- #1
hammer123
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<Re-opening approved by mentor.>
Hi, I've always wondered why when calculating the partition function for a quantum system, we only sum over the eigenstates and not superimposed states. Thus I decided to actually try summing over all normalized states and see what would happen. Feedback is appreciated. (Updated result at the end)
Consider a spin-1/2 system immersed in a constant magnetic field ##B##, with the Hamiltonian
$$ \hat H = - B \hat S_z$$ Consider an arbitrary state decomposed in the eigenstates of ##\hat S_z##
$$| \psi \rangle = \alpha | + \rangle + \gamma e^{i \phi} | - \rangle $$ ##\alpha## and ##\gamma## are taken to be real and greater than zero. The expected energy for this state is $$E_{| \phi \rangle}=\langle \phi | \hat H| \phi \rangle = - \frac {B \hbar} {2} (\alpha^2 - \gamma^2)$$ Now consider the "all-states" partition function of this system, in which all states such as ##| \psi \rangle## are summed over. $$Z = \sum_{| \psi \rangle} \text{exp}[\frac {B \hbar \beta} {2} (\alpha^2 - \gamma^2)]$$ By normalization of ##| \psi \rangle##, this is $$ Z = \sum_{| \psi \rangle} \text{exp}[\frac {B \hbar \beta} {2} (2\alpha^2 - 1)]$$ Explicitly, $$Z = \int_0^{2\pi} \int_0^{1}\text{exp}[\frac {B \hbar \beta} {2} (2\alpha^2 - 1)] d\alpha d\phi$$ The ##\phi## integral is to account for the phase $$Z = 2 \pi ~\text{exp}(-\frac {B \hbar \beta}{2})\int_0^{1}\text{exp}[ {B \hbar \beta}\alpha^2] d\alpha$$ The mean spin in the ##z## direction is then $$\langle m_z \rangle = \frac {1}{\beta} \frac {\partial logZ} {\partial B} = -\frac {\hbar}{2} + \frac {\hbar\int_0^{1} \alpha^2 \text{exp}[ {B \hbar \beta}\alpha^2] d\alpha}{\int_0^{1} \text{exp}[ {B \hbar \beta}\alpha^2] d\alpha} $$ This can be expressed in terms of the ##\text{erfi}(x) = -i \text{erf}(ix)## functions $$\frac {\langle m_z \rangle}{\hbar} = \frac { e^\tau}{\sqrt {\pi \tau}\text{erfi}\sqrt{\tau}} -\frac {1+\tau}{2\tau} $$ where ##\tau = B\hbar\beta##. Can be shown that (according to mathematica)
$$\lim_{\tau \rightarrow \pm\infty} \langle m_z \rangle = \pm \frac {\hbar}{2}$$ Which is good as this corresponds to ## T \rightarrow 0##. However, it appears that $$\lim_{\tau \rightarrow 0} \langle m_z \rangle = -\frac{\hbar}{6}$$ which is not right since this corresponds to the ##T \rightarrow \infty## limit.
So my question is, did I make a mistake or was this entire thing invalid to start with?
Wolframalpha results for the limits:
##\tau\rightarrow-\infty##
##\tau\rightarrow\infty##
##\tau\rightarrow0##
UPDATE:
Okay, so it seems like my initial treatment distinguished the ##| + \rangle## state and that's what caused the non-zero net spin at ##T \rightarrow \infty##. However I'm not really sure why that was so, as far as I can tell I summed over all of the physically distinct states.
Whatever the case, to treat them equally I sum up the expressions for ##Z##, one with ##\gamma## removed and one with ##\alpha## removed$$Z = \sum_{| \psi \rangle} \text{exp}[\frac {B \hbar \beta} {2} (\alpha^2 - \gamma^2)]$$Using ##\alpha^2+\gamma^2=1## $$ =\frac{1}{2} \sum_{| \psi \rangle} \{\text{exp}[\frac {B \hbar \beta} {2} (2\alpha^2 - 1)]+\text{exp}[\frac {B \hbar \beta} {2} (1-2\gamma^2)]\}$$ $$ = \frac{1}{2}\int_0^{2\pi} \int_0^{1}\text{exp}[\frac {B \hbar \beta} {2} (2x^2 - 1)]+\text{exp}[\frac {B \hbar \beta} {2} (1-2x^2)] dx d\phi$$ $$= \pi ~\text{exp}(-\frac {B \hbar \beta}{2})\int_0^{1}\text{exp}[ {B \hbar \beta}\alpha^2] d\alpha+ \pi ~\text{exp}(\frac {B \hbar \beta}{2})\int_0^{1}\text{exp}[ {-B \hbar \beta}\gamma^2] d\gamma$$ The net spin is then $$\langle m_z \rangle = \frac{\hbar}{2}(\frac { e^\tau}{\sqrt {\pi \tau}\text{erfi}\sqrt{\tau}}+\frac { e^{-\tau}}{\sqrt {\pi \tau}\text{erf}\sqrt{\tau}}-\frac{1}{\tau})$$Where as before ##\tau = B\hbar\beta##. This function has the right limits $$\lim_{\tau \rightarrow \pm\infty} \langle m_z \rangle = \pm \frac {\hbar}{2}$$ and, more importantly $$\lim_{\tau \rightarrow 0} \langle m_z \rangle = 0$$ Compared to the net spin derived from only summing eigenstates, ##\langle m_z \rangle = \frac {\hbar}{2}\text{tanh}(\frac{\tau}{2})##, the one derived here converges to ##\pm \frac{\hbar}{2}## more slowly, as expected since the system has more degrees of freedom.Plot of the net spin compared to usual one
