漢克爾變換是指對任何給定函數 f ( r ) {\displaystyle f(r)} 以第一類貝塞爾函數 J ν ( k r ) {\displaystyle J_{\nu }(kr)} 作無窮級數展開，貝塞爾函數 J ν ( k r ) {\displaystyle J_{\nu }(kr)} 的階數不變，級數各項 k {\displaystyle k} 作變化。各項 J ν ( k r ) {\displaystyle J_{\nu }(kr)} 前係數 F ν {\displaystyle F_{\nu }} 構成了變換函數。對於函數 f ( r ) {\displaystyle f(r)} , 其 ν {\displaystyle \nu } 階貝塞爾函數的漢克爾變換（ k {\displaystyle k} 為自變量）為 F ν ( k ) = ∫ 0 ∞ f ( r ) J ν ( k r ) r d r {\displaystyle F_{\nu }(k)=\int _{0}^{\infty }f(r)J_{\nu }(kr)rdr} 此條目沒有列出任何參考或來源。 (2012年7月19日) 其中， J ν {\displaystyle J_{\nu }} 為階數為 ν {\displaystyle \nu } 的第一類貝塞爾函數， ν ≥ − 1 / 2 {\displaystyle \nu \geq -1/2} 。對應的，逆漢克爾變換 F ν ( k ) {\displaystyle F_{\nu }(k)} 定義為 f ( r ) = ∫ 0 ∞ F ν ( k ) J ν ( k r ) k d k {\displaystyle f(r)=\int _{0}^{\infty }F_{\nu }(k)J_{\nu }(kr)kdk} 漢克爾變換是一種積分變換，最早由德國數學家赫爾曼·漢克爾提出，又被稱為傅立葉-貝塞爾變換。 貝塞爾函數構成 正交函數族 權重因子為 r: ∫ 0 ∞ J ν ( k r ) J ν ( k ′ r ) r d r = δ ( k − k ′ ) k {\displaystyle \int _{0}^{\infty }J_{\nu }(kr)J_{\nu }(k'r)r~\operatorname {d} r={\frac {\delta (k-k')}{k}}} 其中 k {\displaystyle k} 與 k ′ {\displaystyle k'} 大於零。 傅立葉變換 零階漢克爾函數即為圓對稱函數的二維傅立葉變換。給定二維函數 F ( r ) {\displaystyle F({\boldsymbol {r}})} ，徑向矢量為 r {\displaystyle {\boldsymbol {r}}} ，其傅立葉變換為 F ( k ) = ∬ f ( r ) e i k ⋅ r d r {\displaystyle F({\boldsymbol {k}})=\iint f({\boldsymbol {r}})e^{i{\boldsymbol {k}}\cdot {\boldsymbol {r}}}d{\boldsymbol {r}}} 不失一般性，選擇極坐標 ( r , θ ) {\displaystyle (r,\theta )} ，使得矢量 k {\displaystyle {\boldsymbol {k}}} 方向指向 θ = 0 {\displaystyle \theta =0} 。極坐標下的傅立葉變換寫作 F ( k ) = ∫ 0 ∞ ∫ 0 2 π f ( r , θ ) e i k r cos θ r d r d θ {\displaystyle F({\boldsymbol {k}})=\int _{0}^{\infty }\int _{0}^{2\pi }f(r,\theta )e^{ikr\cos \theta }rdrd\theta } 其中 θ {\displaystyle \theta } 為矢量 k {\displaystyle {\boldsymbol {k}}} 與 r {\displaystyle {\boldsymbol {r}}} 間夾角。如果函數 f {\displaystyle f} 恰為圓對稱不依賴角變量 θ {\displaystyle \theta } ， f ≡ f ( r ) {\displaystyle f\equiv f(r)} ，對角度 θ {\displaystyle \theta } 的積分可以提出，傅立葉變換寫作 F ( k ) = F ( k ) = 2 π ∫ 0 ∞ f ( r ) J 0 ( k r ) r d r {\displaystyle F({\boldsymbol {k}})=F(k)=2\pi \int _{0}^{\infty }f(r)J_{0}(kr)rdr} 此式恰為 f ( r ) {\displaystyle f(r)} 的零階漢克爾變換的 2 π {\displaystyle 2\pi } 倍。 