3
$\begingroup$

enter image description here

Here we consider the green marked paragraph from Kerson-Huang statistical mechanics book. First of all they say that $\psi$ is a state of the system where $c_n$ is the coefficient of the $n^{th}$ orthogonal basis. So $c_n$ must be a complex number. The orthogonal baciss must be a state of well defined position of all particles of the system. But then in the green paragraph they say that in quantum mechanics we have to consider interactions of the system with the surroundings. So they turn $c_n$ to the wavefunction($1×\infty $ matrics) of the surrounding. My question is how can we multiply 2 wavefunctions and get a wavefunction?

$\endgroup$
2
  • $\begingroup$ ChatGPT can eat the screenshot and spit out the corresponding $\LaTeX$ $\endgroup$ Commented Sep 11 at 9:12
  • 1
    $\begingroup$ @RodrigodeAzevedo I suggest this: simpletex.cn. It is better than ChatGpt. $\endgroup$ Commented Sep 15 at 22:16

2 Answers 2

10
$\begingroup$

By the standard notion that is called the tensor product and usually denoted by $\otimes$

You have actually already seen it before. The wavefunction of a spin-half electron is actually living in the tensor product space of the Schrödinger-like wavefunction and the spin-half-only-part "wavefunction".

Or if you have done the space part of two-electron Schrödinger wavefunctions, such wavefunctions live in configuration space that describes both electrons at once, such that in wavefunction form we write it as $\psi(\vec r_1,\vec r_2)$. It is clearer when we write it in bra-ket notation, where we usually abbreviate what is supposedly $\left|\psi_1\right>\otimes\left|\psi_2\right>$ by omitting the $\otimes$, i.e. it becomes written as $\left|\psi_1\right>\left|\psi_2\right>$ and often also as $\left|\psi_1,\psi_2\right>$


The green paragraph is much easier to understand if they properly introduce Equation (8.1), and I really do not like how it is treated there.

The mathematical and physical meaning that we really want to imbue Equation (8.1) with, and want it to say, is that any complete basis of states $\left|\Phi_n\right>$ can be used to express the identity operator $$\tag1\hat{\mathbb I}=\sum_n\left|\Phi_n\rangle\langle\Phi_n\right|$$ so that a blind application of Equation (1) onto any quantum state $\left|\Psi\right>$ begets $$\tag2\hat{\mathbb I}\left|\Psi\right>=\left|\Psi\right> =\sum_n\left|\Phi_n\rangle\langle\Phi_n|\Psi\right>$$ such that, simply by comparing with Equation (8.1), we get that it must be that $$\tag3c_n=\left<\Phi_n|\Psi\right>$$ in order for all of these equations to agree with each other.

We have only used the completeness of a Hilbert space and that a complete basis must thus exist, and thus this is totally general as long as the quantum postulates hold true.


If you understand the above, then we can proceed to understand the green paragraph. The above only treated the Hilbert space of the system under scrutiny. But any statistical mechanics system that is worth looking at, is also interacting with the external world, as the textbook correctly pointed out.

This means that while $\left|\Phi_n\right>\in\mathcal H_\text{sys}$ continues to only describe the system and live in the system's Hilbert space, we now also have an external world and its external world Hilbert space $\mathcal H_\text{ext}$ that we might express its basis as $\left|\widetilde{\Phi_n}\right>\in\mathcal H_\text{ext}$. The choice of the symbols to describe the external world quantum states to be so close to that of the system's quantum states is useful when we want to describe measurements, otherwise, it is not useful / important to have this similarity.

Then, the appropriate changes are that the total (universe) wavefunction $\left|\Psi\right>\in\mathcal H_\text{uni}=\mathcal H_\text{sys}\otimes\mathcal H_\text{ext}$ has to be expanded in terms of basis states that look like $\left|\Phi_n\right>\otimes\left|\widetilde{\Phi_m}\right>$

When you apply some appropriate replacements to Equations (1), (2) and (3), you will realise that the new $$\tag4c_n=\left<\Phi_n|\otimes\langle\widetilde{\Phi_m}|\Psi\right>$$ which now explicitly depends upon external world, precisely as the textbook stated.

In the measurement case, a good measuring device must have its measurement device quantum states be highly correlated with the system being measured, so that the double sum over $n$ and $m$ (originally being independent), becomes a single sum over $n$ due to imposition of the condition that $m=n,$ which effectively eliminates the sum over $m$

$\endgroup$
1
  • $\begingroup$ Thank for the explanation . I just now have to find the properties of such wavefunction multiplications . $\endgroup$ Commented Sep 11 at 4:22
5
$\begingroup$

My question is how can we multiply 2 wavefunctions and get a wavefunction?

Consider two wavefunctions $\chi$ and $\phi$, which individually satisfy $$ i\frac{\partial \chi(x_1, t)}{\partial t} = -\frac{1}{2m_1}\nabla_1^2\chi(x_1,t) + V(x_1)\chi(x_1,t) $$ and $$ i\frac{\partial \phi(x_2, t)}{\partial t} = -\frac{1}{2m_2}\nabla_2^2\phi(x_2,t) + V(x_2)\phi(x_2,t)\;. $$

Then the product wavefunction $$ \psi(x_1, x_2, t) = \chi(x_1, t)\phi(x_2, t)\tag{1} $$ satisfies $$ i\frac{\partial \psi(x_1, x_2, t)}{\partial t} = (-\frac{1}{2m_1}\nabla_1^2-\frac{1}{2m_2}\nabla_2^2)\psi(x_1,x_2,t) + (V(x_1) + V(x_2))\psi(x_1,x_2,t)\;. $$

That is, the function $\psi(x_1, x_2, t)$ is also a wavefunction for a combined system with a Hamiltonian that is the sum of the individual Hamiltonians for the initial two systems.

So, $\psi$ is a "wavefunction" in the same way that $\chi$ and $\phi$ are "wavefunctions," just for a different (but related) Hamiltonian.


Caveat: The wavefunction of Eq. (1) is a valid wavefunction for distinguishable particles. However, if the particles are indistinguishable then one must consider only totally symmetric (bosons) or totally antisymmetric (fermions) wavefunctions. However, a detailed explanation of this point is beyond the scope of this question/answer.

$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.