Consider an incoming beam of particles with E &gt; U scattering from the potential V ( x ) = { 0 for x &lt; − a U for | x | &lt; a 0 for x &gt; a So Schrodinger's (time-independent) equation reads ψ ″ = { − 2 m E ℏ 2 ψ for | x | &gt; a − 2 m ℏ 2 ( E − U ) ψ for | x | &lt; a Then the scattering states are given by ψ ( x ) = { exp ⁡ ( i k x ) + R exp ⁡ ( − i k x ) for x &lt; − a A exp ⁡ ( − l x ) + B exp ⁡ ( l x ) for | x | &lt; a T exp ⁡ ( i k x ) for x &gt; a where k = 2 m E ℏ 2 , l = − 2 m ℏ 2 ( E − U ) .[The above is from my lecture notes].Now my questions are:1. From the form of the solution in | x | &lt; a it looks like we are assuming that l ∈ R , i.e. that E &lt; U. Why is that? Why can't we have a combination of sines and cosines?2. There's a theorem that says thatAny solution of the time-independent Schrodinger equation with a symmetric potential is a linear combination of solutions of definite parity.We clearly have a symmetric potential here, and for real l, exp ⁡ ( ± l x ) doesn't have definite parity (and as far as I know can't be written as a (finite) linear combination of functions of definite parity). So given this theorem, the 'assumption' that l ∈ R seems even stranger/less justified. Do we just throw this theorem away because we are dealing with scattering solutions?What's going on?

Question

Consider an incoming beam of particles with   E  &amp;gt;  U scattering from the potential  V  (  x  )  =      {                            0                           for           x          &amp;lt;          −          a                                      U                           for                       |                    x                      |                    &amp;lt;          a                                      0                           for           x          &amp;gt;          a                        So Schrodinger&#039;s (time-independent) equation reads      ψ    ″    =      {                            −                                    2              m              E                                      ℏ              2                                ψ                           for                       |                    x                      |                    &amp;gt;          a                                      −                                    2              m                                      ℏ              2                                (          E          −          U          )          ψ                           for                       |                    x                      |                    &amp;lt;          a                        Then the scattering states are given by  ψ  (  x  )  =      {                            exp          ⁡          (          i          k          x          )          +          R          exp          ⁡          (          −          i          k          x          )                           for           x          &amp;lt;          −          a                                      A          exp          ⁡          (          −          l          x          )          +          B          exp          ⁡          (          l          x          )                           for                       |                    x                      |                    &amp;lt;          a                                      T          exp          ⁡          (          i          k          x          )                           for           x          &amp;gt;          a                        where   k  =                    2        m        E                    ℏ        2            ,   l  =      −                  2        m                    ℏ        2              (    E    −    U    )  .[The above is from my lecture notes].Now my questions are:1. From the form of the solution in       |    x      |    &amp;lt;  a it looks like we are assuming that   l  ∈      R  , i.e. that   E  &amp;lt;  U. Why is that? Why can&#039;t we have a combination of sines and cosines?2. There&#039;s a theorem that says thatAny solution of the time-independent Schrodinger equation with a symmetric potential is a linear combination of solutions of definite parity.We clearly have a symmetric potential here, and for real l,   exp  ⁡  (  ±  l  x  ) doesn&#039;t have definite parity (and as far as I know can&#039;t be written as a (finite) linear combination of functions of definite parity). So given this theorem, the &#039;assumption&#039; that   l  ∈      R   seems even stranger/less justified. Do we just throw this theorem away because we are dealing with scattering solutions?What&#039;s going on?

luisjoseblash2 · Accepted Answer

1. You assume in your opening statement that   E  &amp;gt;  U, then you SHOULD actually have a combination of sines and cosines for the middle part. It&#039;s only when   E  &amp;lt;  U is assumed that you have exponentials.2. It&#039;s true that the exponentials on their own do not have a definite parity, however, you can rewrite them as linear combinations of sinh and cosh that do have definite parity.

Consider an incoming beam of particles with E > U scattering from the potential

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