I am trying to compute the (anomalous) transformation law of the free fermion stress-tensor, not with the usual CFT arguments, but by explicit computation.

We can define the classical stress tensor $$T=i\psi^{\dagger}\partial\psi.\tag{1}$$ The quantum counterpart will be divergent, we regularize by point-splitting. The vacuum expectation value is then determined by the correlator

$$\langle T(x,x') \rangle=i \langle \psi^{\dagger}(x)\partial_{x'}\psi(x')\rangle=i\partial_{x'}\langle \psi^{\dagger}(x)\psi(x')\rangle=-\frac{1}{2\pi}\frac{1}{(x-x')^2}\tag{2}$$

where I have used $$\langle \psi(x)\psi^{\dagger}(x')\rangle=\frac{1}{2 \pi i}\frac{1}{x-x'}.\tag{3}$$

Now we define the normal ordered stress tensor by subtracting the divergence

$$:T(x):=\lim_{x' \to x} i\psi^{\dagger}(x)\partial_{x'}\psi(x') +\frac{1}{2\pi}\frac{1}{(x-x')^2}.\tag{4}$$

In a different coordinate system, $x \to y(x)$, we do the same,

$$:T(y):=\lim_{y' \to y} i \psi^{\dagger}(y)\partial_{y'}\psi(y') +\frac{1}{2\pi}\frac{1}{(y-y')^2}$$

From the transformation law of the fermion, $\psi(y)=\left(\frac{dx}{dy}\right)^{1/2}\psi(x)$, I should be able to deduce now that

$$:T(y):=\left(\frac{dx}{dy}\right)^2 :T(x): - \frac{c}{24 \pi} \{x,y\}.$$

with $c=1/2$ for the free fermion. Rewriting gives

$$i \psi^{\dagger}(y)\partial_{y'}\psi(y')=i\left(\frac{dx}{dy}\right)^{1/2}\psi^{\dagger}(x)\frac{dx'}{dy'}\partial_{x'}\left(\left(\frac{dx'}{dy'}\right)^{1/2}\psi(x') \right) $$

The derivative acting on the second term gives

$$i\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx'}{dy'}\right)^{3/2}\psi^{\dagger}(x)\partial_{x'}\psi(x')$$

but what happens to the other term? I tried ignoring it (since I thought it might be zero), but then I don't get the correct transformation law.

Would appreciate any help.

I am trying to compute the (anomalous) transformation law of the free fermion stress-tensor, not with the usual CFT arguments, but by explicit computation.

We can define the classical stress tensor $$T=i\psi^{\dagger}\partial\psi.\tag{1}$$ The quantum counterpart will be divergent, we regularize by point-splitting. The vacuum expectation value is then determined by the correlator

$$\langle T(x,x') \rangle=i \langle \psi^{\dagger}(x)\partial_{x'}\psi(x')\rangle=i\partial_{x'}\langle \psi^{\dagger}(x)\psi(x')\rangle=-\frac{1}{2\pi}\frac{1}{(x-x')^2}\tag{2}$$

where I have used $$\langle \psi(x)\psi^{\dagger}(x')\rangle=\frac{1}{2 \pi i}\frac{1}{x-x'}.\tag{3}$$

Now we define the normal ordered stress tensor by subtracting the divergence

$$:T(x):=\lim_{x' \to x} \frac{i}{2}\left(\psi^{\dagger}(x)\partial_{x'}\psi(x')-\partial_{x}\psi^{\dagger}(x)\psi(x') \right)+\frac{1}{2\pi}\frac{1}{(x-x')^2}.\tag{4}$$

In a different coordinate system, $x \to y(x)$, we do the same,

$$:T(y):=\lim_{y' \to y} \frac{i}{2}\left(\psi^{\dagger}(y)\partial_{y'}\psi(y')-\partial_{y}\psi^{\dagger}(y)\psi(y') \right)+\frac{1}{2\pi}\frac{1}{(y-y')^2}$$

From the transformation law of the fermion, $\psi(y)=\left(\frac{dx}{dy}\right)^{1/2}\psi(x)$, I should be able to deduce now that

$$:T(y):=\left(\frac{dx}{dy}\right)^2 :T(x): - \frac{c}{24 \pi} \{x,y\}.$$

with $c=1/2$ for the free fermion. Rewriting gives

$$i \psi^{\dagger}(y)\partial_{y'}\psi(y')=i\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx'}{dy'}\right)^{3/2}\psi^{\dagger}(x)\partial_{x'}\psi(x') $$

such that

$$:T(y):= \left(\frac{dx}{dy}\right)^2 :T(x):+ \frac {1}{2 \pi}\lim_{y' \to y} \left( -\frac{1}{2}\frac{\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx}{dy'}\right)^{3/2}+\left(\frac{dx}{dy}\right)^{3/2}\left(\frac{dx}{dy'}\right)^{1/2}}{(x(y)-x(y'))^2}+\frac{1}{(y-y')^2}\right)$$

Now when I perform the expansion with mathematica, I get almost the desired result. The right bracket becomes

$$-\frac{1}{24 \pi} \left(\frac{2 x'''(y)}{x'(y)}-\frac{3}{2}\left( \frac{x''(y)}{x'(y)}\right)^2 \right)$$

The 2 in the first term should not be there...

