Energy-momentum tensor for a scalar field

January 5, 2016 phy2403 , ,

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It is claimed in [1] (3.2.1) that the momentum components of the energy-momentum tensor was found to be

\begin{equation}\label{eqn:noetherCurrentScalarField:20}
\Be_n \int d^3 x T^{0 n} = \int d^3 k \Bk a_k^\dagger a_k.
\end{equation}

I don’t see this result anywhere, so let’s calculate it.

First, from the Noether current for the scalar field Lagrangian in question, what is the energy-momentum tensor explicitly?

\begin{equation}\label{eqn:noetherCurrentScalarField:40}
\begin{aligned}
T^{\mu \nu}
&= \Pi^\mu \partial^\nu \phi – g^{\mu \nu} \LL \\
&= \Pi^\mu \partial^\nu \phi – g^{\mu \nu} \inv{2} \lr{ \partial_\alpha \phi \partial^\alpha \phi – \mu^2 \phi^2 } \\
&= \Pi^\mu \Pi^\nu – g^{\mu \nu} \inv{2} \lr{ \Pi_\alpha \Pi^\alpha – \mu^2 \phi^2 } \\
&= \Pi^\mu \Pi^\nu – \inv{2} g^{\mu \nu} g_{\alpha\beta} \Pi^\beta \Pi^\alpha + \inv{2} g^{\mu \nu} \mu^2 \phi^2.
\end{aligned}
\end{equation}

Consider some special cases for the indexes. For \( \mu = \nu = 0 \), the result is the Hamiltonian density

\begin{equation}\label{eqn:noetherCurrentScalarField:200}
\begin{aligned}
T^{00}
&= \Pi^0 \Pi^0 – \inv{2} g^{0 0} \Pi_\alpha \Pi^\alpha + \inv{2} g^{0 0} \mu^2 \phi^2 \\
&= \Pi^0 \Pi^0 – \inv{2} \Pi_\alpha \Pi^\alpha + \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^0 \Pi^0 – \inv{2} \Pi_n \Pi^n + \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \inv{2} (\spacegrad \phi)^2 + \inv{2} \mu^2 \phi^2,
\end{aligned}
\end{equation}

where \( \Pi^2 = (\partial_0 \phi)^2 \ne \partial^2 \phi \). For any \( \mu \ne \nu \) the off diagonal metric elements are zero, leaving just
\begin{equation}\label{eqn:noetherCurrentScalarField:220}
T^{\mu\nu} = \Pi^\mu \Pi^\nu.
\end{equation}

Finally, when \( n \ne 0 \), the remaining diagonal terms are
\begin{equation}\label{eqn:noetherCurrentScalarField:240}
\begin{aligned}
T^{nn}
&= \Pi^n \Pi^n – \inv{2} g^{n n} \Pi_\alpha \Pi^\alpha + \inv{2} g^{n n} n^2 \phi^2 \\
&= \Pi^n \Pi^n + \inv{2} \Pi_\alpha \Pi^\alpha – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \Pi^n \Pi^n – \inv{2} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \Pi^2 + \inv{2} \Pi^n \Pi^n – \inv{2} \sum_{m\ne n,0} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2 \\
&= \inv{2} \sum_{m = n,0} \Pi^m \Pi^m – \inv{2} \sum_{m\ne n,0} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2.
\end{aligned}
\end{equation}

The canonical momenta are

\begin{equation}\label{eqn:noetherCurrentScalarField:60}
\Pi^\mu
=
\partial^\mu
\int \frac{d^3 k}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ a_k e^{-i k \cdot x} + a_k^\dagger e^{i k \cdot x} },
\end{equation}

but
\begin{equation}\label{eqn:noetherCurrentScalarField:80}
\begin{aligned}
\partial^\mu e^{i k \cdot x}
&=
\partial^\mu \exp\lr{ i k^\alpha x_\alpha } \\
&=
i k^\mu \exp\lr{ i k \cdot x },
\end{aligned}
\end{equation}

so
\begin{equation}\label{eqn:noetherCurrentScalarField:100}
\begin{aligned}
\Pi^\mu
&=
i
\int \frac{d^3 k k^\mu}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ – a_k e^{-i k \cdot x} + a_k^\dagger e^{i k \cdot x} } \\
&=
i
\int \frac{d^3 k k^\mu}{(2\pi)^{3/2} \sqrt{ 2 \omega_k }} \lr{ – a_k e^{-i \omega_k t + \Bk \cdot \Bx} + a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx} }.
\end{aligned}
\end{equation}