More information , ... f ( r ) {\displaystyle f(r)\,} F 0 ( k ) {\displaystyle F_{0}(k)\,} 1 {\displaystyle 1\,} δ ( k ) / k {\displaystyle \delta (k)/k\,} 1 / r {\displaystyle 1/r\,} 1 / k {\displaystyle 1/k\,} r {\displaystyle r\,} − 1 / k 3 {\displaystyle -1/k^{3}\,} r 3 {\displaystyle r^{3}\,} 9 / k 5 {\displaystyle 9/k^{5}\,} r m {\displaystyle r^{m}\,} 2 m + 1 Γ ( m / 2 + 1 ) k m + 2 Γ ( − m / 2 ) {\displaystyle {\frac {2^{m+1}\Gamma (m/2+1)}{k^{m+2}\Gamma (-m/2)}}\,} for -2<Re(m)<-1/2 1 r 2 + z 2 {\displaystyle {\frac {1}{\sqrt {r^{2}+z^{2}}}}\,} e − k | z | k = 2 | z | π k K − 1 / 2 ( k | z | ) {\displaystyle {\frac {e^{-k|z|}}{k}}={\sqrt {\frac {2|z|}{\pi k}}}K_{-1/2}(k|z|)\,} 1 r 2 + z 2 {\displaystyle {\frac {1}{r^{2}+z^{2}}}\,} K 0 ( k z ) {\displaystyle K_{0}(kz)\,} , z {\displaystyle z} 可為複數 e i a r / r {\displaystyle e^{iar}/r\,} i / a 2 − k 2 ( a > 0 , k < a ) {\displaystyle i/{\sqrt {a^{2}-k^{2}}}\quad (a>0,k<a)\,} {\displaystyle \,} 1 / k 2 − a 2 ( a > 0 , k > a ) {\displaystyle 1/{\sqrt {k^{2}-a^{2}}}\quad (a>0,k>a)\,} e − a 2 r 2 / 2 {\displaystyle e^{-a^{2}r^{2}/2}\,} e − k 2 / 2 a 2 a 2 {\displaystyle {\frac {e^{-k^{2}/2a^{2}}}{a^{2}}}} − r 2 f ( r ) {\displaystyle -r^{2}f(r)\,} d 2 F 0 d k 2 + 1 k d F 0 d k {\displaystyle {\frac {\operatorname {d} ^{2}F_{0}}{\operatorname {d} k^{2}}}+{\frac {1}{k}}{\frac {\operatorname {d} F_{0}}{\operatorname {d} k}}} f ( r ) {\displaystyle f(r)\,} F ν ( k ) {\displaystyle F_{\nu }(k)\,} r s {\displaystyle r^{s}\,} Γ ( 1 2 ( 2 + ν + s ) ) Γ ( 1 2 ( ν − s ) ) 2 s + 1 k s + 2 {\displaystyle {\frac {\Gamma \left({\frac {1}{2}}(2+\nu +s)\right)}{\Gamma ({\tfrac {1}{2}}(\nu -s))}}{\frac {2^{s+1}}{k^{s+2}}}\,} r ν − 2 s Γ ( s , r 2 h ) {\displaystyle r^{\nu -2s}\Gamma \left(s,r^{2}h\right)\,} 1 2 ( k 2 ) 2 s − ν − 2 γ ( 1 − s + ν , k 2 4 h ) {\displaystyle {\frac {1}{2}}\left({\frac {k}{2}}\right)^{2s-\nu -2}\gamma \left(1-s+\nu ,{\frac {k^{2}}{4h}}\right)\,} e − r 2 r ν U ( a , b , r 2 ) {\displaystyle e^{-r^{2}}r^{\nu }U\left(a,b,r^{2}\right)\,} Γ ( 2 + ν − b ) 2 Γ ( 2 + ν − b + a ) ( k 2 ) ν e − k 2 4 1 F 1 ( a , 2 + a − b + ν , k 2 4 ) {\displaystyle {\frac {\Gamma (2+\nu -b)}{2\Gamma (2+\nu -b+a)}}\left({\frac {k}{2}}\right)^{\nu }e^{-{\frac {k^{2}}{4}}}\,_{1}F_{1}\left(a,2+a-b+\nu ,{\frac {k^{2}}{4}}\right)} − r 2 f ( r ) {\displaystyle -r^{2}f(r)\,} d 2 F ν d k 2 + 1 k d F ν d k − ν 2 k 2 F ν {\displaystyle {\frac {\operatorname {d} ^{2}F_{\nu }}{\operatorname {d} k^{2}}}+{\frac {1}{k}}{\frac {\operatorname {d} F_{\nu }}{\operatorname {d} k}}-{\frac {\nu ^{2}}{k^{2}}}F_{\nu }} Close 傅里葉變換 Wikiwand in your browser!Seamless Wikipedia browsing. 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大於零。
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for -2<Re(m)<-1/2
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可為複數
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