Quantum mechanics ground state

In summary, the ground state of a quantum mechanical system must not have a node, and the first excited state must have 1 node. These requirements arise from the physical and mathematical aspects of quantum mechanics, and are a fundamental property of the system.
  • #1
jasony
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Why must the ground state not have a node? And the first excited state must have 1 node.
 
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  • #2
jasony said:
Why must the ground state not have a node? And the first excited state must have 1 node.

Excellent questions. The first thing to understand is that the energy of a wavefunction is inversely proportional to its curvature ... more sharply curved --> higher energy. Second, if a wavefunction crosses the abscissa (i.e. goes through zero) exactly once, then it must necessarily have higher curvature than a wavefunction that never crosses the abscissa (assuming all the other requirements for QM wavefunctions are satisfied). So, for a given Hamiltonian, the energy of a one-node wavefunction must always be higher than a no-node wavefunction.

Now, that doesn't really answer your first question. To understand that, you need to know that the Hamiltonian is a Hermitian operator, and the eigenfunctions of a Hermitian operator form a complete, orthonormal set. Note also that the ground and first-excited state we are talking about here are eigenfunctions of the Hamiltonian as well. So, in most analytically solvable cases, the mathematical expressions for the eigenfunctions are actually in the form of orthogonal polynomials ... that is:

[tex]\psi_n(x)=N_ng_n(x)\sum_i{c_{n,i}x^i}[/tex]

where n is a quantum number indexing the states, Nn is a normalization constant, gn is a node-less, non-polynomial function such as an exponential or a gaussian, and the cn,i are the expansion coefficients of a polynomial, which are chosen to ensure that polynomials of different order are orthogonal to each other. Clearly, there should be a polynomial of order zero, which will be node-less.

So, given the above points, I hope it is clear that, the node-less quality of the ground state is a fundamental property arising from the physical and mathematical aspects of quantum mechanics.

Note that I have made some simplifying assumptions in the above discussion, for example, if your system has symmetry, then it may not be true that the lowest energy state within a given irreducible representation is node-less, however it should be true that wavefunctions with more modes will have higher energies. I am not actually sure that it holds in *all* cases, although it certainly holds for intro level QM, and I cannot think of any cases where the ground state is *not* node-less.
 
  • #3
Can we always write the wavefunction as [tex]\psi_n(x)=N_ng_n(x)\sum_i{c_{n,i}x^i}[/tex]?

Why?

Is there other simple proof for the nodeless property of ground state wavefunction?
 
  • #4
jasony said:
Can we always write the wavefunction as [tex]\psi_n(x)=N_ng_n(x)\sum_i{c_{n,i}x^i}[/tex]?

Why?

Is there other simple proof for the nodeless property of ground state wavefunction?

that is just the most general polynomial function and the argument is actually the most simplest explanation I have seen. It is just theory of eigenfunctions and eigenvalues, go study functional analysis :)
 
  • #5
jasony said:
Can we always write the wavefunction as [tex]\psi_n(x)=N_ng_n(x)\sum_i{c_{n,i}x^i}[/tex]?

Why?

Since [tex]\sum_i{c_{n,i}x^i}[/tex] is a polynomial and the only requirement of [tex]g_n(x)[/tex] is that it is a non-polynomial, it follows that [tex]\psi_n(x)[/tex] can be any function which satisfies the normalization requirement. Therefore, it can also be any wavefunction.
 

FAQ: Quantum mechanics ground state

What is the ground state in quantum mechanics?

The ground state in quantum mechanics refers to the lowest energy state that a quantum mechanical system can have. It is the state in which a particle or system is at its most stable and has the lowest possible energy.

How is the ground state different from excited states?

Excited states refer to any energy state above the ground state. These states have higher energy levels and are less stable than the ground state. When a system transitions from an excited state to the ground state, it releases energy in the form of radiation or heat.

How is the ground state determined in quantum mechanics?

In quantum mechanics, the ground state is determined by solving the Schrödinger equation for a given system. The solution yields the energy of the ground state and the corresponding wave function, which describes the probability of finding a particle in a certain position.

Can the ground state of a system change?

The ground state of a system can change if the system is subjected to external forces or interactions. For example, when an atom absorbs energy, it can transition from the ground state to an excited state. However, the ground state is always the lowest energy state that a system can attain.

What is the significance of the ground state in quantum mechanics?

The ground state is significant because it serves as a reference point for all other energy states in a system. It also determines the behavior and properties of a system, such as the spacing of energy levels and the stability of the system. Understanding the ground state is crucial for understanding the behavior of matter at a quantum level.

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