Reading from http://quantummechanics.ucsd.edu/ph130a/130_notes/node151.html It says:

This means that the solutions separate into even parity and odd parity states. We could have guessed this from the potential.

Why A or B is zero, then we will have even and odd parity? What does this sentence mean?

And then, from the graph of the link above, it says:

Where the curves intersect (not including the asymptote), is an allowed energy. There is always one even solution for the 1D potential well. In the graph shown, there are 2 even and one odd solution. The wider and deeper the well, the more solutions.

The solution of the graph shows the energy. How do we know a wider and deeper well will have more solutions?


Why A or B is zero ,then we will have even and odd parity? What is this sentence mean?

If you mean that why is it that $A$ or $B$ must be zero rather than $A$ and $B$ it is because although $A=B=0$ is also a valid solution to the equation you are discussing, it wouldn't have a physical meaning i.e. there would be no wave. It is important when working things out to step back and relate your mathematics back to the real problem you are dealing with. If you mean what is even and odd parity then in this context it simply denotes the positive and negative phase components of the wave. Parity in a sense describes the 'handedness' of a function so that a parity transformation would switch the parity of a system which in this case would mean switching between a positive and negative exponential.

the solution of the graph shows the energy. How do we know wider and deeper well will have more solution?

If you look at the equation given that predicts the bound states you can see the right hand side is a function of $\cot(\sqrt{V_{0}} \cdot a)$. If you sketch a graph of y=cot(x) and compare it to y=cot(2x) or any larger function of cot you see that there are more solutions for each value of $x$. This is the same reason you would get more bound states as V0 indicates the depth of your well and a gives the width.

You can think of it more intuitively by remembering that as energy levels are quantized there can only be a finite amount in any range of energy. If you increase this energy range you allow for more possible states.

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The parity operation is inversion through the origin, i.e. $\hat{P}\space f(x,y,z) = f(-x, -y, -z).$

States of well defined parity are eigenstates of this operator. Here we're interested in 1 dimension, so the two possibilities are: $$ \hat{P}\space \psi(x) = \psi(-x) = \psi(x)$$ (even parity) and $$ \hat{P}\space \psi(x) = \psi(-x) = -\psi(x)$$ (odd parity) It should be fairly obvious that $ \sin(x)$ has odd parity, and $\cos(x)$ has even parity. As you can probably tell, parity has to do with mirror symmetry, and since our potential has mirror symmetry, you can bet our wavefunctions will have mirror symmetry too.

(To be more precise, the parity operation commutes with our hamiltonian, and since the eigenstates of the hamiltonian are non-degenerate, these eigenstates must necessarily be eigenstates of the parity operator.)

As for the second part of your question, note the form of the right hand side: $$\tan \left(\frac{2m(E+V_0)}{\hbar^2}a \right) $$ $\tan$ is a periodic function, and as you increase $a$ or $V_0$ (i.e. the height or depth of our well respectively), the frequency increases, and so you get more intersections with the square root on the right hand side (as more of the s shaped tan wiggles creep in from the left as the frequency gets larger). Intersections represent solutions of our equality and hence bound solutions of the Schrodinger equation.

Hope this helps. If anything is still unclear just ask and I'll try to clarify.

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  • $\begingroup$ Thanks. Seem like yours is not derived from the boundary condition as shown quantummechanics.ucsd.edu/ph130a/130_notes/… . I wonder " wave functions will have mirror symmetry" mean? I don't understand The parity operation refers to? Can you please explain more about "(To be more precise, the parity operation commutes with our hamiltonian, and since the eigenstates of the hamiltonian are non-degenerate, these eigenstates must necessarily be eigenstates of the parity operator.)" thanks $\endgroup$ – Outrageous Nov 13 '13 at 4:07
  • $\begingroup$ So: To keep this simple, I'll stick to 1 dimension. The parity operator acting on a function f(x) changes it to the function f(-x) i.e. everywhere in the function you have an x, you replace it with a -x. If you think about this, it means that it reflects the function f(x) about the line x = 0. $\endgroup$ – Jon Nov 13 '13 at 13:13
  • $\begingroup$ Our potential v(x) looks the same when you reflect it about x=0. The shape of our potential determines the shape of our wavefunctions, so you can guess that our wavefunctions will have this same symmetry. (i.e. $\psi(-x) = \pm \psi(x)$) $\endgroup$ – Jon Nov 13 '13 at 13:27
  • $\begingroup$ It sounds like you don't really know enough about quantum mechanics to get what I'm talking about with the commuting, but basically: If you have any operator A which commutes with your hamiltonian H (i.e. AH = HA), then there's a nice result which says that you can find a set of basis states (wavefunctions) which are simultaneously eigenstates of A and H. In this case we have A = P, the parity operator, so our wavefunctions have mirror symmetry. If you're still unclear I recommend reading an introductory quantum mechanics book. I find Schumacher and Westmoreland quite nice. $\endgroup$ – Jon Nov 13 '13 at 13:49
  • $\begingroup$ Thanks.Just one more to ask, for the parity operation ,I have odd and even , you said " the parity operation commutes with Hamiltonian" , Can you please write your parity operator out in this case? Thanks for the recommended book also. $\endgroup$ – Outrageous Nov 15 '13 at 5:03

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