket{m\up\,;\,n\up} \ \ket{m\up\,;\,n\dn} \ \ket{m\dn\,;\,n\up} \ \ket{m\dn\,;\,n\dn} \tag2$$ where labels preceding ";" refer to first particle, those following to the second.

$\def\ket#1{|#1\rangle} \let\up=\uparrow \let\dn=\downarrow \def\PD#1#2{{\partial#1\over\partial#2}}$ There is no repulsion and no unexplained force. I would also add that PEP is an outdated way of describing the matter. In QM you should rather speak of antisymmetry of fermion states. It's only when we build up a many particle state as a tensor product of one particle states that antisymmetry forces us to keep only different states for each single particle. A simple example with two particles will explain this (I hope).

Assume your particles are non-interacting spin 1/2 fermions. Then above expression for energy eigenfunction is to be supplemented by specifying the spin state. Then Dirac's ket notation is preferable: $$\ket{n\up} \quad \hbox{or} \quad \ket{n\dn}$$ both belonging to $E_n$ eigenvalue.

If your system consists of just two particles, a set of base kets would be obtained by taking tensor products, which in Dirac's notation are written just putting two kets one after another. E.g. $$\ket{m\up} \ket{n\up} \quad \ket{m\up} \ket{n\dn} \quad \ket{m\dn} \ket{n\up} \quad \ket{m\dn} \ket{n\dn}$$ for all positive integers $m$, $n$. A shorthand may be used: $$\ket{m\up\,;\,n\up} \ \ket{m\up\,;\,n\dn} \ \ket{m\dn\,;\,n\up} \ \ket{m\dn\,;\,n\dn} \tag2$$ where labels preceding ";" refer to first particle, those following to the second.

$\def\ket#1{|#1\rangle} \let\up=\uparrow \let\dn=\downarrow \def\PD#1#2{{\partial#1\over\partial#2}}$ There is no repulsion and no unexplained force. I would also add that PEP is an outdated way of describing the matter. In QM you should rather speak of antisymmetry of fermion states. It's only when we build up a many particle state as a tensor product of one particle states that antisymmetry forces us to keep only different states for each single particle. A simple example with two particles will explain this (I hope).

Consider particles in one dimension, constrained in a segment $0\le x\le L$ (what is usally called an "infinite potential well"). Energy eigenfunctions (standing waves) are sinusoidal waves vanishing at boundaries: $$\psi_n = \sin {n\,\pi\,x \over L} \qquad (n = 1,2,\ldots)$$ (these aren't normalized, but it's of no consequence for my present purposes.) The corresponding energy eigenvalues are $$E_n = {n^2 h^2 \over 8\,m\,L^2}.\tag1$$ A short derivation follows, which you may skip with no harm.

Assume your particles are non-interacting spin 1/2 fermions. Then above expression for energy eigenfunction is to be supplemented by specifying the spin state. Then Dirac's ket notation is preferable: $$\ket{n\up} \quad \hbox{or} \quad \ket{n\dn}$$ both belonging to $E_n$ eigenvalue.

If your system consists of just two particles, a set of base kets would be obtained by taking tenso products, which in Dirac's notation are written just putting two kets one after another. E.g. $$\ket{m\up} \ket{n\up} \qquad \ket{m\up} \ket{n\dn} \qquad \ket{m\dn} \ket{n\up} \qquad \ket{m\dn} \ket{n\dn}$$ for all positive integers $m$, $n$. A shorthand may be used: $$\ket{m\up\,;\,n\up} \qquad \ket{m\up\,;\,n\dn} \qquad \ket{m\dn\,;\,n\up} \qquad \ket{m\dn\,;\,n\dn} \tag2$$ where labels preceding ";" refer to first particle, those following to the second.

Observe however that if $m=n$ first and fourth expressions are identically zero, whereas second and third are the same apart for sign, thus representing the same state. This is the mathematical form PEP assumes in QM: for $m=n$ just one state exists for two particles, for $m\ne n$ there are four.

First of all let me remark that not fermions alone exert a pressure when confined in a finite volume. Bosons do as well. Radiation pressure is an example, and photons are bosons. So let's compute the pressure exerted by a gas of non-interacting bosons at $0\,$K, when all particles are in the ground state (this isn't forbidden for bosons).