Hi, I've always wondered why when calculating the partition function for a quantum system, we only sum over the eigenstates and not superimposed states. Thus I decided to actually try summing over all normalized states and see what would happen. Feedback is appreciated. (Updated result at the end)
Consider a spin-1/2 system immersed in a constant magnetic field ##B##, with the Hamiltonian
$$ \hat H = - B \hat S_z$$ Consider an arbitrary state decomposed in the eigenstates of ##\hat S_z##
$$| \psi \rangle = \alpha | + \rangle + \gamma e^{i \phi} | - \rangle $$ ##\alpha## and ##\gamma## are taken to be real and greater than zero. The expected energy for this state is $$E_{| \phi \rangle}=\langle \phi | \hat H| \phi \rangle = - \frac {B \hbar} {2} (\alpha^2 - \gamma^2)$$ Now consider the "all-states" partition function of this system, in which all states such as ##| \psi \rangle## are summed over. $$Z = \sum_{| \psi \rangle} \text{exp}[\frac {B \hbar \beta} {2} (\alpha^2 - \gamma^2)]$$ By normalization of ##| \psi \rangle##, this is $$ Z = \sum_{| \psi \rangle} \text{exp}[\frac {B \hbar \beta} {2} (2\alpha^2 - 1)]$$ Explicitly, $$Z = \int_0^{2\pi} \int_0^{1}\text{exp}[\frac {B \hbar \beta} {2} (2\alpha^2 - 1)] d\alpha d\phi$$ The ##\phi## integral is to account for the phase $$Z = 2 \pi ~\text{exp}(-\frac {B \hbar \beta}{2})\int_0^{1}\text{exp}[ {B \hbar \beta}\alpha^2] d\alpha$$ The mean spin in the ##z## direction is then $$\langle m_z \rangle = \frac {1}{\beta} \frac {\partial logZ} {\partial B} = -\frac {\hbar}{2} + \frac {\hbar\int_0^{1} \alpha^2 \text{exp}[ {B \hbar \beta}\alpha^2] d\alpha}{\int_0^{1} \text{exp}[ {B \hbar \beta}\alpha^2] d\alpha} $$ This can be expressed in terms of the ##\text{erfi}(x) = -i \text{erf}(ix)## functions $$\frac {\langle m_z \rangle}{\hbar} = \frac { e^\tau}{\sqrt {\pi \tau}\text{erfi}\sqrt{\tau}} -\frac {1+\tau}{2\tau} $$ where ##\tau = B\hbar\beta##. Can be shown that (according to mathematica)
$$\lim_{\tau \rightarrow \pm\infty} \langle m_z \rangle = \pm \frac {\hbar}{2}$$ Which is good as this corresponds to ## T \rightarrow 0##. However, it appears that $$\lim_{\tau \rightarrow 0} \langle m_z \rangle = -\frac{\hbar}{6}$$ which is not right since this corresponds to the ##T \rightarrow \infty## limit.
So my question is, did I make a mistake or was this entire thing invalid to start with?
Wolframalpha results for the limits:
##\tau\rightarrow-\infty##
##\tau\rightarrow\infty##
##\tau\rightarrow0##
UPDATE:
Okay, so it seems like my initial treatment distinguished the ##| + \rangle## state and that's what caused the non-zero net spin at ##T \rightarrow \infty##. However I'm not really sure why that was so, as far as I can tell I summed over all of the physically distinct states.
Whatever the case, to treat them equally I sum up the expressions for ##Z##, one with ##\gamma## removed and one with ##\alpha## removed$$Z = \sum_{| \psi \rangle} \text{exp}[\frac {B \hbar \beta} {2} (\alpha^2 - \gamma^2)]$$Using ##\alpha^2+\gamma^2=1## $$ =\frac{1}{2} \sum_{| \psi \rangle} \{\text{exp}[\frac {B \hbar \beta} {2} (2\alpha^2 - 1)]+\text{exp}[\frac {B \hbar \beta} {2} (1-2\gamma^2)]\}$$ $$ = \frac{1}{2}\int_0^{2\pi} \int_0^{1}\text{exp}[\frac {B \hbar \beta} {2} (2x^2 - 1)]+\text{exp}[\frac {B \hbar \beta} {2} (1-2x^2)] dx d\phi$$ $$= \pi ~\text{exp}(-\frac {B \hbar \beta}{2})\int_0^{1}\text{exp}[ {B \hbar \beta}\alpha^2] d\alpha+ \pi ~\text{exp}(\frac {B \hbar \beta}{2})\int_0^{1}\text{exp}[ {-B \hbar \beta}\gamma^2] d\gamma$$ The net spin is then $$\langle m_z \rangle = \frac{\hbar}{2}(\frac { e^\tau}{\sqrt {\pi \tau}\text{erfi}\sqrt{\tau}}+\frac { e^{-\tau}}{\sqrt {\pi \tau}\text{erf}\sqrt{\tau}}-\frac{1}{\tau})$$Where as before ##\tau = B\hbar\beta##. This function has the right limits $$\lim_{\tau \rightarrow \pm\infty} \langle m_z \rangle = \pm \frac {\hbar}{2}$$ and, more importantly $$\lim_{\tau \rightarrow 0} \langle m_z \rangle = 0$$ Compared to the net spin derived from only summing eigenstates, ##\langle m_z \rangle = \frac {\hbar}{2}\text{tanh}(\frac{\tau}{2})##, the one derived here converges to ##\pm \frac{\hbar}{2}## more slowly, as expected since the system has more degrees of freedom.Plot of the net spin compared to usual one
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