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I am reading Blundell's Quantum field theory for the Gifted Amateur and stuck at some calculation. In his book p.197, 21.2 Sources in statistical physics, he defined the partition function with the source term by

$$Z(J)=\operatorname{Tr}[e^{-\beta\hat{H}}] := \operatorname{Tr}[e^{-\beta\hat{H_0}+\Sigma_kJ_k \hat{\phi_k}}]$$

Then he calculated the thermal average $\langle \hat{\phi_i} \rangle_t$ (c.f. his book p.197) by

$$\langle \hat{\phi_i} \rangle_t = \frac{1}{Z(J=0)}\frac{\partial Z(J)}{\partial(J_i)}|_{J_i=0} = \frac{\operatorname{Tr}[\hat{\phi_i}e^{-\beta\hat{H_0}}]}{Z(J=0)} $$

And why is the second equality true?

Perhaps,

$$\frac{\partial Z(J)}{\partial(J_i)}|_{J_i=0}=\operatorname{Tr}[e^{-\beta \hat{H_0}} \cdot e^{\Sigma_{k} J_k \hat{\phi_k}} \cdot \frac{\partial}{\partial(J_i)}(\Sigma_k J_k \hat{\phi_k})]|_{J_i=0} = \operatorname{Tr}[e^{-\beta \hat{H_0}} \cdot e^{\Sigma_{k} J_k \hat{\phi_k}} \cdot \hat{\phi_i} ]|_{J_i=0}~? $$

True? If so, how about next step? Can anyone help?

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    $\begingroup$ Use MathJax, not screenshots/images. This was written under your last few question. Could you please follow our policies? $\endgroup$
    – Tobias Fünke
    Commented Jan 14, 2023 at 8:10
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    $\begingroup$ O.K. I edited. Perhaps, can you reopen my question? $\endgroup$
    – Plantation
    Commented Jan 15, 2023 at 5:36
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    $\begingroup$ That looks quite right to me, just set the source term to zero in the last term and then dividing by Z gives you the expectation value written as a trace. You should set all sources to 0, not just the i'th one normally. $\endgroup$
    – Guliano
    Commented Jan 15, 2023 at 6:08
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    $\begingroup$ Thank you. And.. Your comment is somewhat ambiguous to follow, What's the meaning of setting zero to all sources (not just the i'th one) in our calculation ? Perhaps, the author made mistake in some part? $\endgroup$
    – Plantation
    Commented Jan 15, 2023 at 6:47
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    $\begingroup$ I was not sure what your mistake would've been. The notation looks like you only set $J_i$ to zero. Maybe you thought that's not equal to setting all the sources to zero. $\endgroup$
    – Guliano
    Commented Jan 16, 2023 at 7:20

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