Path integral formulation: Difference between revisions

Revision as of 17:20, 13 November 2023

Revision as of 17:23, 13 November 2023

Line 603:

Let's also assume

:<math>\int \mathcal{D}\varphi\, Q[F][\varphi]=0</math>

for any polynomially-bounded functional {{mvar|F}}. This property is called the invariance of the measure., Andand this does not hold in general. (See ''[[anomaly (physics)]]'' for more details.)

Then,

Line 629:

:<math>j^{\mu}(x)=f^\mu(x)-\frac{\partial}{\partial (\partial_\mu \varphi)}\mathcal{L}(x) Q[\varphi] \,</math>

(this is assuming the Lagrangian only depends on {{mvar|φ}} and its first partial derivatives! More general Lagrangians would require a modification to this definition!). We're not insisting that {{math|''q''(''x'')}} is the generator of a symmetry (i.e. we are ''not'' insisting upon the [[gauge principle]]), but just that {{mvar|Q}} is. And we also assume the even stronger assumption that the functional measure is locally invariant:

:<math>\int \mathcal{D}\varphi\, q(x)[F][\varphi]=0.</math>

Line 641:

:<math>q(x)[S]\left[-i \frac{\delta}{\delta J}\right]Z[J]+J(x)Q[\varphi(x)]\left[-i \frac{\delta}{\delta J}\right]Z[J]=\partial_\mu j^\mu(x)\left[-i \frac{\delta}{\delta J}\right]Z[J]+J(x)Q[\varphi(x)]\left[-i \frac{\delta}{\delta J}\right]Z[J]=0.</math>

The above two equations are the '''Ward–Takahashi identities'''.

Now for the case where {{math|''f'' {{=}} 0}}, we can forget about all the boundary conditions and locality assumptions. We'd simply have