Direct Products

A.1.1 Direct Products

We can form a new vector space by taking the direct product of two vector spaces. In quantum mechanics, this arises for instance if we have two particles, or if we have two sources of angular momentum for a single particle. Taking two particles as an example, if the ﬁrst (say a proton) is in the state $| ϕ ⟩$ and the second (say a neutron) is in the state $| α ⟩$ , the state of the whole system is given by $| ϕ ⟩ \otimes | α ⟩$ , where we specify at the start that we will write the proton state ﬁrst, and the symbol “ $\otimes$ ” is really just a separator. All states of this form are in the new vector space, but not all states of the new vector space are of this separable form. As a simple example, $| ϕ ⟩ \otimes | α ⟩ + | ψ ⟩ \otimes | β ⟩$ is a possible state of the system in which neither the proton nor the neutron is in a deﬁnite state, but the two are entangled, such that if we ﬁnd the proton in state $| ϕ ⟩$ we know the neutron is in state $| α ⟩$ , but if the proton is in state $| ψ ⟩$ the neutron must be in state $| β ⟩$ .

Operators can be of three forms: they can act only on one particle, they can act on both particles but independently on each, or they can be more complicated. If $\hat{A}$ acts only on the proton, and $\hat{B}$ only on the neutron, then in the two particle space we have to write the former as $\hat{A} \otimes {\hat{I}}_{n}$ where ${\hat{I}}_{n}$ is the identity operator for neutron states, and the latter as ${\hat{I}}_{p} \otimes \hat{B}$ . Again, the “ $\otimes$ ” acts to separate “proton” from “neutron”. In each case one of the two particles’ states is unchanged. On the other hand $\hat{A} \otimes \hat{B}$ changes the states of both particles, and $\hat{A} \otimes \hat{B} + \hat{C} \otimes \hat{D}$ does too, but in a more complicated way. (The latter expression is only valid if $\hat{C}$ and $\hat{D}$ are proton and neutron operators respectively.) Some examples:

\begin{array}{rcl} (⟨ ϕ | \otimes ⟨ α |) (| ψ ⟩ \otimes | β ⟩) & = & ⟨ ϕ | ψ ⟩ ⟨ α | β ⟩ \\ (\hat{A} \otimes {\hat{I}}_{n}) (| ϕ ⟩ \otimes | α ⟩) & = & (\hat{A} | ϕ ⟩) \otimes | α ⟩ \\ (\hat{A} \otimes {\hat{I}}_{n} + {\hat{I}}_{p} \otimes \hat{B}) (| ϕ ⟩ \otimes | α ⟩) & = & (\hat{A} | ϕ ⟩) \otimes | α ⟩ + | ϕ ⟩ \otimes (\hat{B} | α ⟩) \\ (\hat{A} \otimes \hat{B} + \hat{C} \otimes \hat{D}) (| ϕ ⟩ \otimes | α ⟩ + | ψ ⟩ \otimes | β ⟩) & = & (\hat{A} | ϕ ⟩) \otimes (\hat{B} | α ⟩) + (\hat{A} | ψ ⟩) \otimes (\hat{B} | β ⟩) \\ + (\hat{C} | ϕ ⟩) \otimes (\hat{D} | α ⟩) + (\hat{C} | ψ ⟩) \otimes (\hat{D} | β ⟩) \end{array}

If $| ϕ ⟩$ is the spatial state of a particle and $| α ⟩$ its spin state, the common notation $ϕ (x) | α ⟩$ actually stands for $(⟨ x | \otimes I_{S}) (| ϕ ⟩ \otimes | α ⟩)$ .

The direct product notation is clumsy, and shorthands are often used. If we indicate via a label which space an operator acts on, eg by writing ${\hat{A}}_{p}$ and ${\hat{B}}_{n}$ , or if it is otherwise obvious, we often drop the explict identity operators in the other space and hence just write the operator in the third equation above as ${\hat{A}}_{p} + {\hat{B}}_{n}$ . Even more succinctly, we may use a single ket with two labels to stand for the state of the combined system, for example $| ϕ, α ⟩$ . An example of this would be the two-dimensional harmonic oscillator described in section A.4. Such a labelling though implies that we want to use basis states that are direct products of the sub-space basis states, and that might not be convenient. In the case of addition of angular momentum (section A.2), we are more likely to want to use eigenstates of the total angular momentum of the system as our basis states, and they are not seperable.

References

Shankar 10.1

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