# Why do we multiply 2 wavefunctions in the ground state instead of adding them when considering the total wavefunction?

I'm working through problem 7.3 from Griffith's quantum mechanics 3rd edition:

Problem 7.3: Two identical spin-zero bosons are placed in an infinite square well. They interact weakly with one another, via the potential (where is a constant with the dimensions of energy, and a is the width of the well).

V(x1,x2) = -aVoδ(x1-x2)

(a) First, ignoring the interaction between the particles, find the ground state and the first excited state—both the wave functions and the associated energies.

It made sense to me that since the particles were both bosons (and thus had symmetric wavefunctions), that I would describe the ground state energy as |Ψ0> = |Ψx1,0> + |Ψx2,0>. But the answer gave by multiplying the two wavefunctions instead. Why is this? Don't we express a superpositon as a linear combination? What am I missing here?

• Feb 16 at 19:59

The operator for the multiple particle system acts on the product space, of particle $$I$$ one and particle two $$II$$. The single particle operators must be extended so that they can act on states of the product space, $$\hat H_I\rightarrow \hat H_I\otimes \hat 1_{II}$$ and $$\hat H_{II}\rightarrow \hat 1_I\otimes \hat H_{II}$$. The eigenfunctions of this "sum" of operators can be constructed by multiplying the eigenfunctions of the single particle operators which is why the groundstate of the combined system(without additional interactions) is given by a product of states.

$$(\hat H_I\otimes \hat 1_{II} +\hat 1_I\otimes \hat H_{II})(|n_I\rangle \otimes |m_{II}\rangle )= \hat H_I|n_I\rangle \otimes \hat 1_{II}|m_{II}\rangle + \hat 1_I|n_I\rangle \otimes \hat H_{II}|m_{II}\rangle\\ =E_{n_I} |n_I\rangle \otimes|m_{II}\rangle + |n_I\rangle \otimes E_{m_{II}}|m_{II}\rangle\\ =(E_{n_I}+E_{m_{II}} )|n_I\rangle \otimes|m_{II}\rangle$$

The explicit notation is usually skipped, $$(\hat H_{I} + \hat H_{II})|n_I, m_{II}\rangle =(E_{n_I}+E_{m_{II}})|n_I, m_{II}\rangle$$

Combining particles that come with their own degrees of freedom, extends the dimensionality of your problem while a superposition takes place in one and the same space of states. Superposition state are also generally not eigenfunctions(unless your superposition contains only degenerate states, which is a special case of a superposition).

This follows from separation of variables: if the particles are not interacting then your TISE reads: $$\left(-\frac{\hbar^2}{2m_1}\frac{\partial^2}{\partial x_1^2}+V(x_1) -\frac{\hbar^2}{2m_2}\frac{\partial^2}{\partial x_2^2}+V(x_2)\right)\Psi(x_1,x_2)=E\Psi(x_1,x_2)\, .$$ Assuming a product form $$\Psi(x_1,x_2)=\psi(x_1)\chi(x_2)$$ gives a TISE for $$x_1$$ alone, with solution $$\psi(x_1)$$, and another TISE for $$x_2$$ alone, with solution $$\chi(x_2)$$.

The form of $$\Psi$$ does not hold is there is an interaction as you cannot isolate a function of $$x_1$$ alone, and another function of $$x_2$$ alone.

The total wave function should be the product of the two if you are ignoring interactions. For sake of argument, you can consider the wavefuntions as two independent probability distributions, and following fundamental laws of probability, the probability of two independent events both occurring is the product of their individual probabilities: $$P(a \& b) = P(a) P(b)$$.

Note that the total wavefuntion will be symmetric under exchange, whereas you'd have to be more careful in your construction of the total wave function if you were dealing with fermions: $$\psi_{bosons}(x_1, x_2, t) = \psi_1(x_1, t)\psi_2(x_2, t)$$ $$\psi_{fermions}(x_1, x_2, t) = ( \psi_1(x_1, t)\psi_2(x_2, t) - \psi_1(x_2, t)\psi_2(x_1, t)) = - \psi_{fermions}(x_2, x_1, t)$$

Also, this problem that you are trying to solve is different from the phenomenon of superposition.

The state $$|\Psi_0 \rangle = |\Psi_{x_1, 0} \rangle + |\Psi_{x_2, 0} \rangle$$ that you are considering describes a single particle being in the superposition of two states. But this superposition of states still describes a single particle. It tells you that the particle "is both in the state $$|\Psi_{x_1, 0} \rangle$$ and $$|\Psi_{x_2, 0} \rangle$$", meaning more precisely that if we were to measure it, we would find it with probability $$1/2$$ being $$|\Psi_{x_1, 0} \rangle$$ and with probability $$1/2$$ being $$|\Psi_{x_2, 0} \rangle$$.

If you want to consider a multiparticle state, you should treat the two particles as living in two different vector spaces. Mathematically, we would take the tensor product of two Hilbert spaces, say $$H = H_1 \otimes H_2$$, where $$H_i$$ is the Hilbert space corresponding to the i'th particle. Then we simultaneously describe the two particles with a state being a (tensor) product of two states, for example $$|\Psi_0 \rangle = |\Psi_{x_1, 0} \rangle \otimes |\Psi_{x_2, 0} \rangle = |\Psi_{x_1, 0} \rangle |\Psi_{x_2, 0} \rangle$$ The fact that we want to describe indistiguisable particles, means that the ordering of the tensor product should not matter, there should be no way to dinstiguish the first component from the second. To meet this requirement, we need to either anti-symmetrize or symmetrize the above state (and normalize it correctly). The fact that the particles are bosons tells us to choose to symmetrize the state :

$$|\Psi_0 \rangle = \frac{1}{\sqrt{2}} \left( |\Psi_{x_1, 0} \rangle |\Psi_{x_2, 0} \rangle + |\Psi_{x_2, 0} \rangle |\Psi_{x_1, 0} \rangle \right )$$