Multiplying a wavefunction by a constant number

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dyn
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Hi.
Is the wavefunction for x≤0 , ψ(x) = sinkx - acoskx equivalent to ψ(x) = -sinkx - acoskx where a is a constant ?
Thanks
 
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How to multiply wavefunction with a constant ? Could anyone please tell...thanks
 
Thank you could you give an example? Thanks
 
Just so I make sure I understand this properly and I will use kets instead please tell me which of the following examples are equivalent to the ket
| u > + | v >

1. a ( | u > + | v > ) where a is any number ( real or complex)
2. | u > + ( a | v > )
3 . e-ix ( | u > + | v > )
4. | u > + ( e-ix | v > )
 
They are all different, but the first one is proportional to the initial one. For the third one it depends on what x is. If it refers to a spatial variable then it is not constant, if it is just some constant it is constant.
 
Thanks. Regarding the 1st example if it is proportional to the initial example is it not considered an equivalent wavefunction ?
My confusion arises because I have come across the following statement " multiplying a wavefunction by a constant number or a phase doesn't change the wavefunction"
 
dyn said:
I have come across the following statement " multiplying a wavefunction by a constant number or a phase doesn't change the wavefunction"

Where did you come across this statement? Can you give a specific reference?
 
The exact statement in some lecture notes from MIT is " ψ and αψ represent the same physics for any complex number α different from zero , so
| A > ≅ 2 | A > ≅ i | A > ≅ - | A > where ≅ represents physical equivalence "
 
dyn said:
Is the wavefunction for x≤0 , ψ(x) = sinkx - acoskx equivalent to ψ(x) = -sinkx - acoskx where a is a constant ?
Perhaps you meant: ψ(x) = sinkx - acoskx equivalent to ψ(x) = -sinkx + acoskx
(negative negative = plus)

If you change the overall scaling and phase of everything in the whole system equally, it makes no difference to the system. But if you change the phase of one wavefunction with respect to another wavefunction, this will change the system.
 
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