System of combined observables presented as tensor products

Question

A state can be written as $$| \psi \rangle = \sum c_n | \psi_{nlm} \rangle$$ where $| \psi_{nlm \rangle}$ is the stationary states or eigenstates of the Hamiltonian in three dimensions (spherical coordinates). Hence $\langle \mathbf{r} | \psi_{nlm} \rangle = \psi_{nlm}(r, \theta, \phi)$ is the projection of the Hamiltonian eigenfunction onto the position space and then from a postulate of QM it follows that $|\langle \mathbf{r} | \psi_{nlm} \rangle|^2$ is the probability of a state $| \psi \rangle$ yielding the energy associated with the associated eigenfunction. Similarly we could represent the general state of a spin-$\frac{1}{2}$ particle as a linear combination of eigenstates of $\hat{S}_{z}$ ($z$ component of spin oeprator) hence $$ |\psi \rangle= \begin{bmatrix} a \\ b \end{bmatrix} = a \chi_{+} + b \chi_{-}, $$ where $\chi_{+} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\chi_{-} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ and $|\langle \chi_{+} | \psi \rangle|^2$ is the probability of measuring spin up.

Question: Consider the combined system: An electron in a hydrogen atom occupies the combined spin and position state given as $$\sqrt{\frac{1}{3}}R_{21}Y_{1}^{0} \otimes \chi_{+} + \sqrt{\frac{2}{3}}R_{21}Y_{1}^{1} \otimes \chi_{-}.$$

What is the interpretation of this expression. It does not seem to be a state $| \psi \rangle$ since one side ($R_{21}Y_{1}^{0}$ and $R_{21}Y_{1}^{1}$) of the tensor product in each term is projected onto positional space while the other is the two dimensional vectors $\chi_{1}$ and $\chi_{2}$ which doesn't seem to make sense to talk about position basis projection $\langle \mathbf{r} | \chi_{+} \rangle$. It also doesn't seem like a projection since we have a tensor product rather than a scalar value. How would you interpret exactly what this expression is in terms of states, projections and probability coeffiecients?

See my answer below. Thanks for any help.

Answer 1

To understand the state of the electron, it is convenient to obtain the Hilbert space it belongs to. It can be split into two parts:

• The Hilbert space for the particle without spin is just $L^2\left(\mathbb{R}^3\right)$, the square-integrable functions from the three-dimensional space to the complex plane. A basis of this space is given by the functions $\psi_{nlm}=R_{nl}Y^m_l$.

• The space of states of a particle of spin $1/2$ without taking into account its position is the two-dimensional complex vector space $\mathbb{C}^2$. A basis is just the set $\left\{\chi_+,\chi_-\right\}$.

The Hilbert space of the combined system (the particle in three dimensions with spin $1/2$) is just the tensor product $L^2\left(\mathbb{R}^3\right)\otimes\mathbb{C}^2$. A basis of the tensor product of two spaces if given by all the combinations of the elements of the basis of the original spaces, so any state in our tensor product can always be expressed in the combined basis: \begin{equation} \left|\psi\right>=\sum_{n,l,m,s}c_{nlms}\,R_{nl}Y^m_l\otimes\chi_s \end{equation}

This expression may be interpreted as a superposition of states $R_{nl}Y^m_l\otimes\chi_s$, each of which can be understood as the electron being in the state $R_{nl}Y^m_l$ for its position and in state $\chi_s$ for its spin.

Answer 2

My suggestion is as follows, please give feedback on this: Consider just the first term (for now) $$\sqrt{\frac{1}{3}} R_{21} Y^{0}_{1} \otimes \chi_{+}$$ we can consider this as the projection onto the position basis in the following way:

Consider a general representation of our quantum state $| \psi \rangle$ presented as a linear combination of basis eigenstates in the tensor product space of observables, namely position and spin. Thus for the first term we have something like $| \psi \rangle = | \psi_{nlm} \rangle \otimes | \chi_i \rangle$ where $i = 1,2$. Note that the position ket $| \mathbf{r} \rangle$, which when taken as a bra, is a linear operator and is defined on the tensor product as $\langle \mathbf{r}|_{\otimes} := \langle \mathbf{r}| \otimes I$, where $I$ is the identity operator. Thus when acting on a vector in the tensor product, we have $$\begin{align} \langle \mathbf{r}|_{\otimes} \psi \rangle &= (\langle \mathbf{r}|_{\otimes} \otimes I)(| \psi_{nlm} \rangle \otimes | \chi_i \rangle) \\& =\langle \mathbf{r}| \psi_{nlm} \rangle \otimes | \chi_{i} \rangle \\& = \sqrt{\frac{1}{3}} R_{nl} Y^{m}_{l} \otimes |\chi_{i} \rangle \end{align} $$

Now for the state described in the original question we would have $$| \psi \rangle = \sqrt{\frac{1}{3}} | \psi_{210} \rangle \otimes | \chi_{+} \rangle + \sqrt{\frac{2}{3}} | \psi_{211} \rangle \otimes | \chi_{-} \rangle .$$ Therefore $$\langle \mathbf{r}|_{\otimes} \psi \rangle = \sqrt{\frac{1}{3}}R_{21}Y_{0}^{1} \otimes |\chi_{+} \rangle + \sqrt{\frac{2}{3}}R_{21}Y_{1}^{1} \otimes |\chi_{-} \rangle.$$

If we are interested in finding the probability density of the particle at $(r, \theta, \phi)$ then this would be defined as usual $$ \begin{align} | \langle \mathbf{r}|_{\otimes} \psi \rangle |^2 &= \bigg\langle \sqrt{\frac{1}{3}}R_{21}(r) Y_{1}^{0}(\theta, \phi) \otimes |\chi_{+} \rangle + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta, \phi) \otimes |\chi_{-} \rangle, \sqrt{\frac{1}{3}}R_{21}(r)Y_{0}^{1}(\theta, \phi) \otimes |\chi_{+} \rangle + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta, \phi) \otimes |\chi_{-} \rangle \bigg \rangle. \\ &= \frac{1}{3}|R_{21}(r)|^{2} \big( |Y_{1}^{0}(\theta, \phi)|^2 + 2|Y_{1}^{1}(\theta, \phi)|^2 \big) \end{align}$$

Is the reasoning fine?