This gives
\begin{equation}\label{eqn:noetherCurrentScalarField:120}
\begin{aligned}
\int d^3 x \Pi^\mu \Pi^\nu
&=
-\inv{2} \int d^3 x \inv{(2\pi)^3}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{ – a_k e^{-i \omega_k t + \Bk \cdot \Bx} + a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx} }
\lr{ – a_j e^{-i \omega_j t + \Bj \cdot \Bx} + a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx} } \\
&=
-\inv{2} \int d^3 x \inv{(2\pi)^3}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{
a_k a_j e^{-i (\omega_j + \omega_k) t + (\Bj + \Bk) \cdot \Bx}
– a_k a_j^\dagger e^{i (\omega_j – \omega_k) t – i (\Bj -\Bk) \cdot \Bx}
– a_k^\dagger a_j e^{-i (\omega_j -\omega_k) t – (\Bk – \Bj) \cdot \Bx}
+ a_k^\dagger a_j^\dagger e^{i (\omega_j + \omega_k) t – i (\Bj + \Bk) \cdot \Bx}
} \\
&=
-\inv{2}
\int d^3 k d^3 j \frac{k^\mu j^\nu}{\sqrt{\omega_k \omega_j}}
\lr{
a_k a_j e^{-i (\omega_j + \omega_k) t } \delta^3(\Bj + \Bk)
– a_k a_j^\dagger e^{i (\omega_j – \omega_k) t } \delta^3(\Bj -\Bk)
– a_k^\dagger a_j e^{-i (\omega_j -\omega_k) t } \delta^3 (\Bk – \Bj)
+ a_k^\dagger a_j^\dagger e^{i (\omega_j + \omega_k) t } \delta^3 (\Bj + \Bk)
}.
\end{aligned}
\end{equation}

There are two cases here to consider. The first is \( \nu = 0 \), for which

\begin{equation}\label{eqn:noetherCurrentScalarField:140}
\int d^3 x \Pi^\mu \Pi^0
=
-\inv{2}
\int d^3 k k^\mu
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
}.
\end{equation}

For \( \nu \ne 0 \)

\begin{equation}\label{eqn:noetherCurrentScalarField:160}
\begin{aligned}
\int d^3 x \Pi^\mu \Pi^\nu
&=
-\inv{2}
\int d^3 k \frac{k^\mu k^\nu}{\omega_k}
\lr{
– a_k a_{-k} e^{- 2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
– a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
} \\
&=
\inv{2}
\int d^3 k \frac{k^\mu k^\nu}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.
\end{aligned}
\end{equation}

Here’s a summary of these products

\begin{equation}\label{eqn:noetherCurrentScalarField:300}
\int d^3 x \Pi^0 \Pi^0
=
-\inv{2}
\int d^3 k \omega_k
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},
\end{equation}
\begin{equation}\label{eqn:noetherCurrentScalarField:280}
\int d^3 x \Pi^n \Pi^0
= \int d^3 x \Pi^0 \Pi^n
=
-\inv{2}
\int d^3 k k^n
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},
\end{equation}
\begin{equation}\label{eqn:noetherCurrentScalarField:340}
\int d^3 x \Pi^m \Pi^n
=
\inv{2}
\int d^3 k \frac{k^m k^n}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.
\end{equation}

For the mass term it was previously found that

\begin{equation}\label{eqn:noetherCurrentScalarField:180}
\inv{2} \int d^3 x \mu^2 \phi^2
=
\frac{\mu^2}{4}
\int
d^3 k
\inv{ \omega_k }
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}.
\end{equation}

The Hamiltonian component has been previously calculated, and resolves to

\begin{equation}\label{eqn:noetherCurrentScalarField:360}
\int d^3 x T^{00}
=
\inv{2}
\int d^3 k
\omega_k
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.
\end{equation}

