Why does the system has lower energy if its wave function is symmetric?

In summary, the conversation discusses the relationship between the symmetry of a system's wave function and its energy state. While it is not always true that symmetric wave functions have lower energy than anti-symmetric ones, there is a "node-counting theorem" that states the energies of states are ordered according to the number of nodes in the wave function. This means that a state with zero nodes, which is symmetric, will have the lowest energy. However, it may be difficult to find a reference for this theorem.
  • #1
xfshi2000
31
0
Hi all:
I am confused that in general case, if [H,p]=0 (where H is Hamiltonian of system and P is parity operator), system wave function is either symmetric or antisymmetric. How do we know that system is in lower energy state if its wave function is symmetric by comparing that system is described by antisymmetric wave function? What I said is true or not? If true, what is physical significance behind?

thanks

xf
 
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  • #2
xfshi2000 said:
Hi all:
I am confused that in general case, if [H,p]=0 (where H is Hamiltonian of system and P is parity operator), system wave function is either symmetric or antisymmetric. How do we know that system is in lower energy state if its wave function is symmetric by comparing that system is described by antisymmetric wave function? What I said is true or not? If true, what is physical significance behind?

thanks

xf

It is not generally true that symmetric (wrt parity) wave-functions always have lower energy than anti-symmetric ones. However, there is a "node-counting theorem" (at least for bound states) which basically states the energies of the states are ordered according to the number of nodes in the wave-function. I.e. a state with a wave-function with n nodes has a lower energy than a state with n+1 nodes and so on. Since any anti-symmetric wave-function must have at least one node there usually (always?) exists a symmetric wave-function with zero nodes which has the lowest energy.

At least this is what I can remember, please correct me if I am wrong (I'm too lazy to look it up)
 
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  • #3
jensa said:
It is not generally true that symmetric (wrt parity) wave-functions always have lower energy than anti-symmetric ones. However, there is a "node-counting theorem" (at least for bound states) which basically states the energies of the states are ordered according to the number of nodes in the wave-function. I.e. a state with a wave-function with n nodes has a lower energy than a state with n+1 nodes and so on. Since any anti-symmetric wave-function must have at least one node there usually (always?) exists a symmetric wave-function with zero nodes which has the lowest energy.

At least this is what I can remember, please correct me if I am wrong (I'm too lazy to look it up)



thanks for your answer. By the way, where can I find node-counting theorem? which textbook talk about this theorem?
 
  • #4
Unfortunately I don't know a good reference discussing this. It seems to be very overlooked in QM books. Probably it is more likely to find this in some book on differential equations. Some googling led me to this http://www.emis.de/journals/AM/98-1/kusano.ps" . Not sure if it helps you.

Many people refer to this theorem without giving any references. I remember having a very hard time finding a proof of this theorem and now I don't remember where I found it. Good luck!
 
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Related to Why does the system has lower energy if its wave function is symmetric?

1. Why is the system's wave function symmetric?

The symmetry of a wave function is determined by the underlying physical properties of the system. In quantum mechanics, a symmetric wave function indicates that the system's particles have identical properties and are indistinguishable from one another.

2. How does a symmetric wave function affect the energy of a system?

A symmetric wave function leads to a lower energy state in a system because it allows for more efficient packing of particles. This results in a stronger attraction between particles and a more stable system overall.

3. Can a system have a symmetric wave function and still have a high energy state?

Yes, it is possible for a system to have a symmetric wave function and still have a high energy state. This can occur if the system's particles are not confined to a small space and have enough kinetic energy to overcome the attractive forces between them.

4. What happens if the wave function of a system is not symmetric?

If the wave function of a system is not symmetric, it can lead to a higher energy state. This is because the particles in the system are not able to pack as efficiently, resulting in weaker attraction between them and a less stable system overall.

5. How does the symmetry of a wave function relate to the overall properties of a system?

The symmetry of a wave function is directly related to the overall properties of a system. A symmetric wave function indicates that the system has properties such as conservation of energy and momentum, while a non-symmetric wave function may indicate that these properties are not conserved.

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