Single ket for a product of two wave functions

[Amentia]

Hello,

I would like to write a product of two wave functions with a single ket. Although it looks simple, I do not remember seeing this in any textbook on quantum mechanics. Assume we have the following:

$\chi(x) = \psi(x)\phi(x) = \langle x | \psi \rangle \langle x | \phi \rangle$

I would like to find the expression for $| \chi \rangle$ but how to remove both x representations?

$\langle x | \chi \rangle = \langle x | \left( | \psi \rangle \phi(x)\right)$

This would give:

$| \chi \rangle = \phi(x) | \psi \rangle$

Is there a way to remove the last x? Is this form a correct ket? Also is the situation different if $\phi(x)$ is not a known wave function but only same random function depending on x (say exp) while $\psi(x)$ is a known basis, e.g. the spherical harmonics?

 

[samalkhaiat]

The multiplication of wave functions (in $L^{2}$) is point wise, i.e., for $\chi = \psi \phi$, you have $$\chi (x) = (\psi \phi)(x) = \psi (x) \phi (x)$$ So in the Dirac notations $$\langle x | \chi \rangle = \langle x |\psi \phi \rangle = \langle x |\left(|\psi \rangle |\phi \rangle \right) = \langle x | \psi \rangle \langle x | \phi \rangle$$ So $$| \chi \rangle = | \psi \rangle |\phi \rangle ,$$ just as $$\chi = \psi \phi$$

 

[Amentia]

Hello,

Thank you for your answer. I thought about the tensor product but it is unclear to me because of the inner product.

$$\langle\psi | \chi \rangle = \langle\psi | \psi \rangle |\phi \rangle = \left(\int dx \psi^{*}(x) \psi(x)\right) |\phi \rangle$$ should be true since we have separate subspaces. But by applying directly the usual definition of the inner product to the wave functions, we get:

$$\langle\psi | \chi \rangle = \int dx \psi^{*}(x) \chi(x) = \int dx \psi^{*}(x)\psi(x)\phi(x),$$

so that $\phi$ appears inside the integral. How can we reconcile the two representations?

 

[samalkhaiat]

Your inner product should be in $L^{2} \otimes L^{2}$: Consider the familiar case of atomic ket: $$|l , m ; n \rangle = |l , m \rangle \otimes | n \rangle .$$ So the wave function in the bases $|\vec{x} , \theta , \phi \rangle$ is $$\Psi_{nlm} ( \vec{x} , \theta , \phi ) = \langle \vec{x} , \theta , \phi | l , m ; n \rangle = \langle \theta , \phi | l , m \rangle \langle \vec{x} | n \rangle = Y_{lm} ( \theta , \phi ) \psi_{n} ( \vec{x}) .$$

 

[Amentia]

Thanks, your answer is very clear for this example. But I would like to remove one of the wavefunction by taking the inner product on only half of the subspaces with two functions depending on $\vec{x}$. Is it possible to separate the integration in this case?

 

[samalkhaiat]

Thanks, your answer is very clear for this example. But I would like to remove one of the wavefunction by taking the inner product on only half of the subspaces with two functions depending on $\vec{x}$. Is it possible to separate the integration in this case?

It seems that my notations in #2 have confused you. You probably have been told in QM1 that if particle one is described by $\psi_{1} (\vec{x}_{1}) \in L^{2}(\mathbb{R}^{3})$ and particle two is described by $\psi_{2} (\vec{x}_{2}) \in L^{2}(\mathbb{R}^{3})$ then the combined system of the two particles is described by the tensor product wavefunction $$\chi (\vec{x}_{1}, \vec{x}_{2}) \equiv ( \psi_{1} \otimes \psi_{2} ) (\vec{x}_{1} , \vec{x}_{2}) \in L^{2} (\mathbb{R}^{3} \times \mathbb{R}^{3})$$. In Dirac notation, this is written as $$\left( \langle \vec{x}_{1}| \otimes \langle \vec{x}_{2}| \right) | \chi \rangle = \left( \langle \vec{x}_{1}| \otimes \langle \vec{x}_{2}| \right) \left( | \psi_{1} \rangle \otimes | \psi_{2} \rangle \right) = \langle \vec{x}_{1} | \psi_{1}\rangle \langle \vec{x}_{2}|\psi_{2}\rangle .$$ The important thing to know is the fact that $\chi (\vec{x}_{1})$ and $\chi (\vec{x}_{2})$ are meaningless expressions. Because of the identity $$L^{2}(\mathbb{R}^{n}) \otimes L^{2}(\mathbb{R}^{m}) \cong L^{2} (\mathbb{R}^{n} \times \mathbb{R}^{m}) ,$$ the arguments of the tensor product wave function $\chi$ must be $(\vec{x}_{1} , \vec{x}_{2})$. This you should know from QM1 when treating the so-called exchange symmetry or the Pauli Exclusion Principle. That is, for fermions, you need to consider an anti-symmetric wave function $$\Psi (\vec{x}_{1} , \vec{x}_{2}) = \frac{1}{2} (\psi_{1} \otimes \psi_{2} - \psi_{2} \otimes \psi_{1}) (\vec{x}_{1} , \vec{x}_{2}) ,$$ and for bosons, you take the symmetric wave function $$\Phi (\vec{x}_{1} , \vec{x}_{2}) = \frac{1}{2} (\psi_{1} \otimes \psi_{2} + \psi_{2} \otimes \psi_{1}) (\vec{x}_{1} , \vec{x}_{2}) .$$ Notice that $$\chi (\vec{x}_{1}, \vec{x}_{2}) \equiv \psi_{1}(\vec{x}_{1}) \psi_{2}(\vec{x}_{2}) = \Psi (\vec{x}_{1}, \vec{x}_{2}) + \Phi (\vec{x}_{1}, \vec{x}_{2}).$$

Now, this business of “removing one wavefunction” is straightforward. For example $$\psi_{2} (\vec{x}_{2}) = \frac{\int d^{3}x_{1} \ \psi_{1}(\vec{x}_{1}) \chi (\vec{x}_{1} , \vec{x}_{2})}{\int d^{3}y \ \psi_{1}^{\ast}(\vec{y}) \psi_{1} (\vec{y})} .$$