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So, does it mean DOS is actually a function which depends on T?

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So, does it mean DOS is actually a function which depends on T?

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Can you show me the single-particle spectral function for T=0 and T not zero, and tell me what you think is "different"?

So, does it mean DOS is actually a function which depends on T?

Zz.

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radium

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There is a "lower level" derivation/definition of the DOS when mean-field approximation/Fermi liquid model is valid. If one starts with the single-particle spectral function A(k,w), which is the imaginary part of the single-particle Green's function, then the DOS is the "momentum average" of the spectral function, i.e. you integrate A(k,w) over all possible momentum k and essentially giving you A(w), which corresponds to some form of the density of states.

Zz.

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The definition of Spectral Function need us to calculate the average value of commutator(retard or advance green function) according to all the occupied states. In the T=0 condition, we only need to compute the average value in the ground state. So, the function is independent of T. However, if T is a definite value, we need to include partition function to calculate the average value in all the excited states. So, the Spectral Function include the parameter T.Can you show me the single-particle spectral function for T=0 and T not zero, and tell me what you think is "different"?

Zz.

My question is, the definition of DOS seems like an definite quantity which independent of T. So, if we relate DOS to Spectral Function, there must be some contradiction between these two conception.

In your second response, do you means A(k,w) (momentum average) is dependent on T. And if we integrate all the k, the quantity A(w) is actually accord with the definition of DOS and independent of T?

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