# Finding the state and wave-function for two identical spin-1 bosons trapped in one-dimensional harmonic oscillator

My question concerns the validity of my approach to a problem and wether the answer is correct. I am tasked with writing the state vector(s) and wave-function(s) for when two identical spin-1 bosons are trapped in a one-dimensional harmonic oscillator. The question specifically mentions that a convenient way to write a state is

$$\lvert\psi\rangle = \lvert \text{orbital part}\rangle\otimes\lvert\text{spin part}\rangle$$

and I'm supposed to write the state-vectors and wave-functions when

$$\lvert\text{spin part}\rangle = \lvert 10\rangle = \frac{1}{\sqrt{2}}(\lvert +-\rangle - \lvert -+\rangle).$$

My attempt at a solution involves inspection by way of an exchange operator. As the problem concerns Bosons, the state should be symmetric under exchange from which it follows that

$$\lvert\text{orbital part}\rangle = \lvert 10\rangle.$$

From this I find that the harmonic oscillator level n = 1 and the corresponding eigen-state in the coordinate basis for a harmonic oscillator is

$$\langle x\lvert\psi_1\rangle = 2x\sqrt{\frac{m\omega}{\hbar\sqrt{\pi}}}e^{-\frac{m\omega x^{2}}{2\hbar}}.$$

For two identical spin-1 Bosons in the n = 1 state, I have

$$\lvert 2_1\rangle = \frac{(a_1^\dagger)^2}{\sqrt{2}}\lvert 0\rangle.$$

with $(a_1^\dagger)^2\lvert 0\rangle = 3\lvert 1\rangle = 3\lvert\psi_1\rangle$, I get the orbital part in the coordinate basis to be

$$\langle x\lvert 2_1\rangle = 6x\sqrt{\frac{m\omega}{2\hbar\sqrt{\pi}}}e^{-\frac{m\omega x^{2}}{2\hbar}}.$$

Writing this together with the spin part, I get the final wave-function for the two identical particles to be

$$\lvert\Psi\rangle = 3x\sqrt{\frac{m\omega}{\hbar\sqrt{\pi}}}e^{-\frac{m\omega x^{2}}{2\hbar}}(\lvert +-\rangle - \lvert -+\rangle).$$

Is my approach to solving this and subsequent answer correct?