[<< wikibooks] This Quantum World/Implications and applications/Time independent Schrödinger equation
== Time-independent Schrödinger equation ==
If the potential V does not depend on time, then the Schrödinger equation has solutions that are products of a time-independent function 
  
    
      
        ψ
        (
        
          r
        
        )
      
    
    {\displaystyle \psi (\mathbf {r} )}
   and a time-dependent phase factor 
  
    
      
        
          e
          
            −
            (
            i
            
              /
            
            ℏ
            )
            
            E
            
            t
          
        
      
    
    {\displaystyle e^{-(i/\hbar )\,E\,t}}
  :

  
    
      
        ψ
        (
        t
        ,
        
          r
        
        )
        =
        ψ
        (
        
          r
        
        )
        
        
          e
          
            −
            (
            i
            
              /
            
            ℏ
            )
            
            E
            
            t
          
        
        .
      
    
    {\displaystyle \psi (t,\mathbf {r} )=\psi (\mathbf {r} )\,e^{-(i/\hbar )\,E\,t}.}
  Because the probability density 
  
    
      
        
          |
        
        ψ
        (
        t
        ,
        
          r
        
        )
        
          
            |
          
          
            2
          
        
      
    
    {\displaystyle |\psi (t,\mathbf {r} )|^{2}}
   is independent of time, these solutions are called stationary.
Plug 
  
    
      
        ψ
        (
        
          r
        
        )
        
        
          e
          
            −
            (
            i
            
              /
            
            ℏ
            )
            
            E
            
            t
          
        
      
    
    {\displaystyle \psi (\mathbf {r} )\,e^{-(i/\hbar )\,E\,t}}
   into

  
    
      
        i
        ℏ
        
          
            
              ∂
              ψ
            
            
              ∂
              t
            
          
        
        =
        −
        
          
            
              ℏ
              
                2
              
            
            
              2
              m
            
          
        
        
          
            ∂
            
              ∂
              
                r
              
            
          
        
        ⋅
        
          
            ∂
            
              ∂
              
                r
              
            
          
        
        ψ
        +
        V
        ψ
      
    
    {\displaystyle i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial }{\partial \mathbf {r} }}\cdot {\frac {\partial }{\partial \mathbf {r} }}\psi +V\psi }
  to find that 
  
    
      
        ψ
        (
        
          r
        
        )
      
    
    {\displaystyle \psi (\mathbf {r} )}
   satisfies the time-independent Schrödinger equation

  
    
      
        E
        ψ
        (
        
          r
        
        )
        =
        −
        
          
            
              ℏ
              
                2
              
            
            
              2
              m
            
          
        
        
          (
          
            
              
                
                  ∂
                  
                    2
                  
                
                
                  ∂
                  
                    x
                    
                      2
                    
                  
                
              
            
            +
            
              
                
                  ∂
                  
                    2
                  
                
                
                  ∂
                  
                    y
                    
                      2
                    
                  
                
              
            
            +
            
              
                
                  ∂
                  
                    2
                  
                
                
                  ∂
                  
                    z
                    
                      2
                    
                  
                
              
            
          
          )
        
        ψ
        (
        
          r
        
        )
        +
        V
        (
        
          r
        
        )
        
        ψ
        (
        
          r
        
        )
        .
      
    
    {\displaystyle E\psi (\mathbf {r} )=-{\hbar ^{2} \over 2m}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right)\psi (\mathbf {r} )+V(\mathbf {r} )\,\psi (\mathbf {r} ).}