The other diagonal components, for \( r \ne s \ne t \) are
\begin{equation}\label{eqn:noetherCurrentScalarField:380}
\begin{aligned}
\int d^3 x T^{rr}
&=
\int d^3 x
\lr{
\inv{2} \sum_{m = r,0} \Pi^m \Pi^m – \inv{2} \sum_{m = s,t} \Pi^m \Pi^m – \inv{2} \mu^2 \phi^2
} \\
&=
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
-\inv{4}
\int d^3 k \omega_k
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
} \\
&=
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2 – \omega_k^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
+
\inv{4}
\int d^3 k \frac{(k^r)^2 – (k^s)^2 – (k^t)^2 – \mu^2 + \omega_k^2}{\omega_k}
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
} \\
&=
\inv{2}
\int d^3 k \frac{ (k^r)^2 – \omega_k^2}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}
+
\inv{2}
\int d^3 k \frac{ (k^r)^2}{\omega_k}
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.
\end{aligned}
\end{equation}

This doesn’t have the nice cancelation that killed the time dependent terms in the Hamiltonian. Such cancellation also doesn’t appear in the off diagonal energy-momentum tensor components, which are

\begin{equation}\label{eqn:noetherCurrentScalarField:400}
\begin{aligned}
\int d^3 x T^{n 0}
&=
\int d^3 x T^{n 0} \\
&=
-\inv{2}
\int d^3 k k^n
\lr{
a_k a_{-k} e^{-2 i \omega_k t }
– a_k a_k^\dagger
– a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{2 i \omega_k t }
},
\end{aligned}
\end{equation}

and for \( m \ne n \ne 0 \)
\begin{equation}\label{eqn:noetherCurrentScalarField:420}
\int d^3 x T^{m n}
=
\inv{2}
\int d^3 k \frac{k^m k^n}{\omega_k}
\lr{
a_k a_{-k} e^{- 2 i \omega_k t }
+ a_k a_k^\dagger
+ a_k^\dagger a_k
+ a_k^\dagger a_{-k}^\dagger e^{ 2 i \omega_k t }
}.
\end{equation}

The \ref{eqn:noetherCurrentScalarField:400} result has time dependence that the stated result does not (but is linear in \( \Bk \) as desired)? Did I miss something?

References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL https://piazza.com/utoronto.ca/fall2015/phy2403f/resources. [Online; accessed 02-Jan-2016].

Hamiltonian for a scalar field

January 3, 2016 phy2403 , , ,

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In [1] it is left as an exersize to expand the scalar field Hamiltonian in terms of the raising and lowering operators. Let’s do that.

The field operator expanded in terms of the raising and lowering operators is

\begin{equation}\label{eqn:scalarFieldHamiltonian:20}
\phi(x) =
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i k \cdot x}
+ a_k^\dagger e^{i k \cdot x}
}.
\end{equation}

Note that \( x \) and \( k \) here are both four-vectors, so this field is dependent on a spacetime point, but the integration is over a spatial volume.

The Hamiltonian in terms of the fields was
\begin{equation}\label{eqn:scalarFieldHamiltonian:40}
H = \inv{2} \int d^3 x \lr{ \Pi^2 + \lr{ \spacegrad \phi }^2 + \mu^2 \phi^2 }.
\end{equation}

The field derivatives are

\begin{equation}\label{eqn:scalarFieldHamiltonian:60}
\Pi
= \partial_0 \phi
= \partial_0
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}
=
i
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \frac{\omega_k}{ 2 \omega_k } } \lr{
-a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
},
\end{equation}

and

\begin{equation}\label{eqn:scalarFieldHamiltonian:80}
\partial_n \phi
= \partial_n
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}
=
i
\int \frac{ d^3 k k^n}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
-a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}.
\end{equation}

Introducing a second set of momentum variables with \( j = \Abs{\Bj} \), the momentum portion of the Hamiltonian is

\begin{equation}\label{eqn:scalarFieldHamiltonian:100}
\begin{aligned}
\inv{2} \int d^3 x \Pi^2
&=
-\inv{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\omega_j
\omega_k
\lr{
-a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
+a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
-a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
-\inv{4}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\sqrt{
\omega_j
\omega_k}
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t – i (\Bk + \Bj) \cdot \Bx}
+ a_j a_k e^{-i (\omega_j + \omega_k) t + i (\Bj + \Bk) \cdot \Bx}
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t – i (\Bj – \Bk) \cdot \Bx}
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t – i (\Bk – \Bj) \cdot \Bx}
} \\
&=
-\inv{4}
\int
d^3 j
d^3 k
\sqrt{
\omega_j
\omega_k}
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t } \delta^3(\Bk + \Bj)
+ a_j a_k e^{-i (\omega_j + \omega_k) t } \delta^3(-\Bj – \Bk)
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t } \delta^3(\Bj – \Bk)
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t } \delta^3(\Bk – \Bj)
} \\
&=
-\inv{4}
\int
d^3 k
\omega_k
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}.
\end{aligned}
\end{equation}