If we have $N$ particles, overall energy is given by (1) taken for  $n=1$ and multiplied by $N$; $$E = {N h^2 \over 8\,m\,L^2}.$$ As we are in one dimension we'll speak of force, not of pressure. It's most easily computed by $$F = -\PD EL = {N h^2 \over 4\,m\,L^3}.\tag3$$

For those who find too abstract the above derivation I'll add a semiclassical one. In our box we have free particles bouncing back an forth between boundaries. Their momentum is $p=h/(2L)$. A particle hits one boundary (e.g. the left one) once in a time $${2L \over v} = {2mL \over p} = {4 m L^2 \over h}$$ and every time it exchanges with the boundary a momentum $2p$. Then the momentum exchanged per unit of time, i.e. the force, is $$f = 2p\, {h \over 4 m L^2} = {h^2 \over 4 m L^3}.$$ This holds for one particle. It's only left to multiply by $N$ to get  (3)

$\def\ket#1{|#1\rangle} \let\up=\uparrow \let\dn=\downarrow \def\PD#1#2{{\partial#1\over\partial#2}}$ There is no repulsion and no unexplained force. I would also add that PEP is an outdated way of describing the matter. In QM you should rather speak of antisymmetry* of fermion states. It's only when we build up a many particle state as a tensor product of one particle states that antisymmetry forces us to keep only different states for each single particle. A simple example with two particles will explain this (I hope).

Consider particles in one dimension, constrained in a segment $0\le x\le L$ (what is usally called an "infinite potential well"). Energy eigenfunctions (standing waves) are sinusoidal waves vanishing at boundaries: $$\psi_n = \sin {n\,\pi\,x \over L} \qquad (n = 1,2,\ldots)$$ (these aren't normalized, but it's of no consequence for my present purposes.) The corresponding energy eigenvalues are $$E_n = {n^2 h^2 \over 8\,m\,L^2}.\tag1$$ A short derivation follows, which you may skip with no harm.

Assume your particles are non-interacting* spin 1/2 fermions. Then above expression for energy eigenfunction is to be supplemented by specifying the spin state. Then Dirac's ket notation is preferable: $$\ket{n\up} \quad \hbox{or} \quad \ket{n\dn}$$ both belonging to $E_n$ eigenvalue.

If your system consists of just two particles, a set of base kets would be obtained by taking tenso products, which in Dirac's notation are written just putting two kets one after another. E.g. $$\ket{m\up} \ket{n\up} \quad \ket{m\up} \ket{n\dn} \quad \ket{m\dn} \ket{n\up} \quad \ket{m\dn} \ket{n\dn}$$ for all positive integers $m$, $n$. A shorthand may be used: $$\ket{m\up\,;\,n\up} \ \ket{m\up\,;\,n\dn} \ \ket{m\dn\,;\,n\up} \ \ket{m\dn\,;\,n\dn} \tag2$$ where labels preceding ";" refer to first particle, those following to the second.

Observe however that if $m=n$ first and fourth expressions are identically zero, whereas second and third are the same apart for sign, thus representing the same state. This is the mathematical form PEP assumes in QM: for $m=n$ just one state exists for two particles, for $m\ne n$ there are four.

First of all let me remark that not fermions alone exert a pressure when confined in a finite volume. Bosons do as well. Radiation pressure is an example, and photons are bosons. So let's compute the pressure exerted by a gas of non-interacting bosons at $0\,$K, when all particles are in the ground state (this isn't forbidden for bosons).

If we have $N$ particles, overall energy is given by (1) taken for  $n=1$ and multiplied by $N$; $$E = {N h^2 \over 8\,m\,L^2}.$$ As we are in one dimension we'll speak of force, not of pressure. It's most easily computed by $$F = -\PD EL = {N h^2 \over 4\,m\,L^3}.\tag3$$

For those who find too abstract the above derivation I'll add a semiclassical one. In our box we have free particles bouncing back an forth between boundaries. Their momentum is $p=h/(2L)$. A particle hits one boundary (e.g. the left one) once in a time $${2L \over v} = {2mL \over p} = {4 m L^2 \over h}$$ and every time it exchanges with the boundary a momentum $2p$. Then the momentum exchanged per unit of time, i.e. the force, is $$f = 2p\, {h \over 4 m L^2} = {h^2 \over 4 m L^3}.$$ This holds for one particle. It's only left to multiply by $N$ to get  (3)