I made a few edits: Following Qmechanics comment, I symmetrized the stress tensor. Now I almost get the correct result, but not quite. Can't figure out where the mistake is.

I am trying to compute the (anomalous) transformation law of the free fermion stress-tensor, not with the usual CFT arguments, but by explicit computation.

We can define the classical stress tensor $T=i\psi^{\dagger}\partial\psi$. The quantum counterpart will be divergent, we regularize by point-splitting. The vacuum expectation value is then determined by the correlator

$$\langle T(x,x') \rangle=i \langle \psi^{\dagger}(x)\partial_{x'}\psi(x')\rangle=i\partial_{x'}\langle \psi^{\dagger}(x)\psi(x')\rangle=-\frac{1}{2\pi}\frac{1}{(x-x')^2}\tag{1}$$

where I have used $$\langle \psi(x)\psi^{\dagger}(x')\rangle=\frac{1}{2 \pi i}\frac{1}{x-x'}.\tag{2}$$

Now we define the normal ordered stress tensor by subtracting the divergence

$$:T(x):=\lim_{x' \to x} i\psi^{\dagger}(x)\partial_{x'}\psi(x') +\frac{1}{2\pi}\frac{1}{(x-x')^2}.\tag{3}$$

In a different coordinate system, $x \to y(x)$, we do the same,

$$:T(y):=\lim_{y' \to y} i \psi^{\dagger}(y)\partial_{y'}\psi(y') +\frac{1}{2\pi}\frac{1}{(y-y')^2}$$

From the transformation law of the fermion, $\psi(y)=\left(\frac{dx}{dy}\right)^{1/2}\psi(x)$, I should be able to deduce now that

$$:T(y):=\left(\frac{dx}{dy}\right)^2 :T(x): - \frac{c}{24 \pi} \{x,y\}.$$

with $c=1/2$ for the free fermion. Rewriting gives

$$i \psi^{\dagger}(y)\partial_{y'}\psi(y')=i\left(\frac{dx}{dy}\right)^{1/2}\psi^{\dagger}(x)\frac{dx'}{dy'}\partial_{x'}\left(\left(\frac{dx'}{dy'}\right)^{1/2}\psi(x') \right) $$

The derivative acting on the second term gives

$$i\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx'}{dy'}\right)^{3/2}\psi^{\dagger}(x)\partial_{x'}\psi(x')$$

but what happens to the other term? I tried ignoring it (since I thought it might be zero), but then I don't get the correct transformation law.

Would appreciate any help.

I am trying to compute the (anomalous) transformation law of the free fermion stress-tensor, not with the usual CFT arguments, but by explicit computation.

We can define the classical stress tensor $$T=i\psi^{\dagger}\partial\psi.\tag{1}$$ The quantum counterpart will be divergent, we regularize by point-splitting. The vacuum expectation value is then determined by the correlator

$$\langle T(x,x') \rangle=i \langle \psi^{\dagger}(x)\partial_{x'}\psi(x')\rangle=i\partial_{x'}\langle \psi^{\dagger}(x)\psi(x')\rangle=-\frac{1}{2\pi}\frac{1}{(x-x')^2}\tag{2}$$

where I have used $$\langle \psi(x)\psi^{\dagger}(x')\rangle=\frac{1}{2 \pi i}\frac{1}{x-x'}.\tag{3}$$

Now we define the normal ordered stress tensor by subtracting the divergence

$$:T(x):=\lim_{x' \to x} i\psi^{\dagger}(x)\partial_{x'}\psi(x') +\frac{1}{2\pi}\frac{1}{(x-x')^2}.\tag{4}$$

In a different coordinate system, $x \to y(x)$, we do the same,

$$:T(y):=\lim_{y' \to y} i \psi^{\dagger}(y)\partial_{y'}\psi(y') +\frac{1}{2\pi}\frac{1}{(y-y')^2}$$

From the transformation law of the fermion, $\psi(y)=\left(\frac{dx}{dy}\right)^{1/2}\psi(x)$, I should be able to deduce now that

$$:T(y):=\left(\frac{dx}{dy}\right)^2 :T(x): - \frac{c}{24 \pi} \{x,y\}.$$

with $c=1/2$ for the free fermion. Rewriting gives

$$i \psi^{\dagger}(y)\partial_{y'}\psi(y')=i\left(\frac{dx}{dy}\right)^{1/2}\psi^{\dagger}(x)\frac{dx'}{dy'}\partial_{x'}\left(\left(\frac{dx'}{dy'}\right)^{1/2}\psi(x') \right) $$

The derivative acting on the second term gives

$$i\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx'}{dy'}\right)^{3/2}\psi^{\dagger}(x)\partial_{x'}\psi(x')$$

but what happens to the other term? I tried ignoring it (since I thought it might be zero), but then I don't get the correct transformation law.