For the gradient portion of the Hamiltonian we have

\begin{equation}\label{eqn:scalarFieldHamiltonian:120}
\begin{aligned}
\inv{2} \int d^3 x \lr{ \spacegrad \phi }^2
&=
-\inv{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{ \sum_{n=1}^3 j^n k^n }
\lr{
a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
-a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
-a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
-\inv{4}
\int
d^3 j
d^3 k
\frac{\Bj \cdot \Bk}{ \sqrt{ \omega_j \omega_k } }
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t } \delta^3(\Bk + \Bj)
+ a_j a_k e^{-i (\omega_j + \omega_k) t } \delta^3(-\Bj – \Bk)
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t } \delta^3(\Bj – \Bk)
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t } \delta^3(\Bk – \Bj)
} \\
&=
-\inv{4}
\int
d^3 k
\frac{\Bk^2}{ \omega_k }
\lr{
– a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
– a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}.
\end{aligned}
\end{equation}

Finally, for the mass term, we have

\begin{equation}\label{eqn:scalarFieldHamiltonian:140}
\begin{aligned}
\inv{2} \int d^3 x \mu^2 \phi^2
&=
\frac{\mu^2}{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
+a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
\frac{\mu^2}{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j a_k e^{-i (\omega_k + \omega_j) t + i (\Bk + \Bj) \cdot \Bx}
+a_j^\dagger a_k^\dagger e^{i (\omega_j + \omega_k) t – i (\Bk + \Bj) \cdot \Bx}
+a_j a_k^\dagger e^{i (\omega_k – \omega_j) t – i (\Bk – \Bj) \cdot \Bx}
+a_j^\dagger a_k e^{-i (\omega_k + \omega_j) t – i (\Bj – \Bk) \cdot \Bx}
} \\
&=
\frac{\mu^2}{2}
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j a_k e^{-i (\omega_k + \omega_j) t } \delta^3(- \Bk – \Bj)
+a_j^\dagger a_k^\dagger e^{i (\omega_j + \omega_k) t } \delta^3( \Bk + \Bj)
+a_j a_k^\dagger e^{i (\omega_k – \omega_j) t } \delta^3 (\Bk – \Bj)
+a_j^\dagger a_k e^{-i (\omega_k + \omega_j) t } \delta^3 (\Bj – \Bk)
} \\
&=
\frac{\mu^2}{4}
\int
d^3 k
\inv{ \omega_k }
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}.
\end{aligned}
\end{equation}

Now all the pieces can be put back together again

\begin{equation}\label{eqn:scalarFieldHamiltonian:160}
\begin{aligned}
H
&=
\inv{4}
\int d^3 k
\inv{\omega_k}
\lr{
-\omega_k^2
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}
+
\Bk^2
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
+ a_k^\dagger a_k
+ a_k a_k^\dagger
}
+
\mu^2
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}
} \\
&=
\inv{4}
\int d^3 k
\inv{\omega_k}
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
\lr{
-\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_{-k} a_k e^{- 2 i \omega_k t }
\lr{
-\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_k a_k^\dagger
\lr{
\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_k^\dagger a_k
\lr{
\omega_k^2
+ \Bk^2
+
\mu^2
}
}.
\end{aligned}
\end{equation}

With \( \omega_k^2 = \Bk^2 + \mu^2 \), the time dependent terms are killed leaving
\begin{equation}\label{eqn:scalarFieldHamiltonian:180}
H
=
\inv{2}
\int d^3 k
\omega_k
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.
\end{equation}

References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL lecturenotes.pdf. [Online; accessed 02-Jan-2016].