Would appreciate any help.

I am trying to compute the (anomalous) transformation law of the free fermion stress tensor, not with the usual CFT arguments, but by explicit computation.

We can define the classical stress tensor $T=i\psi^{\dagger}\partial\psi$. The quantum counterpart will be divergent, we regularize by point-splitting. The vacuum expectation value is then determined by the correlator

$$\langle T(x,x') \rangle=i \langle \psi^{\dagger}(x)\partial_{x'}\psi(x')\rangle=i\partial_{x'}\langle \psi^{\dagger}(x)\psi(x')\rangle=-\frac{1}{2\pi}\frac{1}{(x-x')^2}$$

where I have used $\langle \psi(x)\psi^{\dagger}(x')\rangle=\frac{1}{2 \pi i}\frac{1}{x-x'}$.

Now we define the normal ordered stress tensor by subtracting the divergence

$$:T(x):=\lim_{x' \to x} i\psi^{\dagger}(x)\partial_{x'}\psi(x') +\frac{1}{2\pi}\frac{1}{(x-x')^2}$$

In a different coordinate system, $x \to y(x)$, we do the same,

$$:T(y):=\lim_{y' \to y} i \psi^{\dagger}(y)\partial_{y'}\psi(y') +\frac{1}{2\pi}\frac{1}{(y-y')^2}$$

From the transformation law of the fermion, $\psi(y)=\left(\frac{dx}{dy}\right)^{1/2}\psi(x)$, I should be able to deduce now that

$$:T(y):=\left(\frac{dx}{dy}\right)^2 :T(x): - \frac{c}{24 \pi} \{x,y\}.$$

with $c=1/2$ for the free fermion. Rewriting gives

$$i \psi^{\dagger}(y)\partial_{y'}\psi(y')=i\left(\frac{dx}{dy}\right)^{1/2}\psi^{\dagger}(x)\frac{dx'}{dy'}\partial_{x'}\left(\left(\frac{dx'}{dy'}\right)^{1/2}\psi(x') \right) $$

The derivative acting on the second term gives

$$i\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx'}{dy'}\right)^{3/2}\psi^{\dagger}(x)\partial_{x'}\psi(x')$$

but what happens to the other term? I tried ignoring it (since I thought it might be zero), but then I don't get the correct transformation law.

Would appreciate any help.

I am trying to compute the (anomalous) transformation law of the free fermion stress-tensor, not with the usual CFT arguments, but by explicit computation.

We can define the classical stress tensor $T=i\psi^{\dagger}\partial\psi$. The quantum counterpart will be divergent, we regularize by point-splitting. The vacuum expectation value is then determined by the correlator

$$\langle T(x,x') \rangle=i \langle \psi^{\dagger}(x)\partial_{x'}\psi(x')\rangle=i\partial_{x'}\langle \psi^{\dagger}(x)\psi(x')\rangle=-\frac{1}{2\pi}\frac{1}{(x-x')^2}\tag{1}$$

where I have used $$\langle \psi(x)\psi^{\dagger}(x')\rangle=\frac{1}{2 \pi i}\frac{1}{x-x'}.\tag{2}$$

Now we define the normal ordered stress tensor by subtracting the divergence

$$:T(x):=\lim_{x' \to x} i\psi^{\dagger}(x)\partial_{x'}\psi(x') +\frac{1}{2\pi}\frac{1}{(x-x')^2}.\tag{3}$$

In a different coordinate system, $x \to y(x)$, we do the same,

$$:T(y):=\lim_{y' \to y} i \psi^{\dagger}(y)\partial_{y'}\psi(y') +\frac{1}{2\pi}\frac{1}{(y-y')^2}$$

From the transformation law of the fermion, $\psi(y)=\left(\frac{dx}{dy}\right)^{1/2}\psi(x)$, I should be able to deduce now that

$$:T(y):=\left(\frac{dx}{dy}\right)^2 :T(x): - \frac{c}{24 \pi} \{x,y\}.$$

with $c=1/2$ for the free fermion. Rewriting gives

$$i \psi^{\dagger}(y)\partial_{y'}\psi(y')=i\left(\frac{dx}{dy}\right)^{1/2}\psi^{\dagger}(x)\frac{dx'}{dy'}\partial_{x'}\left(\left(\frac{dx'}{dy'}\right)^{1/2}\psi(x') \right) $$

The derivative acting on the second term gives

$$i\left(\frac{dx}{dy}\right)^{1/2}\left(\frac{dx'}{dy'}\right)^{3/2}\psi^{\dagger}(x)\partial_{x'}\psi(x')$$

but what happens to the other term? I tried ignoring it (since I thought it might be zero), but then I don't get the correct transformation law.

Would appreciate any help.