Scalar field creation operator commutator

January 2, 2016 math and physics play, phy2403, scalar field

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In [1] it is stated that the creation operators of eq. 2.78

\begin{equation}\label{eqn:scalarFieldCreationOpCommutator:20}
\alpha_k = \inv{2} \int \frac{d^3k}{(2\pi)^3} \lr{
\phi(x,0) + \frac{i}{\omega_k} \partial_0 \phi(x,0)
}
e^{-i \Bk \cdot \Bx }
\end{equation}

associated with field operator \( \phi \) commute. Let’s verify that.

\begin{equation}\label{eqn:scalarFieldCreationOpCommutator:40}
\begin{aligned}
\antisymmetric{\alpha_k}{\alpha_m}
&=
\inv{4}
\frac{1}{(2\pi)^6}
\int d^3 x d^3 y
e^{-i \Bk \cdot \Bx }
e^{-i \Bm \cdot \By }
\antisymmetric
{
\phi(x,0) + \frac{i}{\omega_k} \partial_0 \phi(x,0)
}
{
\phi(y,0) + \frac{i}{\omega_m} \partial_0 \phi(y,0)
} \\
&=
\frac{i}{4}
\frac{1}{(2\pi)^6}
\int d^3 x d^3 y
e^{-i \Bk \cdot \Bx }
e^{-i \Bm \cdot \By }
\lr{
\antisymmetric{\phi(x,0)}{\inv{\omega_m} \partial_0 \phi(y,0)}
+
\antisymmetric{\inv{\omega_k} \partial_0 \phi(x,0)}{\phi(y,0)}
} \\
&=
\frac{i}{4}
\frac{1}{(2\pi)^6}
\int d^3 x d^3 y
e^{-i \Bk \cdot \Bx }
e^{-i \Bm \cdot \By }
\lr{
\frac{i}{\omega_m} \delta^3(\Bx – \By)
–
\frac{i}{\omega_k} \delta^3(\Bx – \By)
} \\
&=
-\frac{1}{4}
\frac{1}{(2\pi)^6}
\int d^3 x
e^{ -i (\Bk + \Bm) \cdot \Bx }
\lr{
\frac{1}{\omega_m}
–
\frac{1}{\omega_k}
} \\
&=
-\frac{1}{4}
\frac{1}{(2\pi)^3}
\lr{
\frac{1}{\omega_m}
–
\frac{1}{\omega_k}
}
\delta^3(\Bk + \Bm) \\
&=
-\frac{1}{4}
\frac{1}{(2\pi)^3}
\lr{
\frac{1}{\omega_{\Abs{-\Bk}}}
–
\frac{1}{\omega_{\Abs{\Bk}}}
}
\delta^3(\Bk + \Bm) \\
&=
0.
\end{aligned}
\end{equation}

References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL https://s3.amazonaws.com/piazza-resources/i87nj8g7yie7nh/ihdwuk7wva13qq/lecturenotes.pdf?AWSAccessKeyId=AKIAIEDNRLJ4AZKBW6HA&Expires=1451803428&Signature=IF6qOjlKqOYL01FwqT%2FGV6BSDb8%3D. [Online; accessed 02-Jan-2016].

provisional M.Eng study plan

January 2, 2016 Incoherent ramblings

I’m trying to plot out a potential sequence for the remainder of my M.Eng courses. Most of the physics courses only run in the fall, and I still have to take a few more engineering courses to meet the requirements for the program. The following sequence, starting with the courses I’ve taken or am now enrolled on, gets one interesting course into each time slot:

My plan for the rest of the program is currently:

 

I’m hoping that the ECE Introduction to Computational Electrodynamics course will run again, and if it does I’ll switch things up to accomodate it.

Non-interacting particles in a box

December 17, 2015 phy1520 , ,

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Q: [1] pr 4.1

Calculate the three lowest energy levels and their degeneracies for equal mass distinguishable spin half particles in a box of length \( L \). Consider

(a) Two particles.
(b) Three particles.
(c) Four particles.

A: part (a)

The problem statement doesn’t include the dimensionality of the box. The simplest case is the one dimensional box, for which the wave function of one particle is

\begin{equation}\label{eqn:noninteractingParticlesInABox:20}
\psi_1(x) = \sqrt{\frac{2}{L}} \sin\lr{ \frac{n \pi x}{L} },
\end{equation}

and the energy of that particle is

\begin{equation}\label{eqn:noninteractingParticlesInABox:40}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 n^2.
\end{equation}

If the box is two dimensional the energy is

\begin{equation}\label{eqn:noninteractingParticlesInABox:60}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ n_1^2 + n_2^2 },
\end{equation}

and if it’s a 3D box, we have

\begin{equation}\label{eqn:noninteractingParticlesInABox:80}
E = \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ n_1^2 + n_2^2 + n_3^2}.
\end{equation}

Suppose we are considering the 3D box. In statistical mechanics when we are considering particles Fermions, they are indistinguishable, and thus not allowed to share the same spin state at a given energy level. However, for distinguishable particles, that restriction doesn’t exist, and we can have two (or more) such particles in the lowest order energy state. The lowest such energy is

\begin{equation}\label{eqn:noninteractingParticlesInABox:100}
\begin{aligned}
E_{1,1,1 ;1,1,1}
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 6 \times 1^2 } \\
&=
\frac{6}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}
\end{equation}

The particle spin states can be any of \( \ket{++}, \ket{+-}, \ket{-+}, \ket{–} \), so there is a four way degeneracy in the ground state.

the next lowest energy level is

\begin{equation}\label{eqn:noninteractingParticlesInABox:120}
\begin{aligned}
E_{1,1,2 ;1,1,1}
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 5 \times 1^2 + 2^2 } \\
&=
\frac{9}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

where there are \( \binom{6}{1} = 6 \) ways to pick such a state for each variation of spin, for a total \( 6 \times 4 = 24 \) way degeneracy. Finally, since \( 2^2 + 2^2 < 3^2 + 1^2 \), the next lowest energy level is \begin{equation}\label{eqn:noninteractingParticlesInABox:140} \begin{aligned} E_{1,2,2 ;1,1,1} &= \inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 4 \times 1^2 + 2 \times 2^2 } \\ &= \frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2, \end{aligned} \end{equation} with a \( \binom{6}{2} \times 4 = 15 \times 4 = 60 \) way degeneracy for this energy level.

A: part (b)

For three particles (the two particle case wasn’t actually in the problem statement, but seemed an easier starting place), the lowest energy state for a 3D box is

\begin{equation}\label{eqn:noninteractingParticlesInABox:160}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 9 \times 1^2 } \\
&=
\frac{9}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}
\end{equation}

There are now \( 2^3 = 8 \) variations of spin \( \ket{+++}, \ket{++-}, \cdots \), so the ground state is 8-way degenerate. Next up is

\begin{equation}\label{eqn:noninteractingParticlesInABox:180}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 8 \times 1^2 + 2^2 } \\
&=
\frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

where there is a \( \binom{9}{1} \times 8 = 9 \times 8 = 72 \) way degeneracy in this energy level. Finally, the next lowest energy level is

\begin{equation}\label{eqn:noninteractingParticlesInABox:200}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 7 \times 1^2 + 2 \times 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

with a \( \binom{9}{2} \times 8 = 36 \times 8 = 288 \) way degeneracy for this energy level.

A: part (c)

For four particles the lowest energy state for a 3D box is

\begin{equation}\label{eqn:noninteractingParticlesInABox:220}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 12 \times 1^2 } \\
&=
\frac{12}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2.
\end{aligned}
\end{equation}

There are now \( 2^4 = 16 \) variations of spin \( \ket{++++}, \ket{+++-}, \cdots \), so the ground state is 16-way degenerate. For the second level

\begin{equation}\label{eqn:noninteractingParticlesInABox:240}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 11 \times 1^2 + 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

where there is a \( \binom{12}{1} \times 16 = 12 \times 16 = 192 \) way degeneracy in this energy level. Finally, the next lowest energy level is

\begin{equation}\label{eqn:noninteractingParticlesInABox:260}
\begin{aligned}
E
&=
\inv{2 m} \lr{ \frac{\Hbar \pi}{L} }^2 \lr{ 7 \times 1^2 + 2 \times 2^2 } \\
&=
\frac{15}{2 m} \lr{ \frac{\Hbar \pi}{L} }^2,
\end{aligned}
\end{equation}

with a \( \binom{12}{2} \times 16 = 66 \times 16 = 1056 \) way degeneracy for this energy level.

References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.